Mechanics for JEE - Vol.2 [2, 3 ed.]

Citation preview

Study Anurag Mishra with www.puucho.com

Er. Anurag#Mishra

www.puucho.com

Study Anurag Mishra with www.puucho.com

J

www.puucho.com

! ~\I\/~

Study Anurag Mishra with www.puucho.com

E(flIAnE(f for (Main & Advanced)

Vol. II

by:

Er. Anurag Mishra B.Tech(Mech. Engg.)

HBTI Kanpur

SHRI BALAJI PUBLICATIONS ~

~

(EDUCATIONAL.

E':UBLISHERS

&. DISTRIBUTORS)

AN ISO 9001.2008 CERTIFIED ORGANIZATION

Muzaffarnagar

(D.P.) - 251001

www.puucho.com

--.

\

~

-

Study Anurag Mishra with www.puucho.com

~

'6i?

Published by:

SHRI BALAJI PUBLICATIONS lEDUCAT10NAL PUBLISHERS

& OISTRIBUTORSl

6, Gulshan Vihar, Gali NO.1, Opp. Mahalaxmi Enclave, Jansath Road, Muzaffarnagar (U.P.) Phone: 0131.2660440 (0), 2600503 (R) website: www.shribalajibooks.com email: [email protected]

~

First edition

2010

~

Second edition

2011

~

Third edition

2013

~

Reprint

2018

~

@AII rights reserved with author

II!

Price:

'{

400.00

.

II!

l

It•

I

Typeset by : Sun Creation Muzaffarnagar

!

-11"

All the rights reserved. No part of this publication

may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording

or othelWise, without the prior permission of the author and publisher. Any violation/breach

shall

be taken into legal action.

www.puucho.com

I

Study Anurag Mishra with www.puucho.com

prepace I have been involved in teaching Physics for last 14 years. This book is an opportunity to present my experiences. During my interaction with IIT-JEE aspirants.

I

I realised that

most feared topic is mechanics. Some of the reasons put forward by students behind this

I

thought were ;

d

•

No spontaneous

thoughts appear after reading a problem. Mind goes blank. Can not

proceed in a problem. •

How to proceed in a problem? Which law is applicable, that is a given problem will involve conservation of energy or momentum or both?

•

This book will help the students in building analytical and quantitative skills, addressing key misconceptions and developing confidence in problem solving. I sincerely wish that this book will fulfill all the aspirations of the readers. Although utmost

full CJre has been taken to make the book free from error but some errors inadvertently

may

creep in. Author and Publisher shall be highly obliged if suggestions regarding improvement and errors are pointed out by readers. I am indebted Neeraj Ji for providing me an opportunity to write a book of this magnitude. I am indebted to my father Sh. Bhavesh Mishra, my mother Smt. Priyamvada Mishra, my wife Manjari, my sister Parul, my little kids Vrishank and Ira for giving their valuable time which I utilized during the writing of this book and people of Moradabad, who supported me throughout my career. I am also thankful to Mr. T. Kondala Manohar,

Mr. S.P. Sharma,

valuable suggestions

Mr. Sudhir

Rao, Mr. Abhishek Sharma

Sinha (Ranchi),

and Mr. P. Narendra

Mr. Sunil

Reddy for their

in improving the book.

In the last, I also pay my sincere thanks to all the esteemed members of Shri Balaji Publications

in bringing out this book in the present form.

Anurag Mishra

•

www.puucho.com

Study Anurag Mishra with www.puucho.com

..

www.puucho.com

Study Anurag Mishra with www.puucho.com

A Few Worils to the Students How to face the challenge? Following are some doubts which arise in the mind of almost all the students but may face them by taking some care, 1.

I call not solve numerical because my concepts are nat clear. In fact numerical solving itselfis an exercise to learn concepts.

2.

I can not study because I am in depression, I fell into it because I was not studying! Depression is escape mechanism of people afraid offacing failures. Failure is integral part oflearning.

3.

I understand everything in class but can not solve on my own. WRITING work is vital. It is a multiple activity, initially idea comes in mind then we put into language to express it, we are focussed in hand eye coordination, eyes create visual impression on brain which is recorded there. WRITING WORKS ARE EMBOSSED ON BRAIN LIKE CARVINGS OF AJANTA CAVES,

4.

In exams my brain goes blank, but I can crack them at home. Home attempt is your second attempt! you are contemplating about it while home back. You do not behave differently in exam you replicate your instincts. Once a fast bowler was bowling no balls. His coach placed a stump on crease, in fear of injury he got it right. CONCEPTUALIZATION, WRITING EQUATION, SOLVING, THEN PROBLEM GETSTO CONCLUSION!

S.

I am an average student. It is a rationalization used by people afraid of hard work. In their reference frame Newton's first law applies "if I have a misconception I will continue with it unless pushed by an external agent even I will surround him in my web of misconception yielding zero resultant:' AVERAGE IS NOT DUETO CAPACITY LACUANE BUT DUE TO LACK OF DETERMINATION TO SHED INERTNESS,

6.

A famous cliche "I do not have luck in

/1~favour"

PRINCIPLE OF CAUSALITY:

CAUSE OF AN EVENT OCCURS IN TIME BEFORE OCCURRENCE OF THAT

EVENT i.e., cause occurs first then event occurs. SHINING OF LUCKIS NOT AN INSTANTANEOUS EVENT IT IS PRECEDED BY RELENTLESS HARD WORK.

Sow a seed of aspiration in mind, water it with passion, dedication it will bear fruit. luck can give you sweeter fruit.

www.puucho.com

Study Anurag Mishra with www.puucho.com

..• ,

-

.'

Useful tips: 1.

Do not take study as a burden actually its a skill like singing and dancing. I has to he honed by proper devotion and dedication.

2.

Without strong sense of achje~ernent you can't excel. Before entering thtcompetitive field strong counselling by parents is must. Majority do no know what for they are here. No strategic planning, they behave like a tai ender batting in frontofSteyn's bouncers.

3.

Science is not a subject based on well laid down procedures or based OD learning some facts, it' very intuitive and exploratory approad Unless their is desire arid passion to learn you can not discover new ideas .• requires patience and hard work, whose fruits may be tangible later on.

involves

4.

Some students realize very late that they are studying for acquiring skills an~ honing them. Their is a feeling that they can ride at the back of instructor ant achieve excellence. Study comes as torturous exercise enforced on them anI their is some mechanism that can take this burden ofthem.

5.

Science is not about gaining good marks, up to Xth by reading key points goo marks are achieved but beyond that only those survive who have genuin interest in learning and exploring. Self study habit is must.

6.

IF YOU WANT TO GAIN LEAD START EARLY. Majority

of successful student.

try to finish major portion elementary part of syllabus before they entE Coaching Institute. Due to this their maturity level as compared to others j more they get ample time to adjust with the fast pace. They are le~ traumatized by the scientific matter handed over. For those who enter fres must be counselled to not get bullied by early starters but work hardt: initially within first two months initial edge is neutralized. 7.

Once a student lags behind due to some forced or unforced errors his min. begins to play rationalization remarks like I am an average student, my mil) is not sharp enough, I have low IQ etc. These words are mechanisms used II avoid hard work. These words are relative terms a person who has early sta may be intelligen t relative to you. Intelligence means cumulative result of hard work of previous years, th hard work has eve.ntually led to a development of instinct to crack thinjl easily.

www.puucho.com

Study Anurag Mishra with www.puucho.com

Q I. Simple

Harmonic Motion

Equation of Motion (1), Harmonic Functions (2), Phase Diagram (8), General Motion near Equilibrium (22), Energy of a Simple Harmonic Oscillator (24), Equation of S.H.M. with Shifting of Origin (28), Analysis of Two Block System in CM Frame (47), General Motion near Equilibrium

(68), Damped and Driven Oscillations (74),

Subjective Problems (83), Problems: Level-1 : Only One Alternative is Correct (90), Solutions (103), Level-2 : More than one Alternative is/are correct (106), Solutions (113), Level-3 : Comprehension Based Problems (115), MatchingType Problems(121), Assertion-Reason (125), Solutions (127).

:J 2. Wave

&IlL

Motion

Wave Motion (133), The Language of Wave Theory (135), Wave Function (136), Sinusoidal Wave Train (138), The Wave Equation (141), Mechanical Waves (144), Refraction (155), Superposition (157), Vibrations of a Tuning Fork (169), The Velocity of Sound (172), Energy in Sound and Light (179), Standing Waves in Organ Pipes (186), The Decibel Scale (190), Wave Front (207), DopplerEffect

in Case of

Light (221),

Subjective

Problems

(227),

Problems: Level-1: Only One Alternative is Correct (239), Solutions (253), Level-2: More than One Alternative

is/are Correct (257), Solutions (261), Level-3:

Comprehension Based Problems (263), Matching Type Problems (265), Assertion-Reason (267), Solutions (270) .

• .,,',', ,

•

,

'. -,,",,~'~".' .

"

".-

' •. ' ....,..".'.r.., '. ...,:'.: ... ~.

.. -

,

- .... www.puucho.com

.. ~0

'>0

02)' An observer, at t = 0, obse,.,es the particle 'p , at distance ~

moving to the right from mean

Fig. 1E.1 (b)

The reference point can be located at any of the nvo positions G and H as shown in figure for having displacement

~,

position position

a while Q at -Jj A 2 a as shown.

-

b'J.t to move towards mean position it

should be H. Position H corresponds

•Q

to angle 37t,

4 3, ThuS~""4an

•

o

-. p

+ ve A

)(::72

Fig.1E.3(a)

d x=Asm . ( (1lC+ 3rrJ

4

Find,

($, - $,)

Solution: First we solve this problem from equations:

Alternative Mathematical solution Putting t :=: a in equation x ::;A sin((ut + ljI) Weger

moving to the left from mean

YP ",Asin((JJt+$1) at

~=ASin$

t

=0 ~=Asin$1

but at c

12

$=~, 3:t 4 4

Thus, :=:

." Sin,!,]

4

+

~n)

in this solution.

x =

l' p =

since

Find distance traveled by a particle of time period T and amplitude A in time T/12 starting from rest. is at extreme

Fig.1E.3

(b)

Now we have to check which value of 4>]is an acceptable solution. It is decided from sign of velocity. Velocity of particle at t = 0 is given by and

Example

Initially particle +A. Position at time t, x(t)

$1 '" 1t/4,31~l4

Thus

Objective; the ease of using phasor diagram is quite

Solution:

12

a ; 17 < 0 so 2: is rejected x =Asin(mc

evident

=-1

l'

so, Now for Q at

t =

Awcos~]

>0

$1 '" rr/4 is acceptable YQ

:::

A cos((I)( + 4>2)'

0

position

J3A

---

2

www.puucho.com

",ACOS~2

Study Anurag Mishra with www.puucho.com

MECHANICS-II

12 C0592 ::--

:=}

$2

::::>

and

=

J3

Solution: It Sit ZIt It--=wt =)-=-t

2

11- 11/6

= 5rt/6 or

•

6

7.

+"6 = "6

11

vQ < a

Thus

T

t =12

vQ =-Aoosin$2>

since

6

ST

so, Q2 = 511/6 is acceptable

~$= 511/6-1[/4 101t-3n 71t =----=12 12

Example A body executing S.H.M. with time period T and amplitude A.

Example 1\VDbodies are executing S.H.M. with same amplitude A and time period T. When one of blocks is located at right extreme and other blm:k is located at equilibrium position moving towards left. Find the time when both the blocks will have same di.~placementfrom the equilibrium position.

When it is at + A from equilibrium position moving towards 2 right. It receivesan impulse which doubles its velocity without changing direction. Find (a) velocity after the blow (b) Amplitude after the impulse (c) Time taken by the block to reach the right extreme position. Solution: Velocity = ro.J A 2 _ x2

=~JA'-(~)' = Zit

JA2_A2

T

4

2J3"" .J3A

=---=--

zr

Flg.1E.4

Solution: One will have +ve velocity equation one will have -ve velocity but displacement is same

.

Z,/31tA

Veloclty = --

T

T

(doubles)

• •

Angle ralated - + - = rot 2 4 3n:

2)'[

4

T

-=-xt

t=-

;r 8

Example TIvo bodies are executing S.H.M. such that their amplitude is A and time period T when one of the blocks is at right extreme another block is at + A moving towards left. Find ehe rime • 2 after which both the blocks will have same position from equilibrium position. .

Fig.1E.6

z.J31V\ = Zit T T

J3A

., A

Flg.1E.5

www.puucho.com

=

.J A 2 _

JAf -(

x2

~r

Study Anurag Mishra with www.puucho.com

SIMPLE HARMONIC MOIION

13 A2 cose = _.-

Panicle has terned through 2rc. Thus, time

A,

-=-xt 3

ZA, Ax2

2"fi3A

:=

0 =

[

co,-'

_I

../13

Example

Find the distance travelled by the body executing S.H.M. having time period 4 sec starting from + A moving in +ve

.13

_'(

1 )

,/i'3

t =

Example

.

T

2

direction within a inten'al of 15 .•ce. What is average .•peed?

2, t T

=

T T =3

8~

-/13

cos-If~)=0)( cos

t

)

1 =

is

211: 21t

A

cos(J ""--

t

3

Solution: T =- 4sec,t = 15 sec, t= 4x3+ 3 (move a distance 124. in three time period and in remaining 3 sec angle rotated)

COS-l(~I_) -/13

7

A block is connected to a spring such that its time period under normal condition would be T. Block is compressed by a distance A and relcased. An elastic wall is located in front of block at a distance of N2. Find the time period of oscillation of this block. Fig.1E.8

toooooooo~

. -

.•

A Fig. 1E.7 (al

2rr 31I 8=-x3=-

4

Ai2

2

A

S=I24.+ +A+A+

Z

Solution: .. d 2, TIme pena =Velocity =

=

_

x2

=

= 21tJA2_x2

T

A-- 2

15A + A _ .J3A

2

"'

w\/ A 2

(J3AJ

2

(31; J3JA S

Average speed = IS Bleck has reached this

=

Collision occurs here

stale after collision

,

'3JA mls

(31- ..•. 2x 15

A

240.

Find the distance travelled by a body exeCllting S.H.M. in 10 second.Flg.1E.7(b)

if T is

3 see and the body starts from A.J3 from 2

equilibrium position moving in +ve direction?

A2

cos8=- = 60" A

8='::3

www.puucho.com

Study Anurag Mishra with www.puucho.com

MECHANICS.II

14 v = A (l) cos((I)t + lj)) 2mg =--wcos(wt +41) k

mg =A

Let

k OJ

=

liE: m

A

I

2A

2

cose=-=Fig.1E.9

B=~

3

Solution: T = 3x + 1 In 1 sec angle rotated

2B = 21t 3 2. ={O[' 3 '

ZIt 2T1: 8=-xl=-=1200

3

3

=

14A-F3A 2

[1

Flg.1E.l0 (b)

(t =~) (r-~)

21t 2n -=--, 3 7-

=l2A+-~AJ+A

ZT 3

=-

lJ=(I)JA2_x2

Concept: When a block is executing S.H.M. under the influence of string connected to Q spring.

=wJ4A2_A2

=AwJ3

Time taken after string becomes slack v = J3A(I)

v=u+at -F3AOJ= F3AO)- gt 2F3AO) t2 =--g

FIg. 1.17

kxeq = mg; =

Xeq

Total time = tl +t2 ZT 2.13 =-+--x21tx-

7

3

What should be the maximum amplitude so that the block will perfonn

:=

S.H.M.

String gets block if than g.

Thus

= 41t ~

Ik

=

Example A elastic string of constant k is attach to a block of mass m as shown. The block is given an extension of 2mgik from the equilibrium position and released. Find the time period of of the block.

Solution:

From previous

but it has given extension

concept Amax > of 2~g

,:g

g

rm + 4J3../f. x

2x 21t

acceleration of block becomes more

0)'

oscillation

T

'iT

A=-L

A

+ 4-./3

.Jm

Tn

k

rm

Vk

4fifCHF3)

1Woparticles executing S.H.M. with same angular frequency and amplitudes A and 2A all same straight line with same mean position cross each other in opposite direction at a distance A/3 from mean position. Find the phase difference in the two S.H.M s. Solution: Fig. 1E.ll show phasor diagram. Let particle P is going up and particle Q is going down. From the figure

www.puucho.com

Study Anurag Mishra with www.puucho.com

SIMPLE HARMONIC MOIION shown, the respective

15

phase differences

of panicles

m 3m -/1=-"

P' and Q'

are

2

2

"

or

t'=3

The block executes S.H.M. and initially at , , , '

.-

..

2k

,

velocity ~ . If amplitude of oscillation is A, we have 3

Fig.1E.11

.. ,(1) "3

3

~[A'_(m

[Phase angle of P']

>C9 = 3m

or an

" .. , d '+'2 = Sill

_(mg)'2k

A2

~=(I)

l)): :::SIO

..

1t - SIO

,( 1)

'6

0 block is

at a distance x = mg above its mean position and having a

,, "

t =

fPhase angle ofQ']

g )'] 2k

Note that for a new spring block system has Thus phase difference in the two S.H.M. is

'"

" .. , .. ,(l]'6 - .. ,(l)3"

u'+' = O2 - '1'1 = Sin

Example

TC -

Sin

"'=~

Sill

A=)m6~2 +(~:r

Thus

12~

A spring block pendulum is shown in Fig. lE.12 (a). The system is hanging in equilibrium. A bullet of mass m/2 moving at a speed tJ hits the block from downward direction and gets embedded in it. Find the ampliwde of oscillation of the block now. Also find the time taken by the block to reach its upper its extreme position after hit by bullet.

Time taken by particle to reach the topmost point can be obtained by phasor as shown in figure. This figure shows the position of block P and its corresponding circular motion particle Po at the t = O.

_.-

..

,,-

h. f "I",d

., .. I

p

._._._~

,

length

,, ,, ,

'0

.1-

,

.

'

Fig.1E.12 (a)

Solution: In equilibrium position mg = kxo when bullet of mass ml2 gets embedded in the block, the equilibrium

Fig.1E.12 (e)

8 cos-1 (x'A) t=-=-=~~

,,' I

k \ 3m 2

will change because new mass becomes 3m/2

position

and now the new mean position of the block will be at x from previous mean position then, we have 3

-mg

=k(xo+x)

-g

= k(xo + x)

2 3m 2

or

---1-,"tOml2

Flg.1E.12

ExampJ,e

)~7cos-

l

13

(::)

_

(b)

2k

of momentum

=

,

x= mg

Apply conservation velocity

.. .

cose=~

,

to get common

Fig. 1£.13 (a) shows a spring block system, hanging in equilibrium. The block of system is pulled down by a distance x and imparted a velocity v in downward direction as silOwn in Fig. 1E.13(b). Find the time it will take to reach its mean position.

www.puucho.com

Study Anurag Mishra with www.puucho.com

..".,------------------------::-;;;-16

MCCHAHICS.II

smooth Fig.1E.14 (a)

Flg.1E.13

(b)

Solution: As shown in Fig. lE.13 (b), when the block is pulled down by a distance x the velocity of block v at a displacement x from its equilibrium position as v or

=wJA2 k

VZ =

..Solution: Due to an inelastic collision, the velocity of combined block is reduced by half that is u/2 from conservation of momentum. Now at equilibrium position the velocity of block can be written as ~=AOJ

2

_x2

IJ=A

CAl_x:.!)

2

m ~V2

or

For combined block new angular frequency orS.H.M. is

2

A ='-k-+X I

See phasor diagram at t = 0, block P is at a distance x from its mean position in downward direction and it is moving downward P will reach its mean position when particle Po reaches position A by traversing an angle e. Thus it will take a time given as e 7t-sin-1(x/A)

if

t=;;;=

~

\ 2m

or

w=J2~ A=U&

Since oscillation starts from equilibrium position, particle has to cover a phase of It/2 radians, to reach extreme, thus time taken by particle to reach its extreme position is

B

. .....

C

... .•• ..

".

"

.... ..

.. ... .• ,

"

'---_

pO""

•••

>:;.

P

II

ro

A

. .-' D

Solution: When block is passing through the position of half amplitude Ao/2 , its velocity can be given as

Fig. 1E.13 (e)

D'

t

Consider a spring Q block system. If block is pulled toward right by a distance Ao and released, when the block passes through a point at a displacement Ao/2from mean position, another block of same mass is gently placed on it which sticks to it. Find the new amplitude of oscillation and find the time now it will take in reaching its mean position and extreme position on left side.

=R ["_Sin'l ~] .

V=IO~A2_(A20

--+x k

v = [k

JA~_

f;

Fig. 1£.14 shows a block of mass m resting on a smooth horizontal ground attached to one end of the spring of force ccnstam k in naturallengrh. If another block of same mass and moving with a velocity v toward right is placed on the block which stick to it due to friction. Find the time it will take to reach its extreme position. Also find the amplitude of oscillations of the embined mass 2m.

r

A~ = .J3 A o 4 2

[k

V;

When another block of same mass sticks to it, due to velocity of combined block becomes half and OJ of oscillation becomes get

www.puucho.com

J 2mk . At a distance Ao/2 from mean position, we

Study Anurag Mishra with www.puucho.com

17

SIMPLE HA~MONIC MOTION

0'

Example

A fE I JA'2 _ At

,,13 o == k 2 \Im \2m

Fig. 1£.16 (a) shows a block P of mas$ m resting on a smooth

,

~AJ

+

is

A

2",0.

horizontal surface, attached co a spring of force constant k which is rigidly fIXed on the wall on left side, shown in Fig. 1 £.16 (aJ. At a distance I to the right of block there is a rigid

Ao =~AJ

8

A' =

4

8

wall. If block is pushed toward lift so thut spring is compressed by a distance

0

If at t "" 0, the second mass is added to the oscillating block. Fig. lE.15 show the corresponding position of particle in phasor diagram at c = a in II quadrant. The time taken by P to reach mean position from a position of Ao/2 from mean

51 and released, it will start its oscillations. If 3

collision of block with the wall is considered to be perfectly elastic. Find the time period of oscillations of the block.

b~~

position is given as B

.... ... ... " '- cr, the bob will oscillate above the point of suspension with an angular frequency.

The equation of motion for an object executing 5.H.M. may be written as x(t) = A cos(wt + $)

,

1. The time period of a simple pendulum

I

> mg

Fbuoyam

Concept: In the foregoing whenp=oandp>o.

1. For p = cr, T

=g(l-~)

When p = cr, the bob moves in uniform circular motion as its time period is infinite.

W=JH1-~)

We have

gelf

;;;:;

Substituting keq in

motion

or

< mg

is given as :

The velocity functions derivative of x(t).

are

found

by taking

first

sin(wt + 41) The kinetic energy of the particle can be written as v(t) = -Aw

c__ ~ --h-_~-_-_~-_-_-~_-_-~_-~- ~- -_ --- --------- ..'-_.

------ --_ -----------.--... .._._ ... ----------------.----------.-._-------------------. --- ---_ .."._-_._--------------------- -.--..... _--_. --------------------- ---.-. __ ._.-----------------.-----------------_ .. ----_ .._-_ .._------.-.-.-.-----.-.----.-.-.-.-.- .•.~._':. --.-----.-.-.-.-- -.-.-.-.-.-.-.-.-.':.':.

I 10,_.. ""....

-.""-. ~

.------_..•••----

.-.

"

~=-.-.-=.-. ~ -"".~-.""-.-_-_:-d-.l .-- ••......-- •.•..

K

-. -. -. __ "w"w"w -. - •••• "w"w -. -. -. -. -.-.

po

2 sin2(wr + 0)

2

The potential energy is U =

.! kx2

=

.!kA 2 cos2(wt

2 2 So the total mechanical energy becomes E

Fig. 1.23

=.! leA2[sin2(wt 2

+ $) + cos2(wc

+ ell)

+ 41)]

=.!1eA2 2

As, k = mw2,

where

the entire energy is kinetic

P =

2

A

2

2 The equation indicates that the total mechanical energy of an undamped simple harmonic oscillator is constant. -- At the equilibrium position potential energy is zero and

(i)Ifp cr,

(~1eA

2 )-

T= ""

T=2XJg(~1-1)

--

2. When p < cr, the bob oscillates below the point of. suspension because weighting is greater than buoyant force.

At x = fA the velocity is zero and the entire energy is potential. As the mass m oscillates between x = ::I:Athere is an exchange between kinetic and potential energy of the spring; however the sum (K + U) is constant (Fig. 1.24).

www.puucho.com

Study Anurag Mishra with www.puucho.com 25

SIMPLE HA~ONIC MOTION

. parabola; consequently for displacement x, of the aLOm to atom distance from equilibrium, the potential energy

Energy

function is..!. kx2, where k values describe bond strength. 2 If x I and x 2 are the position of the masses nI 1 and tn2 on the x-axis, the extension or compression in the spring is x=(X2-x1)-1

-,

From Newton's second law, IF '" nla

o

-A

... (1)

d'x

i')

+kx '" nil __ 1

de

m 1'1=0

-kx = m

...(2)

2

d2x

nl2 __ 2 eft 2

... (3)

We multiply first equation by m2 and the second by ml, and subtract the first equation from the second to obwin

X" +A

v=O m

d2(xl-Xl)

-(m k 1+11l2)x=11l1m2

)(=0

(

... 4)

1

de

Since l is constant, the derivative on the right hand side can be written as (')

efl(Xl

dtl

Fig. 1.24

Physical

quantity

\Equili~r~um \ \

NO.\ 1.

Potential energy I U=-m(1)x

2

2

POSItIon

position Maximum

Minimum U min.

::

0

2 1 X==-/1IW

2

A

2 2. K

~

::

2

sin (WI +0) Maximum

"".!. mo)2 A2

K max,

cos2

=~

2

nw)2

Total energy E=U+K 122 =-/11(1)

A

)dlX _ del--

where the term

Minimum Kmin, == 0

A2

Constant

Constant

1 " E=-mw.A. 2

E=.!.I1W)2A2

eqn. (4) can be written as

is defined as the reduced mass

+ml

2

Il

Thus, angular frequency = Jk/Il The expression for the relative velocity and acceleration benveen the masses can be obtained by differentiating equation (1) w.r.t. time. x=(x2-x\)-l dx l' '" ~ = l'Z de dl'

a

Example.

" ....•

dl2

dx k --+-x=o 2

2

Illustration 4. Consider twO masses connected by a spring as "-< shown in Fig. 1.25. A k ffi1 1112 diatomic molecule can be ~ modelled on this pattern. ~I al The spring represents the Fig. 1.25 bonds between the two atoms. The potential energy function of the diatomic system is V-shaped. For small vibrational amplitudes the bottom of the U is shaped like a

d2x

kx

11111112 tnl

de

Kinetic energy

2

ma.

m\ml ( ml+ml

-1 /11(1)-" A

2

2

3.

Extreme

-x\)-I]

de2

-

After some rearrangement,

Table 1.2

s. \

d2[(Xl

~Xl)

= ~

de

'" Q2 -

-

VI

a1

25 _

Fig. 1£.25 (aJ shown two /flasses Tnl and fill connected by a spring of force constant k and the system is placed on a smooth horiwntal surface. If the block are compressed slightly and released. Find the time period of such oscillations.

www.puucho.com

k

~~~,~~~~~;;;;B" Fig. 1 E.25 (a)

Study Anurag Mishra with www.puucho.com 26

_

MECHANICS-II

Solution: In the 1 absence of any external force present, the centre of mass of the two block system remain at rest during oscillations. In Fig. lE.25 (b) thus point C Fig. 1E.25 (b) will remain at rest and with respect to this point nIl and nt2 will oscillation independently. Now we split the spring in two pans of length I] and 12 which are given by

1---/ --t;--'2----..j

,-

K:~::C,:::a

m,l

II =

ntl +m2

mil

12 =

and

+ m2

nIl

System behaves as if these two springs are be fixed at point C separately. The respective force constants can be given as

PROBLEM SOLVING STRATEGY Force and Torque Method 1. Determine the equilibrium position of the particle or rigid body undergoing simple harmonic , motion. 2. Now consider a small linear displacemem in case of a body in translation and angular displacemem in case of a rigid body undergoing S.H.M. 3. Draw a free body diagram of the body in the displaced position. 4. Identify the net force or net torque that tries to restore the body to its equilibrium position. 5. Apply Newton's second law to calculate acceleration or angular acceleration IT and

k] = kl = kent] +m1)

11 2

6. If acceleration is directed towards equilibrium position or angular acceleration tries to restore the body to equilibrium configuration the motion is simple harmonic. Compare the expression for acceleration or angular acceleration with the standard expressions

nI2

12

nIl

=-IC

~1

k2

0=_(002)x

/2---;

1*-1,

angular

frequency

and

time

period

of

The total energy of a simple harmonic oscillator is constant. Determine kinetic energy (KE) and potential energy (PE) in displaced position. As [mal energy is constant,

m

..

Fig. 1E.66 fa)

a any time

Solution: The horizontal position of the rod is in equilibrium position. We will determine the static displacement of mass m and apply

mg

v=_1

Equilib~um Position

27[ (b)

(-)

In the equilibrium position,

Flg.1E.65

is Xo = a80' Torque of spring force and weight balance each other, i.e., ... (1)

In a slightly displaced position equations of motion are mg - T = rna = mra

and

(2)

Tr-ka2(8o+S)=Ia

(3)

or

••••

1.••

.

Fig. 1E.66 (b)

Stretch in the upper spring 01 = mgl kJb If a point on the rod at a distance b from the left end is displaced byo1' the right end is displaced by an amount0z, where

02 = l.. mgl = mg ( b bkl

+8) = 10.

k1

!)'

lb

(from similar triangles).

ka'

a =-----9 (I +mr2)

Then total static displacement, 0 = O2 + mg

k,

Comparing above eqn. with necessary condition of S.H.M., i.e., a = --0

m} KI _>_ m2 K2

or 2

so

The above inequality is the desired condition.

and (c) Assuming ~

Example In the shown arrangement, both the springs arc relaxed. The coefficient of friction between mz and m, is ).l. There is no friction between m} and the sUlfaee. if the blocks are displaced

Fini

K2

of. m2 with ml and m2 together displaced towards right can be shown a's given in Fig. lE.89 (e). 1f A be the oscillations of m2'

slightly, they together perfornt simple harmonic motion. Obtain:

N

!S., the F.B.D.

>

mz

K,

>0

K2A -

f

amplitude

r-m,

or

f

As

f

=:;

of

= mzoo2A

A(K2

-

(b) The condition ifthefrictionalforce on block rn2 is to act in the direction of its displacement from mean position. (c) If the condition obtained in (b) is met, what can be the maximum amplitude of their oscillations?

Solution: (a) For the oscillations of the blocks together, the equivalent force constant equals KI +K2 and 1 K +K hence the frequency I , 2rr m} + m2

or Here Hence,

=_

www.puucho.com

mzff(2) ~m2g

A

82 Thus we get

the

www.puucho.com

,13MJi2

.!ce2 2

Example

and stationary

time period will change because I

3 , Inew =-MR 2

V

-

of C = (l)2(2R) = 12 g(1 - coseo) 5

(e) In this case moment

position

II2gROi

- sg

acceleration

X

80

24 ROa _ /12gR8a "2 - \ 5

_

a

a=--MR'

2

because 80 is small so sin 80

tV'listed develops

t=C8=/a I , C8=-MRa 2

cos80)

\ 5,

when

=

2 ncwUl new

1 new OO~ew 2

00

Study Anurag Mishra with www.puucho.com

MECHANICS.II

80

)+

(d) -(~ kx5 - ~ kx2

Example The axle of a pulley of mass m = 1kg is attached to the end of a spring of spring constant k = 200N/m whose other end is fixed co the ceiling. A rope of negligible mass is placed on the pulley such that its left end is fixed to the ground and its right end is hanging freely from the pulley which is at rest in equilibrium. We begin to pull the endpoint A at the right end of the rope by a constant vertical force of F = lSN. Friction

--

1

2

mg(xo - x) + 2F(xo - x) = 0

k(xo + x) + mg + 2F = 0

+ x = 2(mg + 2F) __ 2[~1_O_+_3_0o]= _8_0 = 40 em

x

k

o

200

200

= 35em

Xo

Displacement of point A is 2(xo - x) = 60em

can be negleeted between the rope and the pulley.

Example 14.rod of mass Tn and length I is pivoted at a point 0 in a car whose acceleration towards left is ao. The rod i.~free to oscillate in vertical plane. In the equilibrium state the rod remains horizontal when other end is suspended by a spring of stiffness k Find the time period of small oscillations of the

m A

rod in given by T = 2~. Find the value of C. [Gh'en value

F

C,,3

k

= 20Njm,

ao

= lOm/s2•

m

(a) What is time period of S.H.M. of system ? (b) What is amplitude of S.H.M. ? (e) What is velocity of pulley when pulley is at equilibrium position?

displacement of point

(d) Find tile maximum applying F. Solution:

Thus,

a

= -km x,

(b) In the absence

k

a

A after

Fig.1E.108 (a) rna

Solution: In equilibrium, taking moments of the forces about 0,

= V-; (k

w

mg(I/2) = k,l

of F equilibrium

,= (mg/2k)

is,

e ~ initial extension in the spring

Initially pulley is at equilibrium t

=

1 k& I = 1m.]

(a) At equilibrium kx.q = 2F + mg

After a displacement of x -{k(x.q + x) - (2F + mg)} =

at

=

•'0

Fig.1E.107 (a)

a

k,

mg x=-=5cm k

with k(x+x,q)

F

new

mg

Fig. 1E.107 (b)

equilibrium

is

at

= mg + F

or x~ = 15cm Thus initially pulley is equilibrium position. Hence amplitude A = 15 em. (c) v=w.JA2-X2 1J

= /200 .J152 \ 1 =

mg X'q

= 15cm

away

from k(e - {sine)

mao (Pseudo) _

a

a

3W2m

mg

Fig.1E.108 (b)

www.puucho.com

Study Anurag Mishra with www.puucho.com 81

SIMPLE HARMONIC MOTION

=

5kR'

0:=-

(MR-+l)

MR2

where]

8,

.,

=-2

5kR'

=

0:=---8 ~MR2 2

It WIt Compare ""h

T

=

a = --(I).'8Th us

OJ2

10k =3M

21t pM \ 10k

Example

11O....-?'

The shown container contains liquid of variable density which varies as d;:; do( 4- ~~) where ho and do are constants and y is measured from the bottom of container. A small solid

block density is Sdo/2and mass m is releasedfrorn the bottom of the container. Show that the block will execute S.H.M. and find its time period.

Example

109

_

A solid uniform cylinder of mass M performs small oscillations due to the action of two springs. each having stiffness k. Find the period of these oscillations in absence of any sliding (springs have their nuwrallength initially).

---_ .._--------------_ .._---_ ..._------_.-----_ ----_. __ .---_ --Fig.1E.110

Solution: First of all lets find the volume of the body :==> p = m/V ==> Sdo/2 = m/V ==> V = 2m/(Sdo)

Fig. 1 E.1 09 (a)

Solution: For small angular displacement of cylinder. The energy of system angular displacement e is 1.,121212 E "" - k(2R8)- + - k(R8)

+ - 1\1" + -}O) 2 2 2 2 where t' is the velocity of centre of mass and 00 is the angular velocity of cylinder. Since E is constant. ~

Buoyant force = 2m x do(45do

Fig.1E.109

J

Now point of stability ==>

==>

)g

mg = 2m (4 _ 3Y 5 h,

Y = ho/2 the mean position = ho/2 when body is at height ho/2. force acting on it is

i.e., zero Now, if we replace Y by x+ ho/2we get the equation of force acting upwards as

k

M

1x g

and weight = mg

I.e.,

k

3y ho

2: [4_

(b)

dE = 0 dr

3

h

2m 3x --x-g

(

x + h; )}

-

mg

0

6mgx

2 de 2 de dv dv dm 4kR A.-+ kR 0- + Mv- +1\111-+ 1(1)- = 0

=>

SkR2S+MR2o:+Ia

Comparing with F = -kx, we find k = 6mg 1(5ho)

de

de

de

=0

de

de

www.puucho.com

5

ho

=> ---

5ho

Study Anurag Mishra with www.puucho.com

~8;:::2:::::=:':~::::::::"------------------------\MiiiEiFcHiiiAiiNF.i1cts.mu T =

f!ti \ k

2rr

=>

2,

Tn

x 5ho

=>

6mg

rho6g

2,--

Two blocks of equal masses m are connected by a relaxed spring with a force constant k. The blocks rest on a smooth horizontal table. At t = 0, the block on the left i.s given a sharp,

impulse

J

towards the right, and the blocks begin to slide

along the table. Find the velocity a/the

left

block as afunetion

of time.

Fig.1E.111

Example Three identical springs each of force constant k have been joined to the three identical balls (each of mass mY, as shown in the Fig. 1E.112 (a), which are at the three vertices of an equilateral triangle. In the shown arrangement, each of the spring is in its natural length. What all three balls are simultaneously given small displacements of equal magnitude along the directions as shown in the figure, the oscillation 'frequency for the blocks will be .

(a)

-"'-

Solution: When impulse acts on block A, initially velocity of block A and Bare uA = JIm and Us = 0 respectively. Velocity of eM of system is JI2m m k Fig.1E.112

fig.1E.111 (b)

uB = 0,

llA

= Jim,

vCM = J/(2m)

We win solve this problem in eM frame because in eM frame motion is purely simple harmonic. Initial velocity of block in eM frame is given by

u' A = U' A -VCM = J/(2m) U'S=U'B-VCM

=-J/(2m)

In eM frame blocks are doing S.H.M. \Vith angular frequency ro::; At

x A.'CM

::;

I

.. k

V mm,,(m

+ m)

=

'.' Solution:If blocks.

be the magnitude of the displacement of

=

y xcos8 => Change in the length of each of the spring = 2y.: 2xcos6 ~ Net force F on each of the ball \ViII have magnitude 4kxcos6cos6

(2k

V-;;;

initial instant spring is relaxed A sin rot and XD,CM ::; A sin(rot + It)

Thus velocity of A in CM frame v' A::; U. A cosmt J =cosme 2m and velocity of A in ground frame ~ ~ ~

therefore

Flg.1E.112

(b)

~ ~ F = - (4k cos2 e) x hence v (oscillation frequency)

VA:g ::; vA,,'Of+vCM

Thus

X

(a)

e

2

= ~ (4kcos 2lt\ m

VA =v'A+VCM

J 2m

= -(1 + cosmt)

=_1 f3k(B=300) 2lt

V-;;;

www.puucho.com

Study Anurag Mishra with www.puucho.com 83

SIMPLE HARMONIC MOTION

1. The displacement of a block att 0). [Ans. 0.284

2.

5,

0.866

S,

1.16 S, 1.74 s.]

A block of mass In ::::O.5kg is attached ro a horizontal spring of spring constant k = SON/m. At t = 0.1s, the displacement x = -O.2m and the velocity (':::: +O.Sm/s. Assume that displacement of block is given by x A sin«!lt + ljJ), (a) Find the amplimde and the phase constant. (b) Write the equation for x(r). (e) When does the condition x = 0.2 m and l' = -O.Sm/s

-'Fig. 1.5

=

(b) Show that for small displacement F

occur for the first time

x

? [Ans. (a) 0.206 m, -2.33 fild. (b) 0.206 sin (10 1 - 2.33) Ill, (c) 0.414 s.)

3.

6.

cycle, (b) the maximum \'elocity. 4.

5.

(b) 0.531 m/s.]

In the Fig. 1.4 shown, there is no friction between block m A B and ground. The coefficient of friction for M B the contact surfaces of the two masses is .u5 and the Fig. 1.4 coefficient of kinetic friction is Uk' The system is set into simple harmonic motion \•... ith amplitude A. Find the maximum amplitude of oscillation such that top block A will nor slip on block B. [Ans,

_!:..-x~ zig

and the oscillations arc not simple harmonic.

A particle performing simple harmonic motion has a frequency of 1.2 Hz and a maximum acceleration of 4m/s2• Find (a) the total distance travelled in one [Ans. (a) 0.282 m,

=

x,

[Aos. (a) 1.30 Hz. (0) At the topmost point of the o5cillmion, (e) At the lowermost point of the 05eillation, (d) 0.147 m]

7.

11,em + '''''')g] K

A spring with spring constant k is "ttached to a mass m i.e., confined to move along a frictionless rail "s shown in the Fig. 1.5. Natural length of the spring is 10 when mass m is at origin of coordinate system. (a) Show that the x-component of the force that the spring exerts on the mass is

In the Fig. 1.6 shown, a mass III hangs by a string attached k = 50.0 Nlm to mass M. (a) What is the frequency of oscillation? (b) At \•... hat point in the oscillation is the tension in M=0.50kg the string a minimum? (c) At what point in the oscillation is the tension in the string a maximum ? (d) What m = 0.25 kg amplitude of oscillation Fig. 1.6 ensures the tension in the string is zero at some instant during the motion?

Show that if

31.

[Ans.2rcffff]

32.

33.

{Ans. Error is greater when the clock is elevated.]

Y

f¥]

\3.

28. Prove that if a physical pendulum having mass M and moment of inertia [ about its pivot is repivoted at its centre of oscillation, the original pivot becomes the new centre of oscillation and the oscillation period is unchanged. 29. A dumb-bell has two small masses m} and m2 attached to the ends of a rod of negligible mass as shown in the Fig. 1.29. Find an expression for this physical pendulum. Express your answer in m, terms of the parameters given in the figure. 2 [Ans. 2rc ml/j + m21~]

34.

).j-(4-T f-m-L-)]

The block of mass m1 shown in Fig. 1.33 is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the Fig. 1.33 spring in the equilibrium position. (b) The blocks are pushed a further distance (21k)(ml + m2)g sine against the spring and released. Find the position were the two blocks separate. (c) What is the common speed of blocks at the time of separation ?

A bullet of mass m strikes a block of mass M (as shown in Fig. 1.34). The bullet remains embedded in the block. Detennine Fig. 1.34 (a) the velocity of the block immediately after collision and (b) the amplitude and frequency of the resulting simple hannonic marion. [Ans. (a) nwj(m + M) (b) mv/ .jk(m + M)f:: (l,'2rc).jkj(m + M)]

35.

30. A child playing with his bicycle turns it upside down. Each wheel has moment of inertia 0.125 kg m2 and radius 32.0 em. When the child attaches a small mass to the edge of one of the wheels and rotates the wheel through a small angle, it is found to oscillate with a period of 2.00 s. How much mass did the child place on the wheel edge? [Ans. 0.58 kg)

f ~(If"

[Ans. (a) (ml+m2)gsin8 k (b) when the spring acquires its natural length]

Fig. 1.29

m~12-mlgll

Fig.1.31

A particle of mass m is attached to mid-point of a wire of length L and stretched between two fixed points. If T be the tension in the wire, find the frequency of lateral oscillations. [Ans.

27. A unifonn rectangular plate of sides x and y is hung from a pivot at the middle of one of the sides of length x. (a) If x is fixed and y varied, for what ratio xfy will the period of oscillation be a minimum? (b) What will be the period ? [Ans. (a) ~ = 2, (b) 2rc

A particle of mass m is attached to three springs A, Band C of equal forO F = { -2kx forx Tz (b)T1=!: "V = P(Ax) AL

PA Mg (M)A=--x=--x

T=211: [c] The system is equivalent mj

K

+k

x'

T

x

L

(,~g

12

(_2x_x3)dx=x2+o 4 Since there is no loss of energy, so x4 2 1 2 -+x =E. = -f1lV =3 4 '2 x4 +4x2-12=0

49. [c] U=f

m

Net restoring force = -

7.

Q2 -411 =-

So,

F'

L

L-x

A A' smO, -=

46. [bJ

=-!L(I-cost)

V

1 j3)and set free. Assuming the wall to be perfectly elastic, the period of such pendulum is:

19:

A particle moves such that its acceleration is given by: ,0;

-~Cx- 2)

Here: b is a positive constant and x the position from origin. Time period of oscillations is :

2.H

(a) 2.$

, Ce) 16. A body of mass m is attached to a spring of spring constant k which hangs from the ceiling of an elevator at rest in equilibrium. Now the elevator starts accelerating upwards with its acceleration varying with time as a = pt + q, where p and q are positive constants. In the frame of elevator: (a) the block will perform S.H.M. for all value of p and

20.

t.

71

2.J~+ 2

21.

(d) 2.~ 1 ~+2

The potential energy of a particle of mass 2 kg moving along the x-axis is given by Vex) = 16(x2 - 2x) joule. Its velocity at x= 1 m is 2 mls. Then: (a) the particle describes uniformly accelerated motion (b) the particle describes oscillatOl)' motion from Xl O.Sm to x2 105m (c) the particle executes simple harmonic motion (d) the period of oscillation of the particle is It I 2 second On a smooth inclined plane a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has a force constant k, the period of oscillation of the body is: (assuming the spring as massless)

=

q

(b) the block will not perform S.H.M. in general for all value of p and q expect p = 0 (e) the block will perform S.H.M. provided for all vaiue of p and q expect p = a Cd) the velocity of the block will vary simple harmonically for all value of p and q 17. A linear harmonic oscillator of force constant 2x106NI In and amplitude 0.01 m has a total mechanical energy of 160 J. Its: (a) maximum potential energy is 100 J (b) maximum kinetic energy is 100 J (c) maximum potential energy is 160 J (d) maximum potential energy is zero 18. A particle is suspended by two ideal strings as shown in the figure. Now mass m is given a small displacement perpendicular to the plane of triangle formed. Choose the correct statement(s).

(b)

Cal2.~ ec)

2lt,/M

=

:q

2.~v:

Cb) (d)

2lt)2M ~ine

.1

22.

IIII~:IIIII

M

(a) The period of oscillation of the system is 2lt

J~l

(b) The period of oscillation of the system is 2lt~

A simple pendulum has a time period T. The bob is now given positive charge : (a) if some positive charge is placed at the point of suspension, T will increase (b) if some positive charge is placed at the point of suspension, T will not change (c) if a uniform downward electric field is switched on, T will increase . Cd) if a uniform downward electric field is s.••• itched on, T will decrease

www.puucho.com

Study Anurag Mishra with www.puucho.com

SIMPLE HARMONIC MOTION 23.

109

A coin is placed on a horizontal platform which undergoes horizontal S.H.M. about a mean position O. The coin does not slip on the platform. The force of friction acting on the coin is F. (a) F is always directed towards 0 (b) F is directed towards 0 when the coin is moving away from 0 and away from a when the c?in moves towards a (c) F = a when the coin and platform come to rest momentarily DL

1. The new angular frequency of the system will be : (a) 10 rad/sec (b) 15 rad/sec (c) 20 rad/sec Cd) none of these 2. The total energy of system after collision at any moment of time is :

www.puucho.com

L Horizontailloor

Study Anurag Mishra with www.puucho.com 118

MECHANICS.II

1. At the instant the speed of block is maximum, the magnitude of force exened by the spring on the block is : (a) mg (b) zero

(a) y = -A sin[wt + 5] (b) y = -A cos[wr + 5] .~~~.~ = ASin[Wf +5+%]

(e) mg

Cd) none of these 2 2. As the block is coming down, the maximum speed anained by the block is :

J~gL

(a)

(e) -J3gL

(b)

jiL

(d)

:!. jiL

(b) • -+

-510

e • -+ 2g

-sm 4g

2g

(d) •

4g

-+

22g4g

is the amplitude of oscillation, 0 = sin

•

--0 (b) _2_

(a) ~

(e)

00

• -0

(d) • - 0

an

00

2

~

3

-1

11

PASSAGE

I _ 3

-sm -1 -2 3

PASSAGE A block of mass m is connected to a spring of spring constant k and is a[ rest in equilibrium as shown in figure (a). Now, the block is displaced by h below its equilibrium position and imparted a speed Vo towards down as shown in figure (b). As a result of the jerk, block executes simple harmonic motion about its equilibrium position. Based on above information answer the following questions:

A plank of mass M is placed on a smooth horizontal surface. Two light identical springs each of stiffness k are rigidly connected to struts at the ends of the plank as shown in Fig. When the springs are in their unextended position the distance between their free ends is 31. A block of mass m is placed on the plank and pressed against one of the springs so that it is compressed by 1.To keep the blocks at rest it is connected to the strut by means of a light string, initially the system is at rest. Now the string is burnt. :

SIring

;:t;:;;*~ ///77ll/

/////

1, Maximum displacement of plank is :

E"m",mI1... Position

(a)

"

--"!!.-

(b) ~

+M

m

11

(-)

m+M

(e) ~

(d)~

m+M

m+M

2. Time period of oscillation of block:

(bJ

(a) (2tt +3) ~

V~

1. The amplitude of oscillation is :

----

(a)

h

-t[!!-], A

3. Find the time taken by block to cross the mean position for the first time :

00

( a) -a -+ -sm -1 -I 4g 4g 3 -1

•'~:.-~ereA ..

oo=g

2 3. Till the block reaches its lowest position for the first time, the time duration for which the, s'pring remains . compresse d IS:

Jf Hi . . Hi Hi. () Hi Hi Hi Hi'

=Asin[(l)t+o+~]

. ""(d}y

(b)

ln~~ + h2

(d) none of these 2. The equation for the simple harmonic motion is :

(b) ( 0

(b) Velocity is negative

I

II

Extreme position

(q)

Mean position

4

12. A panicle of mass 2 kg is moving on a straight line under the action of force F = (8- 2x) N. The panicle is released from rest at x= 6m. For the subsequent motion, match the following (all the values in the column U are in their 5.1. units) :

"

(a)

Acceleration is negative

(e)

O 0

ax

a

dP :::_ BiJy

Thus positions where i)y > 0

ax

>

~l. A wave given by YI = A sin (oot - lex) is sent down the string. Upon reflection the intensity is reduced to one-fourth of the incident wave. Show that superposition of the tlVO waves in the region x < 0 can be written as a combination of a standing wave and a travelling wave.

=

in phase of

amplitude

Reflected wave

Since the reflection is from a denser medium

of reflected

(amplitude)

Transmitted wave is travelling in the same direction as the incident wave, frequency is unchanged, wave number is changed due to change of medium. The reflected wave travels in the opposite direction to incident wave. The medium of incident and reflected waves is same, hence same wave number. The equation of reflected wave is

A sin (mf + kx) 2

of superposition, Y=Yl+Y2

- kx)+(

~ )sin ("1/ - kx) - (~ )sin(wt

We have split up the expression equal parts. We can split a wave parts. Any wave function can superposition of nvo or more than second and third term b~ formula have

Y = ( ~ ) sin

«(I)t -

Yi+Yr =Yr A,. +Ar = Ar dyi aYr Oyr

+ kx)

of incident wave in nvo equation in symmetrical be assumed to be a t\vo waves. After adding for sum of two sines, we

-+--=--

and -A;k\

COS(l)t

...(ll

ax

COS(O[

+ Ark!

= -Ark2 =

COSCllt

-Ark2

•••

Solving egns. (1) and (2) simultaneously, A

t -

2k1

k1 + k2

we get

A i

,

k = 21tv in the expression

we obtain reflected and incident wave velocity in the two mediums.

www.puucho.com

(2)

= k1 - k2, A'k ' 1 + k2,

A -

By substituting

Two wires of different linear mass densities are joined. consider the junction to be at x = 0. An incident wave Y i = Ai sinew - k1x) is travelling to the right from the region x :::;O.At the boundary the wave is partly reflected and partly transmitted. Find the reflected and transmitted amplitudes in terms of tile incident amplitude.

+ Ark[ -Aik\

kx) - A sin /exCOS(lJt

29:.----

ux

ax

In the resultant expression, first term is the travelling wave and second a standing wave.

Example

Y r = Ar sin(wt + k1x) the displacement and slope of the string Thus at the boundary, x = 0

At the boundary must be continuous.

= A sin (mt - kx) - ( ~ )sin (wt + kx) ~)sin(wt

wave is

Y r = AI sinew! - k2x)

wave is A/2

2

From principle

of transmitted

2

Y2 = A [sin(wc +h+1t)] =: _

--1-- pO Fig.2E.29

change Equation

intensityoc

Wave numt>er "2 Transmitted amplitude ~

Wave number "1 Incident amplitude A, Reflected amplitude A,

an additional

wave

Wire 2

«0

wave undergoes

Transmitted

Wire 1

Equation of reflected v'/ave is

=(

,

,,~

"

It.

As

in the positive x-direction)

When a wave is transmitted from one medium to other, the frequency of wave is unchanged, wave velocity (I'), wavelength 0..) and wave number (21t/k) change.

Fig.2E.28

Cu' > 11) the reflected

wave is

- k]x)

Ai sinew

(travelling

YI~~YI Incident wave

'I Solution:

of incident

amplitudes

A r = .'-''---'-"-A." A t 1'1

+

l!2

21'2

for Ai and Ar in terms

A,

of

Study Anurag Mishra with www.puucho.com

,

.•. The expression for transmitted amplitude is always positive, hence there is no additional phase change during transmission. Reflected amplitude Ar is positive when V2 > VI' Wave velocity on a string l.s iven by the expression •..

Ii

=!.,

where

11 is

~ Yi

= A,

sin ((,)1 - k,x)

~ y,=A,sin(wl+k1x

+ It)

'" -Po, sin(oll + k1x)

linear mass density.

Transmitted ~

wave

Y, =At sin(wl - k2x)

~

> 1l2' v2 > VI and III < J.12' VI < 112 when a wave travels from a string of higher mass density to that of lower, the wave velocity in second medium is higher and the amplitude of reflected wave is positive. Similarly when a wave travels from a string of lower mass density to that of higher, the wave velocity in the second medium is lower and the amplitude to reflected wave is negative. .•. Medium in which wave velocity is higher is referred as rarer medium and when wave velocity is lower the medium is denser. .•. When a wave is reflected from a denser medium (fixed boundary, rigid boundary), the reflected wave suffers an additional change in phase of :t. •.. Fig. 2.40 (a) shows the reflection of a pulse on a rope with a fixed end point. When the pulse arrives at the boundary it exerts a vertical force on the fixed point. which in tum exerts an equal and opposite force on If

III

,

(,)

Incident wave

~ Yi

= A, sin

((01 - k,x)

Refleded

wave

~

Transmitted

wave

~ Yl = ~

y, '" A, sin(rot + k1x) Denser

sin(rot - kzX) Rarer

(b)

Fig. 2.41

•.. When a wave travels from one medium to the other having different physical characteristic there will be a redistribution of energy. A part of the incident . energy is reflected at the boundary and a part of the incident energy is transmitted. Intensity, defined as the energy transmitted per unit area per sec is given by I = ~p"ro2A2 2 I "" A 2

Hence the amplitude of reflected and transmitted wave is different. (,)

Example

(b)

Fig. 2.40

the string. When the string exerts upward force on the end point,. it exerts a downward force on the string. This downward force on the rope generates an upside down reflected pulse travelling in the opposite direction. •.. Fig. 2.17 (a) shows the reflection of a pulse on a rope with free boundary, the ring at the end can slip smoothly on the rod. The free end rises until all the energy of the end segment is stored as elastic potential energy. The free end comes to rest at a maximum vertical displacement twice the height of the crest. Carried up by its inertia, the end segment pulls upward on the rope, generating a reflected wave pulse that travels back toward the source, right side up and reversed in direction.

Two waves of equal amplitude, both travelling along the positive x direction and wavelengths A) = A,A2 = A +.6.A superpose on a string. Discuss the shape of the combination at t = Q compare it with the beats phenomenon. Determine the distance from peak to peak of the amplitude modulating factor and number of wavelengths contained between successive zeros of the modulating envelope. Solution: The two superposing waves can be described as

www.puucho.com

YI

=Asin[~~(X-vt)]

Y2 = ASin[~: (X-N)]

Study Anurag Mishra with www.puucho.com

WAVE MOTION In terms of wave number k = l/t. (number of complete wavelengths per unit distance, it need not be an integer), we have, from principle of superposition, Y z = A{sin[2rrkl (x If we substitute t = 0, then y = A[sin2rrk]x+ sin2nkzxl

y = YI +

= 2A cos[n(k1

-

Total kinetic energy at time t is

- l't)] + sin[2rckz (x - vt)]} o

x)

mA2w2 sin2wt 21

2

2

2

rordx

i\.2

Jo

sin2 wt

mA2w2 Kmax. = ---

4

A,\1 = 2A cos[n(k1 - kz)x]

The distance between points where amplitude modulating factor becomes zero is defined by the change of x corresponding to an increase of 7t in the quantity n(k\ - kl)x. Denoting this distance by D, we have

o

4rr2mA ~_ 2y2

4

f

2n

Ttl!

,w

(K)o_o

(b)

_

f:n,w

It

D=

2

A w sin kxsin

4 sin2wt=1

For KEma •. ,

The amplitude modulating factor is

or

"2

= .!.mA2w2

kz)x] sin( 2rc k1 ~ k£

n(k] - kz)D =

l(m)I

I=i\." 0 -

J

K=

dt

1

kl

-

kz o

=

==

i'l'.2

').2

)'2-1..1

L\A

0

Number of wavelengths contained in this distance is D I" --~. A bol,

Example

31

_

A string of mass m is fixed at both ends. The fundamental frequency of excitations is v and maximum displacement amplitude is 4. Find (a) the maximum kinetic energy of the string. (b) the mean kinetic energy of the string averaged over one oscillation period.

8

VIBRATIONS

. )ij( .

d~"'--Arosinkxsinror

.

at

Consider a differential element dx of th,e string at distance x from fixed end. The differential element has mass m dm = ~dx

OF A TUNING FORK

The transverse vibrations of the prongs of a tuning fork are similar to a rod free at both ends (see Fig. 2.42). The free ends of the prong are antinodes and the respective nodes at the base. Both the prongs are displaced simultaneously from their equilibrium positions are in the same phase. Imagine a straight rod shown in the figure kept at suppons P and Q bent to give it a U shape. In this situation the nodes get close to each other. Midway benveen them lies an antinode where

Solution: (a) The equation of a stationary wave in a string fixed at both ends is ~ = A sin kx cosrot

P A ";

'N ,

,

A

N

.QA

• NAN

•

1-)

(as)' at

-. : ..

A'

: "A :: "': ""

:' ::

NAN

N

". P,'-, -- -••••Q ,.A ,.Jl ••• _ •• __.- ••••• A

1 The kinetic energy of this element is 1 dK="2dm

_

I (0)

fig. 2.42

=.!. m A2ro2sin2 kxsin2wt 2 1

As the string vibrates in its fundamental mode, the length of string is I = ,.j2.

a handle is fixed. During the oscillations of prongs the antinode rises up and down thereby setting a longitudinal wave in the handle. When the vibrating handle is kept on a table, a loud resonant sound is produced. When a prong of a tuning fork

www.puucho.com

Study Anurag Mishra with www.puucho.com 170

MECHANICS-II

is hit by a rubber hammer, it oscillates in its fundamental mode, thereby giving a pure fundamental harmonic. For a tuning fork of rectangular cross-section, the fundamental frequency is given by

v=

A)!'. p

~~~~ i~

••••••

String at rest: lowermost

position

(.)

t

-----r---- •• I III ••

P~'g'

/'

at rest

where Y = Young's modulus p = density of material of tuning fork

~

String moving: ma~ velocity upwards

(0)

= thickness

of tuning fork 1 = length of tuning fork When a tuning fork is waxed, its frequency decreases due [0 increased inertia. t

••••.

'~

Prongs~ moving ou~

.~._ ~ String at rest uppermost

posifion

(e)

When a tuning fork is field, its frequency increases.

Prongs

Melde's Experiment 1. Transverse mode: In this mode of vibration, the prong's velocity is perpendicular to the length of the string

at rest

String moving: rna:>:velocity downwards (d)

T

..

FIg. 2.44

p

E

p

s M

2. Longitudinal mode: In this mode of vibration the tuning fork's prong moves parallel to the string. The frequency of the transverse vibration of the string is half that of the tuning fork. Figs. 2.45 (a-d) illustrate the mode of vibration, the time in which the prong completes one vibration, the string completes only half. In longitudinal mode,

•

Fig. 2.43

Strtng at rest: lowermost position

[Fig. 2.43]. Due (0 vibration of fork, transverse vibrations are set up in the string of the string vibrates in p loops, we have N = (v

p

(OJ

P

s

E

)crans.

= Pl~ns.

P T

:u.

:+tttt-H+ String moving : ma~ velocity up (e)

JTt~ns,

• where N is the frequency of the tuning fork. Figs. 2.44 (a-d) show how the vibrations proceed. The frequencies of the tuning fork and the string are the same, but the string's vibrations lag in phase by Tef2 with the vibrations of the tuning fork.

String at rest: uppermost position (d)

(.)

Fig. 2.45

N

2" = (Vp)long.

--21

_ Plong. JTIOng'

www.puucho.com

-~

Study Anurag Mishra with www.puucho.com

171

WAY! MOllON Model for Sound Waves in a Gas When sound propagates through an elastic material, the propagating disturbar:ee is a variation in ambient density

due to bulk shift in the positions of the partic:es of the material from their equilibrium position. Air has elastic properties to feel it take an air pump, "I'!tcn we press on the plunger, a restoring forcc is quite apparent. Consider the wall of a speaker that oscillates bClCk and forth simple harmonically. As it moves to the right it accelerates the air pankles near it. The volume of a thin layer near the \\'all is compressed because air particles beyond this layer have not moved. The compression is represented by dense dots near the wall in Fig. 2.46. The density in this region is greater than the ambient density. The additional momentum transferred to the pan:ides by the moving wall is transferred LO the other particles to the right by collisions like this a compression pulse propagates through the gas.

Concept: As the wall is oscillating in S.H.M .• the air particles also e.xecute S.H.M. about their original equilibrium positions. The speed of the particles is zero at the maximum amplitude and mw.imum elf the mean position. The maximum density occurs at the centre of a compression where particles have the greatest velocity in comparison to i/.~ neighbours. Fig. 2.47 shows the velociry of particles by different arrow lengths. The particle at the middle of the compression tries to break away from particles at the left and catch up with those at tl1eright. As a result, the compressional pulse moves fonvard.

-

Centre of rarefaction is at equilibrium position of particle's simpie harmonic oscillation with particle moving to the left

-..-.+-

Magnified view of region around centre 01 rarefaction

Fig. 2.47

EqUilib

1

~~~i:i~~.~I.~~~I 'um ••• ~

In a similar manner, at tl1e position of rarefaction, the vmiation in densiry has its maximum value below the ambient density. The particles at the middle of rarefaction are moving faster than those on the right and left as shown in Fig.

Wall

................. t :_(V) +Vz) PIV} +PzVz PmiJc. = V +V j

z

Velocities in the respective gases are '} =

and

www.puucho.com

Cz =

f? f?

\ p,

\ p,

Study Anurag Mishra with www.puucho.com

178

MECHANICS.II

where Y is (he Young's modulus of its material and pits density. The speed of sound in a fluid is

X(x,t) = Asin(kx-rot) Show that the pressure variation in the medium is given by

v=J!

2

pv MJ= - 2n -,-A (

where B is bulk modulus and p ilS density. For adiabatic compression of a gas, B = yP. For air, Y "" 1.4. Hence we have vair

1.44PPAI

t'A!

YPair

--=

with

Palm.

= l.Ol3x 10-5 N/m

2

, ,

Y=7.05xlOlON/m2 PAl =2.7x103kg/m3

Displacement t

5000

(b) Suppose the bar is held at a distance x from the struck end. We have

v

X=-=4 4v =

The change in pressure of a gas is related to changes in its volume by bulk modulus, which is defined as

v

3

4v

Area

and

time

=A

...

M' B=--~ViV

""

.! m from the struck

end. If it is

MJ = -Bad'=

-=-

Fig.2E.39

"V V

AM --. Mx

dX = -Bad Ax

or v = 1875Hz [not the same as before] (c) If the bar is struck transversely, me wave generated is transverse and the velocity of propagation is

dX

If we take limit as lix approaches to zero, -

ax, a partial

v=~

-Bad

....

Front edge of disturbed region

Thus pressure change,

3 so held but struck at the other end, we should have

2

x + Ax

~v= A&"

5000 1 =-ffi 4x 3750 3

Hence the bar is held at

at

Volume change associated with this displacement,

= 341m/s

,

molecules

The displacement of air molecule, M = X(x+ Ax,t) - X(x,t)

YPair X

of

'= X(x+Ax,t)

Vair=VAI

= 6.83 X 10-2

cos(kx-rot)

Solution: Consider a piston that can oscillate back and forth in an air-filled tube. As the piston moves to the right it compresses the air and sends a compression down the tube. When the piston moves toward the left, the air to the right of it is expanded; at that time a rarefaction is sent down the tube. The volume enclosed by the cross-sectional area A and the length Ax is AAx. Displacement of molecules at x and time t = X(x,t)

Pair = 1.16Skg/m3 1,44PPAI

)

ax

where G is the shear modulus. As the shear modulus of a solid is generally smaller than its bulk modulus, v is now smaller. And as

derivative, as t is kept constant. !::JP =

becomes

""

ax

-Bad -

ax = -Bad l. [A sin(kx - wt)]

ax

v

v=21 the frequency generated is lower.

= -BadAkcos(kx-wt)

The bulk modulus is related to the velocity as 2

Bad = pv

Example So Consider a longitudinal (compressional) wave of wavelength A travelling with speed v along x.direction through a medium of density p. The displacement of the molecules of the medium from their equilibrium position is

MJ = -provA cos(kx - rot)

,

2n:pv2 = ---Acos(kx-w)

~ Note that volumetric strain produced in air as a longitudinal wave travels through it is given by

www.puucho.com

Study Anurag Mishra with www.puucho.com

179

WAVE MOTION

Power in Sound Waves To understand energy transport by sound, we return to the cylinder piston model one-dimensional wave in a tube. As the piston oscillates it does work on the air in contact with it. Energy is transferred to the air and transmitted down the hibe by the sound wave. The amount of work done per second on an elementary slice by the air to its left is: ~~ Power = F.v = A[P(x)l'(x)]

Piston motion

-~: -=::: - ~: - -::=:g;:::j ~

.

= A [Po

x Displacement

= A[PoOlScos(1exo

.. . ...... .. "fd-P::

Pressure

.

change

, ,

"V = ax v ax

mt)

1

'" -Akcos(kx-ffit)

X = A sin((I)f - kx)

%)

ENERGY IN SOUND AND LIGHT Wave Fronts and Rays Waves moving at speed l'w in two or three dimensions spread out from a point source of energy in ever-eXpanding circles or spheres called wave fronts. Concept: With the origin at the wave source, the wave

phase is: $=kr-rot+l1>o It has a fixed value on each wave front. As time passes,

each wave front expands at the wave speed, but its phase remains constant:

de;> dr -c=ok--ro

(Ill)

+ P. wSo cos (lex-

1

wt)]

1

I'sP.2

=

IP.2 -0)1\--

2

'(kPo

P.2

=A--=A-2yPo 2povs

Power is proportional to the square of the wave amplitude P•. The average power is also proportional to A, the area of our slice. The intensity of a wave is the average power transmitted per unit area of water front.

M = BAkcos(rot - h)

COs( wr - k:o: +

cos(1ex2

= A(O+-P.wso) = -AwP.so 2 2

Maximum volumetric strain produced", Ak •.. it is important to recognize that the variation' of pressure M is relative to atmosphere. •.. The pressure and displacement wave arc Te/2 rad, out of phase from c [2(1 + cos2rrt)][-4rrsin2rrt]

=

1 => 1 + cos 2m :: 0, cos2nt =-2

From (1),

a

2nt = mt

"[

t="2

v

o

zr

dT

where t.v is the frequency difference induced in the string as a result of a change in tension L'l.T. In other words, L'l.v is the number of beats observed if the string's tension is changed by an amount t.T. Using the given data,

3

- -~3""] t ---"2'_31 [t -=31.56'

dv

"v=(2~01C~ol

1 3 ]

1=0''2,1'2''''

= 1 cycle

Example fort

.... 1ntenslty IS minimum

,154

and

Imin.

3 6 3

9 16

... etc.

a string are 75, 125, 175 Hz.

... ]

(b) What is the fundamental (c) To which harmonics

116[ at t = ~, ... ] t=~'~I~>

the string is fixed ar one end or at both

(a) State whether ends.

[au=O,l,

=o[at

70 _

Three successivejrequenciesfor

lort = -,-,-,

=-

I mnx. =

S-l

= O,~,l, .... , etc.

Intensity is maximum

[max.

4~T(~rI2

\' t.T L'l.v""-2 T

Hence,

sin 2rrt = 0

2m = mt:t 2n

)-"

dt - 4LT Tp2

frequency

?

do thesejrequencies

correspond?

(d) Taking the speed of the transverse wave on the string as 400 mis, determine the length of the string.

J

...

In one second, intensity becomes two times maximum and two times minimum. Hence beats frequency is 2 Hz.

Solution: (a) The three successive frequencies are in the ratio 3: 5 ; 7, the odd harmonics are present, therefore, the string is fixed at one end. (h) The fundamental frequency is f = 25 Hz (e) The given frequencies are the third, fifth and seventh harmonics.

Two identical wires arc stretched by the same tension of 100 N, and each emits a note of frequency 200 cycles/sec. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.

(d) f =-v => 41

I

v 0

-

4f

=--400 = 4m 4x25

Example21-~

Solution:

The frequency of the fundamental note emitted by each •••.. ire before the tension change occurs is

v= ~

(~r2

...

(l)

A tuning fork and an organ pipe at temperature 88°C produce 5 beats per second. When the temperature of the air column is decreased to Sloe the two produce 1 beat per sec. What is the frequency of the tuning fork?

If T changes, v will also change. We can find the relation between these 2 changes by taking the derivative of (l) w.r.t.

r

www.puucho.com

Study Anurag Mishra with www.puucho.com 200

MECHANICS-II

Solution: For a one end dosed organ pipe,

First overrone of dosed organ pipe = ~

= (211-I)v

f

41,

41 For open organ pipe,

f

Wave velocity in air,

v = )~

f

We can see that and Hence we can write

.

oc

"..

~-~=2.2

4l~ 210

.fi

or

= K.ff

~-~=2.2 4l~

3x110- 330 = 2.2

If the frequency of the tuning fork is f' . two expressions for beat frequency can be written !,-K.ff = 5

I, or

330_3xllO=2.2 I,

K.ff-!'=5

t"

With decrease in temperature the frequency of organ pipe decreases. If we choose the first expression, as K JT decreases, the beat frequency must increase contrary to given condition.

21,

,

210

f

DC

•

These -two produce beats frequency of 2.2 Hz when sound~dtogether, the expressions for beat frequency are

v

oc

v

Firs( overrone of open organ pipe = ~,

= ;;

Hence the second expression

which on solving for

f'

(1) (2)

'Find the temperature at which the fundamental frequency of an organ pipe i5 independent of small variation in temperature in tenns of the coefficient of linear expansion (a) 01 the material of the rube.

18

Solution: Fundamental

yields

f

1'=71 Hz

= 1.0067m

10 = D.9937m

, 'Example

is permissible.

Kh73+88-!,=5, !,+5=19K Kh73+51-!'=I, !,+I=I8K So dividing eqn. (1) by (2), 1'+5 19 --=f'+l

or.

frequency of an organ pipe,

= ~. Length I of the organ pipe at temperature 21

T

1=1,[I+u(T-T,)] Velocity of sound, The first overtone of an organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find (he lepgths of. rhe.,pipes.

t: r:,

'--....,---

41

J~

,

h

--~---=-21,,[l+a(T-T,,)]

T

•... [I +

Open organ pipe, nv

where n = 1,2,3, ...

10 =2i' ,

-41, = 110 Hz = --4x

l~

110

1+.!. (T - To 2 T,

.!. (T m

(T;,T, JJ

= 1+a(T-T)

'

n =1+u(T-T,)

For small (T - T" )we may use binomial approximation,

Fundamental frequency of dosed organ pipe, v 330

21"

Jt,

3

• w eren=l,2,

lY~T

We have to find the temperature at which f(T) = f(To) for small (T - To)

Solution: The frequency in nth mode of vibration of one end closed and an open organ pipe are given by Closed organ pipe, , = (211- I)v

v=

2

= 0.75m

- T, T,

J = 1+ a(T - To)

J = u(T - n' '

I u=-orT=-

ZT ,

www.puucho.com

I

'2a

Study Anurag Mishra with www.puucho.com

201

WAVE MOTION

Example

. 2"(5.)

P::APoSill-

74y--

),

Mh

the maximum

amplitude of pressure

o sin( 5; )

!>R :=:

(c) At the free end, X:=:0, P =0 0, because it is a node, so Pma.~.=Pmin. =Po

(a) Find the length L of the air column.

(d) At [he closed end,

of pressure variation at the middle

of the column? (e) What are the maximum and minimum pressure at the open end of the pipe ? (d) What are the maximum and minimum pressure at the closed end of the pipe ?

X=L=-

-5" = 440Hz 4L L= 5x330=~m

4x400 16 (b) The displacement function and pressure variation function differ by a phase of Tri2, at the position of displacement aminode there is pressure node and vice-versa. The amplitude of pressure variation of standing wave with x :=: 0 at its open end will be

N A

N

r

x"'O

A N

,n P=llroSill

N A N

1

Example

A

x'" L

Standing wave in terms of pressure

Fig.2E.74

. (2") T

APo sm

T

32 m

"oooo

:::~

:~i

T

'2 m

t

24 m

2.0m

~U

1 :::~!!~! 1 }~

...

t

'6 m :~~

~~~~

:::~ 1

I

2.8

m

fa.

m ~~:~

Fig. 2E.75

x

Solution: The frequency of one end closed organ pipe is given by (2n

For the second overtone,

,n =uro

75..---

p:=: APo sin kx

=0

--

A 3.6 m long pipe resonates with a source offrequency 212.5 Hz, when water level is at a certain height in the pipe. Find the heights oflVater level from the bottom of the pipe at which resonances occur. Neglect end correction. Now the pipe L~filled to a height H( =- 3.6m} A small hole is drilled very c/o.,e to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of \vater level in the pipe as a junction of H. if the radii of the pipe and the hole are 2 x 10-2 m and 1 x 10-3 m respectively, calculate the time interval between the occurrence of first two resonances. Speed of sound in air is 340 m/s andg:=: 10m/52.

N

Standing wave in terms of displacement

(2"), SA) 4

prnin. = Po -ilPe

and

mf

A

.

Pmax. = Po + Me

04

l

SA

4 The amplitude of pressure wave,

Hence, Solution: (a) Fundamental frequency of a one end closed organ pipe is ,,/4L, only odd harmonics are present. The second overtone is fifth harmonic.

A

}!

-12

variacion. (b) What is the amplitude

8

=0 AP

The air column in a pipe closed at one end is made to .vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed a/the sound in air is 330 m/s. End corrections may be neglected. Let Po denote the mean pressure at any point in

the pipe and

-J.

+ l)t'

X=o!::.=oSI,

v = .... --, where n = 0,1,2,3, ... " 41 The first five harmonics are v 3v SV VI =-, V3 =-, Vs =-, v, =-,7v '1'9 =-9v 411 412 413 414 4ls

2

The exciting source is the same for all the harmonics,

L

=0 51_ 4

At the middle of the column, 8

Pressure amplitude at the middle,

www.puucho.com

Study Anurag Mishra with www.puucho.com

MECHANICS.II

202 VI

"'.....!::'.... = 212.5, 41,

II =

\'2

=~=212.5,

1=

340 = DAm 4(212.5) 340x3 =1.2m

412

2

4{212.5)

V3=~=212.S, 413

1=

340xS

3

4{212.S)

v 4 = l.!:!.... = 212.5, 414

I

m

=2.0m Flg.2E.76

4

;;;; 340 x 7 = 2.8 m 4{212.5)

n=~{TI A

w

15= 340x9

s =~=212.5,

V

415

-3.6m

4 x 212.5

The heights of water level at these resonances are hI = (3.6- OA)m = 3.2m,

hz = (3.6-1.2) m = 204m

h3 = (3.6- 2.0)m = 106m, hs = (3.6- 3.6)m = Om

h4 = (3.6- 2.8) m = a.8m

where r is tension in string, m total mass of wire, L length of wire. A wavelength of vibrations. Wavelength of the transverse vibration will remain unchanged, since distance benveen pulley and wall is constant. Consequently length of resonating part of the wire remains unchanged. Change in length of wire, dL := La dr. As

Velocity of efflux through small orifice is

tlw

)2gh From equation of continuity, (a)2gh)dt = -Adh 0

(337 -IJQ: 327

att = 1

Solution: (a) y = 3sin2

rs

7

As Q is heading rowards the wall, we can assume its reflection ro be emitting the sound, Q' is the acoustic image ofQ.

jCx-I)

103 ~

: ~[2sin(sOlt)sint

7

P:Q+-orP:Q-2

v':

c = 1mls

Example

Wall

Fig. 2E.103 (a)

seen, the string \,,-ilIvibrate in its second harmonic. 2L \M

~ ~ ~ ~

Fig. 2E.102 fb)

(b)

.

,

Q

Let P and

rr

Q be the

rr

frequencies of tuning forks P and

" "". ",

Fig. 2E.104 (a)

(b) A sonometer wire of 2.20 1/1 is fixed between two ends. Where should the two bridges be placed to divide the wire into three segments .~uchthat the fundamental frequencies of the second and third segments be in resonance with first and second overtone of the first segment respectively.

www.puucho.com

Study Anurag Mishra with www.puucho.com

224

MECHANICS-II

Solution: (a) As P and Q are the positions of the steamers, the directions propagation of sound is along PQ LHPQ = 45' Velocity of steamer P along PQ = 10/.fi

ml' = 7.07 ml'

Likewise, velocity of steamer Q,aiQng PQ

10/.fi ml' = 7.07 ml'

=

v=(~lvo

bottom of the string =

= (332+7.07)"300

33Z-7.07)

.

v

(j!) '. ".

"

"" .... '.

:,

Then, I} + '2 + 13 = 2.2m.

v=2.zl'I;;;iT =- "212 =- 0313

as tension and mass per unit length are not changed 1/n1:1/n2:1/n3

Now, "2is in resonance with the second overtone of segment one. =>

"2

and, "3 segment one.

~

=-

2n1

But

',=~G=~G=~

Where k is a constant ::::) 6k+3k+2k=2.2 k ::::)

= 2.2/11 = O.2m

11 = 1.2m, 12

=

O.6m, 13

for a body to fall from rest through a r(l-X) g

t(x) IS

=>

rCI-X) g

=

zIT vi -zl¥\g

x=lorl/g

x = I represents the situation at the instant the pulse and the body are released, so they meet again at a distance l/g from the bottom of the string.

Exp.rnple ,= .'_. - . iTwo particles of a medium disturbed by the wave propagation are at Xl = Oand Xz = 1 em. The displacement of the particles is given by the equation:

YI = (2sin 31tt) cm y 2 = 2sin(3Jtt - n/8) cm (t is in seconds)

=

1,+12+13=2.2rn

~t

dx =Z(l/g)v'-Z(x/f),n

,JiX

Determine the frequency, wavelength, wave velocity and the displacement of the particles at t 1 sec and x 4cm.

311}

/\:12:13 =- 1:1/2:1/3 ',:12:/3 =- 6:3:2

~

The time . I'- x dlstance

is in resonance with the second overtone of

"3 =-

= JiX

Solving the above equation, we get

where 11 is the fundamental frequency of a segment, Tis tension and m is mass per unit length

11:12:13::::

Mil

Equating T with t, we get the value of x for which the pulse and the body meet.

Let the lengths of the segment be 11.12 and 13

n1'1

T=J

:

Fig. 2E.104 (b)

~

Mgx/l

,

...... ~a

(b)

=

If T be the time taken for the pulse to travel top to the point P, then

:,

'.

Mf

Speed of transverse wave at this point is

Hz

~ "

A uniform inextensible string of length 1and total mass M is suspended vertically and tapped at the top and so that a transverse impulse runs down it. At the same moment a body is released from rest and falls freely from the top a/the string. How far from the bottom does the body pass the impulse? Solution: Tension at a point P at a distance x above the

c -vp

=- 313

.Exarnple

=

Solution: As the motions of particles are simple harmonic, amplitude = 2cm CJ) (Angular frequency) = 3 IT rad/see. As the phase difference of panicles separated by 1 em is iV8 8 A =-

"

= OAm

www.puucho.com

X 21t

= 16cm

Study Anurag Mishra with www.puucho.com

225

WAVE MOIION Hence, wave velocity = ~

2.

The wave equation,

Example

x A = 24cm/s

then would be

Consider the shown arrangement. The system is released from the state of rest at t = O.At t ;; 0, a transverse jerk i5 produced at point A (see figure). If the block m2 is at a di5tance lfrom the pulley, at t = Q Obtain the time required for the jerk to reach the pulley. A~sume mass per unit length of the string is 11. {Ignore friction. Mass of the string is much less than the mass of the blocks}

J

y = (20aO) sin[ ~~ (24t - x) em. Putting

t

= 1 sec and x = 4cm, we get y

Example

=

2cm

107...--

A

A cylindrical volume of cross-section areQ A and length L contains a uniform mixture of three gases H2,Ol and CO2" If ellCir ratio by volume be 1: 1: 3. Find the time taken by a sound to travel this length L if the temperature varies linearly from TIC= SlOe) to T2(= 88°C) across the length of cylindrical volume.

(~R)+(~R)+(3R

Solution: Cv(mix) = _2

2

108...>

m,It-------C7~ ,.~p

x 3) _

5 Fig.2E.108

= 21 R = 2.1R 10 3.8 'mix = 28

Solution: Acceleration of m2 =

Speed of sound at a point where temperatures equals

,IY~Twhere M is mass, of one mole.

Here,

is T

T1 = dt

dx.lii = ~

2 mIg

mjm2g

(mj +m2)

m] +m2

T =mlg---~~~

In grounds reference frame, speed of transverse jerk de

(TTT)

== (T] +mx)==

{!;. = v, say

\~

. ) dl = v + at ( at time t as, - = v + at

as

"lyRT T(x)==Tj+

=

"" a, say

where 11is mass per unit length and T is tension in string.

Moz +MHz +3Mcoz M = ---------

5 Time taken to travel distance dx at distance x from end at temperature

Speed of transverse wave

m]g (m] + m2)

, f(v+at)dt=vt+-at2 o

x

1' [~

r

==m

]

1

2

=

2l-2vt-at

=

at

=:}

-2v +.J4v2 + Sal r = ------(rejecting the -ve value)

2

2

=0

+ 2vt - 21 = 0 2a

=

Iv

2

+ 21 _ ~

\ 02

Examgle

,ay Sohing, we get { = 3.511sec.

a

a

109 ~

In the shown arrangement, the end A of string i5 fixed to the rigid support and end B is kept free. A transverse pulse i5 produced at P at [ = 0, which moves towards right. if [he shape of [he string changes periodically, find its time period. Tension in the string = BON. Mass per unit length of string is

0.2 kg/m.

www.puucho.com

Study Anurag Mishra with www.puucho.com

226

MECHANICS.II As the phase at S, is ahead

Ap30cm

-p-----~:B -

2:;:2,

.1

I 100cm

Fig. 2E.109 (a)

Solution: Speed of transverse jerk on the string = v, say = ~

a

Shape of string at

t

=

Shape of string at

t

= 2(PB)

,

a!le~'="="=':::::::::::::::::::.o~

to that at So by the disturbances reaching at P from SO S, and So will be in phase. FIg.2E.110(b) Similarly, it can be shown that all the disturbances reaching at P will be in phase. H~~ce, amplitude of resultant disturbance at P = 7Ao (If Ao be amplitude of individual disturbance). As intensity < (Amplitude)2 ,J p = 4910,

ALP __ -----B

v) Shape ofSldng att = 2(P:

2(P:)

Shape of string at

t

=

Shape of string at

t

= 2 -;-

PA)

(

r-v

A~

B

A~-------

B

A~C--

B

ALP- __

B

r-v

14 hollow spherical body has inner and outer radii as R1 and R2 respectively. Thermal conductivity of the material of the body is k.. There is a nafTOWpassage in the body as shown in the figure. At t = 0, temperature of air in the spherical cavity is T while outside temperature is To( < T). lfno be the number of moles at time t. Consider air to be largely consisting of diatomic gases.

Fig. 2E.109 (b)

Hence

the

required

time

.

period, T = 4AB. Putting the values, we get T = Q.2sec.

v

Example A number of sonic sources emitting notes of same frequency x are located at points y = ina on y-axis where n = 0, I, 2 and 2

3. If the phases of sources at

t =

0 be 0 for So, (20 bi""" ") for 51

and 5'1; ~~1t for 52 and S' 2; ~~1t: for S3 and S' 3 respectively, find the resultant intensity of sound at a point on x-axis at a distance b from a (b » a), if the intensity due to single source is 10, y

Fig.2E.111

Solution: Thermal resistance of the spherical body

= _1_7dx2 4Jtk

"

~

x

R]]

= _1_[R2 4nk R2R,

r

= r say '

'" (4)

nR

,

0= R(ndT +Tdn) ndT = -Tdn From egns. (iv), (v) and (vi) CprPoV Pov

0

5', 5', Fig. 2E.110 (a)

, Solution: Path difference SIP -SoP = asin8

= ~

b

corresponding phase difference =

2

~~-dn = (To--)dt nR nR where Po is the atmospheric pressure.

J nRT

dn

o - PoV Integrating we get, "0

=)dt

a C prPo V

a (2n)

b,

11

www.puucho.com

,... (2) ,., (3)

PV=nRT:::::)(T=PV]

5,

... (1)

r

dQ = nC pdT nC prdT = (To - T)dt

Also

5,

,

1

dQ = (T, -T) =>dQ = (T, -1') dt

5,

5, 5'

R

= noV [1+ (To _1)e7~~r] RTo T

". (5)

",(6)

Study Anurag Mishra with www.puucho.com

227

WAVE MOTION

DESCRIPTION 1.

OF WAVE

Show, for a sinusoidal transverse \\'ave travelling on a string, that the slope of the string at any point x is

5.

Show that if damping is ignored the amplitude A of circular water waves decreases as the square root of the distance r from the source: A /3 I, = I, = [, >

(b) (d)

I, < I, < I, I, = I, + I,

2 A listener is at rest with respect [0 the source of sound , A wind starts blowing along the line joining the source and the observer. Which of the following quantities do not change? (a) wavelength (b) frequency ee) time period Cd) velocity of sound 25. A sound source S starts falling from a height of 180 m from horizontal surface under gravity and emits 5500 Hz sound continuously. The distance between . the source and observer at an instant, when the - .frequency heard by the observer is also 5500 Hz, is (Speed of sound in air is 330 m/s and collision of source with surface is perfectly inelastic) : 24.

-2tr"102Pa

Source

PO'

Observer

O!.

. :

SSm

• P. (e)

.Q

440m

0

0.1

(a) SsJ80m (e)

-Sn> L\.n2

Cd) 2-/2 "0

Ag

(b) L\.nl = L\.n2

,'(e) L\.nl < L\.n2 (d) none of these 42. .Which of the following functions of x and t represents a progressive wave : (a) y = sin(4t - 3x)

(b) y = ~~

4

Ag

37. A source of sound is moving with • S 12 • • A B velocity 11 / 2 and two observers A and B are moving with velocity '1/' as shown. Find ratio of wavelength received by A and B. Given that velocity of sound is lOu : Ca) 19/21 Cb)21/23 Ce) 17/23 (d) 17/21 38. When beats are produced by two progressive waves of nearly the same frequency, which one of the following is correct : (a) The particles vibrate simple harmonically, with the frequency equal to the difference in the component frequencies. (b) The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves. (c) The frequency of beats depends upon the position, where the observer is, (d) The frequency of beats changes as the time progresses. 39. There are cases when an explosion at a point A will be heard at A point B that is far away from A while in a certain region located much closer to A than to B, the explosion is not heard due to obstruction. This will be possible if : (a) air temperature increases with altitude (b) air temperature decreases with altitude (c) air is blowing from B to A (d) air is blowing from A to B 40. A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork. if their oscillation frequency is 6BO Hz and the velocity of sound in air is 340 m/s : (a) 2 mls (b) 1 m/s (c) 0.5 m!s (d) I.S mls 41. When a source approaches a stationary observer with velocity v then apparent change in frequency is dnl,

•

"

•

•

"

1

(c) y = --4t + 3x 43.

If

equation

y = xo cos 21t( nt -

1

--

+(41-3x)2

(d) All of these of

f).

transverse

wave

is

Maximum velocity of particle

is twice of wave velocity, if A is : (a) (c)

(b) 2ltxo

ltX

/2xo (d) It Xo 44. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is Y I = A sin (It x / L) sin 0)t and energy is 'E1 and in another experiment its displacement is , y 2 = A sin (21t x / L) sin 20lt and energy is E 2' Then It

.' '(a) £2

= £1

(c) £2 = 4£1

45.

A small amplitude progressive wave in a stretched string has a velocity of 10 mls and a frequency of 100 Hz. What is the phase difference, in radians, between two points 2 em apart in the string? ia) ~ 8

•

46.

=

(b) £2 2£1 (d) £2 = 16£1

(b) 3" 8

Ce) ~ Cd) ~ 4 2 A transverse wave of amplitude S mm is generated at one end (x = 0) of a long string by a vibrating source of frequency sao Hz. At a certain instant of time, the displacement of a particle A at x 1m is -5mm and that of particle B at x = 2 is +5 mm. The wavelength of the wave is :

=

(a) 2 m

47.

48.

(b) 3 m

Ce) 4 m Cd) 5 m Two speakers 2 m apart playa frequency not in same phase. Along a line 3 m in front of the speakers, the intensity is heard as a minimum immediately in front of each speaker and there is only one maximum between these points. The frequency of the note is (Take velocity of sound as 330 mls) : (a) 256 Hz (b) 128 Hz (c) 512 Hz (d) 275 Hz A supersonic jet plane mov~s parallcllO the ground at speed v = Q75 mach (l mach = speed of sound). The

www.puucho.com

Study Anurag Mishra with www.puucho.com

245

WAVE MOTION frequency of its engine sound is 10 = 2 kHz and the height of the jet plane is h = 1.5 km. At some instant an observer on the ground hears a sound of frequency f = 2io- Find the instant prior to the instant of hearing when the sound wave received by the observer was emitted by the jet plane. Velocity of sound wave in the condition of observer = 340 m/s : (a) 5.9 sec . (b) 1.9 sec (e) 9.5 sec Cd) 12.2 sec A stone hangs from the free end of a sonometer wire 49. whose vibrating length, in resonance with a tuning fork, is II' When the stone hangs wholly immersed in water, the resonant length is reduced [0 12, The relative density of the stone is :

(b}l-c~r

(a)l-~~ 1

(e)

I -

,

(d)_I_

c:)

1-

•

54. A wave represented by the equationYt"" A cos (kx - rot) is superimposed with another wave to form a stationary wave stich that the point x = 0 is a node. The equation of the other wave is : (a) Y2 = - A sin(kx - rot) (b) Y2 = - A cos(kx + rot) (c) Y2 = A sin(kx + rot) (d) Y2 = A cos(kx + rot) An open organ pipe of length 1 is sounded together 55. with another open organ pipe of length 1 + x in their fundamental tones. Speed of sound in air is v. The beat frequency heard will be (x « l) : 2

It

vx 412

(b) vI

2x 2

quantities are in 51 units. The phase difference between the points are in 51 units. The phase difference ber...veen the points separated by a distance of 10 m along x-axis is :

•2

(d) amplitude of wave is greater than ~

(a)

50. The equation of a travelling wave is given as y = 5 sin IOn (l - 0.01 xl, along the x-axis. Here, all

(a) -

(b) amplitude of wave;s greatec than!:c 2. (c) amplitude of wave is less than ).

(b) •

(cl ..::::..

2(1.

4 51. The equation of a wave disturbance is given as ; Y == 0.02 cos('::' +5Orct) cos(lO rtx), where xandy

,2

are in meters and t in seconds. Choose the wrong statement : (a) Antinode occurs at x == 0.3 m (b) The wavelength is 0.2 m (c) The speed of the constituent waves is 4 mls (d) Node occurs at x == 0.15m 52. Which of the following is not the standard form of a sin wave? (Here, symbols have their usual meanings) (a) y ==Asin 2"(~ (b)y == A sin(vt

(c) y ==A sinw(t

-

f)

2f

It

-~)

(d) y "" A sin k(vt -x) 53. In a phase progressive harmonic wave particle speed is

always less than the wave speed if : (a) amplitude of wave is less than

.!:...

(d)

f

v, v

57. A transverse wave Y = 0.05 sin (201tX - 50rte) in meters, is ptopagating along +ve X-axis on a string. A light insect starts crawling on the string with the velocity of 5 emls art = o along the +ve X-axis from a point where x = 5 cm. After 5s the difference in the phase of its position is equal to : (a) ISO. (b) 250. (c) -245" (d)-5" If 1\ and 1 are the length of air column for the first and 58. 2 second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the distance of the antinode from the top end of the resonance tube is : 1 (a)

2(/, - I,)

(b) -(21, -12)

2

(d)12-/} 12 - 3/1 2 2 A sonometer wire resonates with a given tuning fork 59. forming standing waves with five antinode between two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinode for the same positions of the bridges. The value of M is: (e)

-kx)

21

«
--l

~I'cm

____

0

(d)

63. In the above problem 62, shape of the wave at time t = 35 if a is a fIxed end will be :

o

•

o

(b)

/111cm --.. ,=

and $2 respectively for the two waves.

h is :

(a) I

(b) ~ 6

(e) ~

(d) ~

.4

66.

(a)

$}

$,

1an

(e)

and

"'~

3k

o

(b)

XI

7

The equation for the vibration of a string fixed at both , ends vibrating in its third harmonic is given by; Y '" 2cm sin [(0.6 cm-I)x) cos [(500Tts-1)tJ

The length of the string is : (a) 24.6 cm (b) 12.5 em (c) 20.6 cm (d) 15.7 cm 67. Source and observer both moving simultaneously from origin one along x-axis and the other along y-axis with speed of source = 2 (speed of observer). The graph between the apparent frequency observed by observer (f) and time (t) would be : f

to -------------(a)

..

www.puucho.com

Study Anurag Mishra with www.puucho.com 247

WAVE MOTION

f

o

.•••••

(a) 400 Hz (b) 200 Hz (c) 600 Hz (d) 300 Hz 71. The fundamental frequency of a sonometer wire of length I is fa. A bridge is now introduced at a distance of iiI from the centre of the wire (iii « I). The number of beats heard if both sides of the bridges are set into vibration in their fundamental modes are: (a) 8f,"/ Cb) f,"/ / / Ce) 2fo"/ Cd) 4f,"/ / / 72. A detector is released from rest over a source of sound 3 0: frequency fo = 10 Hz. The frequency observed by the detector at time l is plotted in the graph. The speed of the sound in air: (g = 10m/s2)

---_ ••••

Cb)

(c)

fo

Cd)

"

f (Hz) \

2000

(Here, fo = natural frequency of source) 68. An observer starts moving with uniform acceleration Q towards a stationary sound source of frequency fo. As the observer approaches the source, the apparent frequency f heard by the observer varies with time t as:

Cb)

(a)

1000

t(s)

30

(a) 330 m/s (b) 350 m/s (c) 300 m/s (d) 310 mls 73. A standing wave is maintained in a homogeneous STring of cross-sectional area a and density p. It is formed by the superposition of two waves travelling in opposite directions given by the equation; Yl = a sin (cot ~ h) and Y2 = 2a sin (wt + h) The total mechanical energy confined between the sections corresponding to the adjacent antinode is ; rrspm2a2

3rrspw2a2

Ca)

Ce)

(c)

69. Equations of twO progressive waves at a certain point in a medium are given by Yl = a sinew + 411)and y 2 = a sin (rot + l!l2)' If amplitude and time period of resultant wave formed by the superposition of these n\lo waves is same as that of both the waves, then 1-$2i5: rr

Ca) 3 rr (e) -

(b) 2rr

3

Cd) ~ 4 A string fixed at both ends is vibrating in the lowest 70. mode of vihmtion for which a point at quarter of its length from one end is a point of maximum displacement. The frequency of vibration in this mode is 100 Hz. V to to) +

only

%)

t

= to and

Jt

3

=

2

~

47td\ dz

23.

[a]

Use Doppler's effect.

24.

[bJ

Source observer

f' = c < to. and is the

satisfies

the

above

l'sound - va

14.

(d]

f

25. [b] Source will be at rest ,after collision on the surface at P.

y = A sin kx cos rot

513 = A

sin k x 10

SJ3=Asinkx20

e

y = A sin CUlt + 8)

+ v", + I' ••.

frequency will not change.

29. [c]

.,;a

.Jij

(,)

,IP\P2 cos -3

f=f

equation

conditions. 1 12. [dJ Assume-r = A cos e . - 1 = A sm

dx

v

vsound - t'

=

Also, (~)

d] ::: 0.5

-

=6W/m2

we see that

y = amplitude

d2

Intensity at C : dx f,hCL-x)

1

L\x =

I. 300

J,

From the graphs,

Path difference =

= Zn ill" = 2n-v

Hence total time ist = de = ~==== 2/-:: 8. [eJ

J~

19. [dJ At open or at holes pressure node will form.

where p is the mass density i.e., mass per unit length of the cable wire. L'x

v:

after solving k = ~ . A =

..

la + b \ ab

The functions /2 [,:

or

.'

.-,-

S ••••••••••••••

[b)

V.

v the speed of sound

=-

.

36.

I(_V

is the original frequency of source

JO

F I2 - JO

Thus,

=

III

[J-

2 0..35 - 0.3

= 0.025

- ~v -

1m••

311

-

e=~-~

e

=

2

Lt = 2(l.-t + e)

-

I.

for

)..

Jg~2;~ 2~VO =

www.puucho.com

< 0

Study Anurag Mishra with www.puucho.com

255

WAVE MOTION 100. [aJ

95. [bJ

101. [bJ

=[ 96.

..

]'=4/9

2

,f(j.J.z /~l) +1

[aJ

102. [c]

.

-Asmwt

w = 2rrv = 21t (l00) = 200n rad/s v = w ~ K = ~ = 271:(100) = 20re m-I

v

K

Y

(T+dT)+jlgdz-T=O

dr = -J.1gdz T =

=

y, + Y 2 = -2A sinwc (antinode)

I.F:=-O

also

Atx = a ------andy2 y, = Asin(--wt}

=

30

ssin( 2007t.t

_

3

2~.x)

...(1)

At

2

t=3s,x=lrn,

)lV

y

dT = djlv2 + 2vdv dJl As v is independent of z

ssin( 600rc_ 2~)

=

=

5Sin( 2;)

=

4.33m

2

Let us examine d y

dv = a

d,'

dT= t,2dll

•... (2)

d' --..l...

=

--(l)"2

dt2

From equation (1) and (2), we get

A sin (wr

-](x)

.

~Jd~= -.!...v f'dz Z

"

1.1.

Jl =lloe~[g/v2~

97. [d] Pressure variation = _BdV = _BAS

V

th,

in

medium

""

if we then

103. [d]

(B = Bulk modulus of gas]

98. [aJ

The (c) choice seems to be appropriate, start with y = -4.33 an, 4 2 a = +171 x 10 em/5

For points A and H, path difference = 0 constructive interference at point A and B.

where 1 and 2 be the points on the rope at the base and the top. The m or mass per unit length of the rope = 6/12 = 0.5 kg / ffi. At !.he base the tension is 20 N and the [OP it is 80 N.

At point C, path difference = SOm =~ 2 (A = 100m)

destructive interference at point C. 99. [b] At time

=>

t

0.06 0,5

1'= (/ - v,) = 1./4 Fundamental frequency

10

~

c

104. [dJ

10=--4(1-vc)

C!1

4/2

=

2.)80 A2

0.5

~y:

A2 = a.12m

= 41

c

df -c -=----(-v)=de 4(l-vt)2

_1_)20 c=

Both the gases are diatomic, so their y is same. So the speeds are dependent o.n temperature and molecular weight '

www.puucho.com

Study Anurag Mishra with www.puucho.com • US

=>

~

I ;

-.a:.l1

,

yR(273 + 15) = ~y R[T) 14 16 T=329K

Time gap between beats = 60/40 s.

= (329- 273)OC = 56°C

105. [d] If the echo comes back, whim the next is on, then it won't be clearl, SBmuh, trR~S. (x + x) = 2x, distance

c(:)=

2x

Similarly, when man goes closer to mountain,

x = 270m

www.puucho.com

Study Anurag Mishra with www.puucho.com 257

WAVE MOTION

More than One Alternative is/are Correct

ePevel 1. A simple harmonic particle

progressive

wave, in a gas, has a

of y = a at time t =

displacement

!:. at

4 origin of the wave and a p,utic1e velocity of y = the same

instant

but at a distance

x =

3.

the I'

!:. from 4

at

the

origin where T and I. a[c the periodic time and wavelength of the wave respectively. Then for this wave: (a) the amplitude A of the wave is A = 2a (b) the amplitude A of the \vavc is A :; a (c) the equation of the wave C'1 '" (l X 10- ) sin (5J!x - 25 x lO rrt)

kinetic

over one periodic

(d) the equation of the wa\'c can be represented by a

string

2

(e) The mean

x]

x]

the

EK = -1 mA "(J)

t'

,,[ y=2ucos-t-:'"

its ends. The and it has an displacement

5.

_D_

Figure shows nvo isotropic. • 5, 52 Point sources of sound, S, and 52' The source emit v,'aves in phase at wavelength 0.50 m; they are separated by D '" 1.75 m. If \\'c move a sound detector along a large

www.puucho.com

Study Anurag Mishra with www.puucho.com 258

MECHANICS.II

circle centered at the midpoint betw"een the sources. Choose the correct options : (a) Number of points waves arrive at the detector when waves are exactly in phase is 14 (b) Number of points waves arrive at the detector when waves are exactly out of phase is 14 (e) Number of points waves arrive at the detector when waves are exactly in phase is 7 Cd) Number of points waves arrive at the detector when waves are exactly out of phase is 7 6. A source S of sound wave of fixed frequency N and an abselVcr 0 are located in air initially at the space points A and B, a fixed distance apart. State in.which of the following cases, the observer will not see any Doppler effect and will receive the same frequency N as produced by the source. :' (a) Both the source S and observer a remain stational)' but a wind blows with a constant speed in an arbitrary direction (b) The observer remains stational)' but the source S moves parallel to and in the same direction and with the same speed as the wind (c) The source remains stational)' but the observer and the wind have the same speed away from the source (d) The source and the observer move directly against the wind but both with the same speed 7. lWo identical wave A and B are produced from the origin at different instants t A and ts along the positive x-axis, as shown in the figure. If the speed of wave is 5 m/sthen: y(mm)t

A

-10 x(m)

0.5

-5 -10

(a) (b) (c) (d) 8. The

the the the the

wavelength of the waves is 1m amplitude of the waves is 10 mm wave A leads B by 0.0167 s wave B leads A by 1.67 s (x,y) coordinates of the comers of a square plate

are(O,m,~m(~L)a~(QLlThe~~ofthe~~ clamped and transverse standing waves are set-up in it. If u (x,y) denotes the displacement of the plate at the point (x.y) at some instant of time the possible expression(s) for u is(are) : (a = positive constant). (a) a cos (1tX /21) cos (1t)' / 2L) (b) a sin (1tX / L) sin (1t)' / L) (c) a sin (1tX / L) sin (2ny / L) (d) a cos (2ru / L) sin (1t)' / L)

"9. The speed of sound in a cenain metal is V. One end of a long pipe of that metal of length L is struck a hard blow. A listener at the other end hears two sounds, one from the wave that has travelled along the pipe and the other from the wave that has travelled through the air. v is the speed of sound in air. Suppose that t = 1.00s and the metal is steel. Choose the correct options: (a) Time interval t elapses between the arrivals of the two sounds is L(V + v)/ vv (b) Length L is 432 m (c) Time interval t elapses between the arrivals of the two sounds is L(V - v)/ vv (d) Length L is 364 m 10. A transverse sinusoidal wave of amplitude a, wavelength A. and frequency f is travelling on a stretched string. The maximum speed of any point on the string is v / to, where v is the speed of propagation of the wave. If a = 1O-3m and v = 10 m / s. the A.and f are given by : (a) A. = 2x x 1O-2m (b) A. = 1O-3m 103 Hz Cd) f = 10' Hz 2. 11. As a wave propagates : (a) the wave intensity remains constant for a plane wave (b) the wave intensity decreases as the inverse of the distance from the source for a spherical wave (c) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave (d) the wave intensity decreases as the inverse of the distance fro a line source 12. A standing wave of time period T is set up in a string clamped between two rigid suppons. At t = 0 antinode is at its maximum displacement 2A : (a) The energy of a node is equal to energy of an aminode for the first time at t = T / 8 (b) The energy of node and antinode becomes equal after eveI)' T12 second (c) The displacement of the panicle of antinode at (e)

f

=-

t =

:!: is .J2A

8 (d) The displacement of the particle of node is zero 13. y(x,t)

=

0.82

[(4x + St)

represents a moving pulse + 5]

where x and yare in merers and t in second. Then: (a) pulse is moving in positive x-direction (b) in 2s it \ViIItravel a distance of 2.5 m (c) its maximum displacement is 0.16 m (d) it is a symmetric pulse

www.puucho.com

Study Anurag Mishra with www.puucho.com 259

WAVE MOTION 14.

(a) The number of waves striking the surface per

In a resonance tube experiment, a close organ pipe of length 120 em resonates when tune with a tuning fork of frequency 340 Hz. If water is poured in the pipe then: (given l' air = 340 m / sec) (a) minimum length of water column to have the resonance is 45 em (b) the distance between two successive nodes is 50

second is f

c(e -v)

(b) The wavelength of reflected wave is --fCc +v) (c) The frequency of the reflected wave

em

(c) the maximum length of water column to create the resonance is 95 em Cd) none of the above 15. The equation of a wave travelling on a string is given by y 8 sin [Sm-1)x _(4x-l)t]. Then:

=

(a) velocity of wave is 0.8 m/s (b) the displacement of a particle of the string at t = 0 and x = ~ m from the mean position is 4 m 30 ee) the displacement of a panicle from the mean position att::

0, X =....'!-m is 8 m 30 Cd) velocity of the wave is 8 m/s 16. A wave equation which gives the displacement along the y-direction is given by y = 10-4 sin (60t +2x) where x and yare in metres and t is time in seconds. This represents a wave: (a) travelling with a velocity of 30 mls in the negative x-direction (b) of wavelength It m (c) of frequency 30/ It Hz (d) of amplitude 10-4 m 17. A wave pulse moving to the right along the x-axis is represented by the wave function y(x,t) = 2.0 , where x and yare in (x - 3.0t)2 + 1 centimeters and r is in seconds. (The maximum pulse height is defined as maximum displacement along y-axis). Then: (a) The speed of the pulse is 0.33 COlIs (b) The maximum pulse height is constant with time (c) The speed of the pulse is 3.0 cm/s (d) The maximum pulse height is decreasing with time 18. A sound wave of frequency f travels horizontally to the right. It is reflected from a large vertical plane surface moving to left with a speed v. The speed of sound in medium is e :

(c+ v)

---,

. IS

(e +v)

f---

(c -v)

(d) The number of beats heard by a stationary listener to the left of the reflecting surface is _"I_

e - v

19. Standing waves can be produced: .. ', (a) on a string clamped at one end free at the other (b) on a string clamped at one end free at the other (c) when incident wave gets reflected from a wall (d) when two identical waves with a phase difference of It are moving in the same direction 20. In a standing wave on a string rigidly fixed at both ends: (a) All the particles must be at their positive extremes simultaneously once in half of the time period (b) All the particles must be at their positive extremes simultaneously once in a time period (c) In one time period all the particles are simultaneously at rest twice (d) All the particles are never at rest simultaneously 21.

22.

A travelling wave pulse is given by y =

2+

(x

6 +

3t)2

where symbols have their usual meanings, x and yare in m and r is in s : . (a) The pulse is travelling along negative x-axis with velocity 3 mls (b) The pulse is travelling along negative x-axis with velocity 1.5 mls (c) The amplitude of the wave pulse is 3 m (d) The pulse is a symmetric pulse In figure. a point source 5 of sound waves lie near a reflecting wall AB. A sound detector D intercepts sound ray RI travelling directly from S. It also intercepts sound ray R2 that reflects from the wall such that the angle of incidence ei is equal to the angle of reflection 6r• Assume that the reflection of sound by the wall causes a phase shift of 0.500A. Find the lowest frequencies at which there is maximum constructive interference of RI and R2 at D.

www.puucho.com

Study Anurag Mishra with www.puucho.com 260

MECHANICS.II

A

(c) The fundamental frequency of A is equal to the third ovenone of B Cd) The velocity of wave in B is half that of velocity in A 24. Standing waves are produced on a stretched string of length L with fixed ends. When there is a node at a distance L / 3 from one end, then: (a) minimum and next higher number of nodes excluding the ends are 2, 5 respectively (b) minimum and next higher number of nodes excluding the ends are 2. 4 respectively (e) frequency produced may be v! (3L) (d) frequency produced may be 3v! (2L) [v = velocity of waves in the string] 25. Two pulses travelling on the same string are described

2.50m

"

" B

by

5

o

Yl =-----(3x-4t)2

(a) 39.3 Hz (e) 78.6 Hz

-5

+2

andY2 =------(3x+4t

_6)2 +2

Marks the correct statement(s). (a) The direction in which each pulse is travelling, Yl is in positive x-axis, Y 2 is in negative x-axis. (b) The time when the two waves cancel everywhere is 0.75 sec. (c) The point where the two waves always cancel is

(b) 59 Hz (d) 118 Hz

23. The length, tension, diameter and density of a wire B are double than the corresponding quantities for another stretched vvireA. Then: (a) Fundamental frequency of B is 1/2./2 times that of A (b) The velocity of wave in B is 1/'./2 time that of velocity in A

x= 1m.

(d) Amplitude is different for two waves

AN9WERS

1.

(b,c)

2.

(a,b,C)

3.

(a,d)

4.

(c,d)

5.

(a,b)

9.

(c,d)

10.

(a,c)

11.

(a,c,d)

12.

(a,c.d)

13.

(b,c,d)

(h,c)

21.

(a,c,d)

17.

(b,c)

25.

(a,b,c)

18.

(a,b,c)

19.

(a,b,c)

20.

www.puucho.com

.

6.

(a,d)

7.

(a,b)

8.

(h.c)

14.

(a,b.c)

15.

(a.b)

16.

(a,b,C,d)

22.

(a,d)

23.

(c, d)

24

(a,d)

Study Anurag Mishra with www.puucho.com

----

261

WAVE MOTION

=

Level-2: More than One Alternative

,. ,.

Is/are Correct

1. [b, c] Let

y

the

= A

equation

sin(2.

to

the

waves

be,

l~-f)+ $]

Let the displacement o.f the string be given by

where A is the amplitude of the wave and ljl phase angle. It is given that y = a, when x = Oand also that y =

IJ

A when x = - and 4

l

t

y = v = 2~

=

!.and

cos[ 2~ (~-~)

Q) + ljl]

v=--cosQ T Putting A =a

v

= --2rtA T

(cos

OJ

amplitude

c) =

cos (we + 0)

ay

=

or

sin

-ffiA

1tX

L

sin (we + 15)

The maximum kinetic energy is equal to the string's total energy of oscillation. Note that all points of the string achieve their maximum kinetic energy at the same instant of time, where y = o for all x. Since dm = ~dxwhere

J.I = (~)

The maximum kinetic energy, EK =

27ta =-

T

=

Hence the equation to the wave is

=

. v[ TX] y=asm~t-T

=

y=asm~e--;;

. v[

nil.

1tX

where 0 is a phase factor. So the transverse velocity is given by

2. v = T Q

Where v = gas.

= A sin

is the mass per unit length of the uniform string.

Q = 0, y = a = A, so that

Also,

t)

L

y(x,

2xA

:)

y (x,

4

T =4

Substituting in (1), y =a = A sin (~ + and

3. [a, d]

max.or[Mli;)'dm] max or[~H~)' dx]

The maximum value of (~) sin2(rot

2,

occurs when

+ 15) = 1

x] ,

is the velocity of the wave in the The integral

J sin o

2 1tX

dx over the half.cycle has

L L

the average value of 2

www.puucho.com

Study Anurag Mishra with www.puucho.com

262

MECHANICS.1i

The mean kinetic energy of the string averaged over one periodic time is obtained from (1) by integrating the time dependent factor sin2(roc + 0) over one period, 0 to T. T

Now since

f sin

2

(rot

+ o)dc =

o

!.. 2

The mean kinetic energy of the string averaged over one periodic time is

E.

< EK > = ----!..

T =

f0 sin T

2

.!.[.!.mA2(I)2] 2 4

E

(ror + 8)dc = 2.. 2 =

.!.mA2(02 8

them, approaching or separating from the other. Effectively, it is the medium that moves in each of these cases. The received (apparent) frequency differs from the emitted frequency if and only if the time required for the wave to travel from the source to observer is different for different wave fronts. With a uniform steady motion of the medium, past the observer and source, the transit time from source to observe is the same for all wave fronts. Hence it follows that apparent frequency = true emitted frequency. Thus [here is no Doppler effect. In cases (b) and (c), Doppler effect will be observed as the source and observer have a relative speed and so either approach or recede from each other. 24. [a, d] 3'=L

4. [c, d]

2

Since the first wave and the third wave moving in the same direction have the phase angles 4l and (41+ x). they superpose with opposite phase at every point of the vibrating medium and thus cancel out each other, in displacement, velocity and acceleration. They, in effect, destroy each other out. Hence we are left with only the second wave which progresses as a simple harmonic wave of amplitude A. The velocity of this wave is the same as if it were moving along. 6. [a, d]

~

Minimum loops"" 2

•• ~U-3~'~'--2-U-3--.

Fundamental v = -

3

v

2L

CJCJ20JJ .U3 ••

In both cases (a) and (d), the source and observer are relatively at rest, thus neither of

www.puucho.com

2U3

•

Next higher loops"" 5

Study Anurag Mishra with www.puucho.com 263

ePevel PASSAGE

Q2

Comprehension Based Problems

v = 400 mls , 100:pascal-----------. ,, ', ', ,, ,, ' ' ,, ' ' , Pressure

1

A source S of acoustic wave of the frequency v 0 = 1700 Hz and a receiver R are located at the same point. At the instant t = 0, the source starts from rest to move away from the receiver with a constant acceleration ro. The velocity of sound in air is v = 340 ml s. 1. If (l) = 10m / 52 , the apparent frequency that will be recorded by the stationary receiver at t = las will be : (a) 1700 Hz (b) 1.35kHz (e) 850 Hz (d) 1.27 kHz 2. If ()) = 10m/52, for O lOs, the apparent frequency recorded by the receiver at t = 15s. (n) 1700 Hz (b) 1310Hz (e) 850 Hz Cd) 1.23 kHz 3. If ro= 10 m / S2 the apparent frequency that will be recorded by the stadonary receiver just at the instant when the source is exactly 1 km away from the receiver will be : (b) 1310 Hz (a) 1700 Hz (d) 1.26 kHz ee) 850 Hz

A plane pressure pulse triangular in shape approaches a rigid wall along normal at a speed of 400 mls. At time t = 0, situation is shown in the Fig. The peak pressure is 100 Pa. By the wall pulse gets reflected and pressure near the wall gets doubled. Height of the wall is 2 m and width is also 2 m. A detector on the wall records a minimum excess pressure of 16 pascal.

80m

100m

1. When for the first time detector will record the pressure pulse? (a) 31.5xl0-2s (b) 21xl0-2s (c) 22xl0-2s (d) none of these 2. For how much time the detector will record the pulse? (a) 31.5xlO-2s (b) 34.5xl0-2s (c) 37.5xl0-2s (d) none of these 3. What is the maximum force applied by the pulse on the wall ? (al 200 N (b) 400 N (c) 800 N (d) none of these 4. Total impulse imparted by the pulse on the wall will be: (a) 150 Ns (c)

750 Ns

(b) 300 Ns

(d) none of these

3

PASSAGE

PASSAGE

•

A steel rod 2.5 m long is rigidly damped at its centre e and longitudinal waves are set up on both sides ofe b~ rubbing along the rod. Young's modulus for steel = 2 x 10 1 N / m 2. density of steel = 8000 kg / m 3 1. If two antinodes are observed on either side of e, the frequency of the mode in which the rod is vibrating will be : (a) 1000 Hz

(b) 3000 Hz

7000

(d) 1500 Hz

(c)

www.puucho.com

Hz

Study Anurag Mishra with www.puucho.com

rr.:2:6:;6==-----------=. \ (a)

(b)

Column-I

-

\

AGI-r_~.,

(p)

4r.,

I

Column-II

7.

Column I represents the standing waves in air columns and string. Column II represents frequency of the note. Match the column-I with column-II. (v =: velocity of me sound in the medium)

Y =-A rosCh-we)

(q) y=A

cos(ffit-kx)

(r)

y",A

sin(rot-kx)

(s)

y=A sin(h-cllt)

V

o

t/\

(el

\

--==:--------=====;;;MiiiE;CHiiiANiiiIC;;S.;;;;iil

~' (d)

t/\

'V

-,

\2

5. Match the following:

\

\

Column-I

(al In refraction

\

•

Column-II

(p) Speed of wave does no'

change (b) In reflection

(el

In refraction medium

(q) Wavelength is decreased

in a denser (r)

Frequency doesn't change

\

Cd) In refraction from a denser (5) Phase change of medium place

It

takes

6. Three travelling sinusoidal waves are on identical strings having same tension. The mathematical form of the waves arey, = A sin(3x -6t), Y2 = A sin(4x

-80,

and YJ = A sin(6x -12r)

Column-I

(a)

8. 'tWo waves are propagating along a taut string that coincides with the x.axis. The wave function for the cwo waves are Y) =Acos[K(x -l't)] and Y2 =Acos[K(x+ vt)+S] Column I represents the condition on 0 for constructive interference and destructive interference and the position x of the stationary points for constructive and destructive interference. Match the following:

Speed of each wave is

Column-II

Column.1

(p) Stationary point

(b)

o

(q)

Constructive interference

(e)

x=

(,)

Destructive interference

= (2n+l)n: (2n+l)n:/2 K

(d) x = nrr/ K

(r)

(d)

(,) Not a srationary point

9. Symbols have their usual meanings. Column-II

Column-I String slope versusr at (p)

x =0 (e) Yzisbestrepresentedby

Column~1l

0= 2mt

(a)

(q) 2 mls

\

(a)

(p)

(b) y1isbestrepresentedby

\

for

y=Asin(kx+Wl)

(b)

Instantaneous veloc- (q) icy of particle versus t at x=O for y=Asin(kx-(J)()

(e)

Displacementofparti(r) cle \'ersll.~t at x = 0 for s = So sin«(J)t - kx)

Y3 is best represented (s) by

www.puucho.com

I

Study Anurag Mishra with www.puucho.com

267

WAVE MOTION (d)

Displacement of part ide versus t at X = 0 for p:. Po cos (kx -(IJ{) where p represents pressure variation

(s)

2.

10. Column-] shows different sets of standing waves in a string of length L whose ends are fixed or free according to respective figure and Column-II shows possible equations for lhem where symbols have usual meaning. Column-II

Column-I

"-

(,)

~x=L

X=O (bl

"- "(p)

~

3.

(ql

~

,0

Y =

A .

y= A.

x=l

4.

rrx

Sin TCOS(J)f

"'" 510-'","" L

5. (,)

(c)

t

rrx y=AcOS-COSWl

:zL

~,oL

x=O (,)

(d)

~ = T + Vd(g + el)

M+m M

---"-=-~-

T = Fh

m=(~~-I)M ~ (S.4 -1)(600) 7.0

of equilibrium,

-

Vd(g + el)

=(g+a)v[~P2+~Pl-d]

= 120kg

= 15x 1O-3[~x 1500+~ x 1000~6N

www.puucho.com

sooJ

the

Study Anurag Mishra with www.puucho.com

Example

(P2

What volume of helium is needed to lift a' load of 180 kg (including weight of empty balloon)? Density of air. Pair = 1.29 kg/m3 • density of helium Phelillm = O.18kg/m3 •

Solution: From Newton's second law, for the equilibrium of system,

F,

Depth submerged in mercury is 32 mm and in water 28 mm.

180

=

... (2)

which on solving for x yields x = o.032m.

Solving for V,we find

180

~

Piron Virong = Pwg(l- x)A + PHggxA (7.7 X 103)(0.06)3 = (1.0 x 103)(0.06- x)(0.06)2 +(13.6 x 103)x(0.06)2

= (PHeYg + 180)g

Pair - PHe

PHggxA

weight of displaced mercury

=> Archimedes' principle can be used for two or more than two liquids for any shape of object. From eqns. (1) and (2),

The buoyant force is equal to weight of displaced air.

V =

.

weight of di.placed water

Fb = mHeg + mloadg

Pair Vg

Pi) = P",g(l- x)A +

-

(1.29 - 0.18)

= 162.16mJ

This is the volume required to keep the system just in equilibrium, a slighdy greater volume is needed. As the density of air decreases with altitude, volume required is still more.

A hollow sphere of outer radius 9.0 cm and inner radius 8.0 cm floats almost half submerged in a liquid of density 800

g/cm3• (a) What is the mass of the sphere? (hi What is the density of. the material of the sphere?

m_, _ •••• ..,Flg. JE.42

, Solution:

(a) In equilibrium Fb = w

mg = p'-':,g A cube of iron 6 em along each edge is in equilibrium at the interface of two liquids, water and mercury. Find the length

at

the cube in each liquid. The densities of iron and mercury are 7.7x103 kg/m3 and 13.6x103 kg/m3 respectively.

Thus m = P'-':, The submerged volume is half the volume enclosed by the outer surface

14,

v:.

=-x-1tro

2

m=

3

(~1t

)c0.090)3(800)

= 1.22 kg

Air, if any, inside the cavity has been neglected. (b) Density of material =

!2: V

41t33

V=-(ro

3

-rj)

= 4. [(0.09)' 3

Fig.3EA3

= 9.09

Solution: Let x be the submerged depth in mercury, the (0.6- x) length protrudes in water. If P and P2 are the pressures on the upper and lower faces of the cube, the net upward force = (P2 - PI )A, where A is the cross-section area of cube. In equilibrium it must be equal to weight of the cube.

.. e enslty Thd

IS

j

W = (P2 PI

-

PI)A

...(1)

= p ••.gh +P

8Iffi•

P2 = Pwgh + Patm. + P••.g(l- x) + PHggx

X

_ (0.08)']

10--4 m 3

1.22__ = --9.09x 10--4 1.3x 103 kg/m3

= Example

An iron casting has a number of cavities in it. It weighs 6000 N in air and 4000 N in water. Determine the lotal volume of all the cavities in the casting. The density of iron (without cavities) is 7.87 g cm 3 .

www.puucho.com

I

Study Anurag Mishra with www.puucho.com

309

SOUDS AND FLUIDS Solution: Volume of the cavities Vcav can be determined by taking difference between the volume Vcast of the casting as a whole and the volume of the iron in the casting,

W = .!.L(Lsine)(Lcose)pg

2 The sum of vertical forces yields Fb = Fe cose+ W Pb{2

or

case =

PeL2

W is weight of ca~ting

Pb = Pc +.!:.pgLsinO

or

2 As L -) 0, we obtain Pb = Pc> which equals Pa.

Effective weight of casting in water, Wcff = W - PwgVcasl Vcast Thus

x case +.!:.pgL2 sine case 2

W - Weft

:= --~~

Pwg _ W - W.,ff.

Vca\' -

=

---

W

Pwg gPiron 6000- 4000 3

(0.998 X 10 )(9.8)

6000 (9.8)(7.87

Fig. 3E.47 shows a triangular plate submerged in a liquid of density p. What is the hydrostatic force on one face of the plate?

X 103)

= a.127m3

:n "

A

E

Example

46~

,

Y

h(y)=y+ho

~yl_'qdY

Fig. 3£.46 shows a wedge-shaped volume of fluid where the average pressure on each a/its/aces is Pa,P" and Pc, Lengrlt of the sides of the top face is L and it is inclined at an angle e with the bottom. Show that Po.= PI> = Pc in limit as L -) Q

b----~.I

C

Fig.3E.47

Solution: We consider a differential area clement of thickness dy and length l(y) at a distance y from the top of the plate. Pressure en the position differential area element, dP = pg(y + ho)'

Force on this element, dF = pg(y + 110)I(y)dy. Total force is obtained if we integrate this expression when y varies from 0 to Q. F:

Fig.3E.46

Solution: This problem shows that the pressure within a fluid is independent of orientation. Area of left face, Aa = L(L sine) Area of top face A, = L2

pg(y

... (1)

+ h,)ICy)dy

We can integrate the expression for F only when we express l(y) as a function of y. We apply properties of similar triangle on triangles ABC and ADE, we have ICy)

y

/(y):

~y

--:b "

e

Area of bottom face, Ab = case The corresponding forces on each face are Fa = PaLl sine

Thu'

...(2)

a On substituting expression ICy) in eqn. (1), we obtain

Fb = PbL2 case

F:

2

Fe = PeL

Since the fluid element is static, the sum of horizontal forces must sum to zero, we obtain Fa = Fe sine PaL2 sine = PeL2 sinij Thus Pa = Pc The weight of the fluid volume,

J dF: J:

J dF: J: pg(y

+ h,,)(~y

ldY

: pgb[ a; + h;" ] ~

Note that we had to use calculus because the different elements of triangular plate are at different depths; so pressure varies as y, the depth of elements, varies.

www.puucho.com

Study Anurag Mishra with www.puucho.com 310

MECHANICS.II

Example During a flood, the water in a basement rises to a height ho above a windows as shown in Fig. 3E.48. What is the force on the window due to water of density p? ~

h,

t

, J.d,

•j

-

-b

Fig. 3.47 Some properties equilibrium are

T

Fig. JE.48

Solution: The pressure due to water at depth y on a differential area element, as shown in Fig. 3E.48, ispgy. The force on this element is ',' dF (pgy)b dy

=

f

dF =

fZ.'

f".,

=pgb"

The pressure is the same at all points on a horizontal line through a connected liquid in hydrostatic equilibrium. y

the tube supports tile "extra" liquid

1

ho

Normal force of the wall of the tube

, Force due to the pressure of the liquid on the right

Weight of the liquid

Fig. 3.48

I , = -pgb(a + 2aho)

2

CONCEPTUAL

Ii Th'W'''~

_ I P [y'] ".,

-28

Focus on this piece of tile liquid

pgby dy

ydy

of a liquId In hydrostatic what you might expect

not

If PI = P2. Figure shows that the weight of this extra liquid is supported by the wall of the tube. Only the liquid that's directly above point 1 needs to be supported by the pressure at point 1.

The total knee on the window will be the sum of all the forces on strips covering the entire window. Hence F=

(b)

I') Is this possible

POINT

Hydrostatic Pressure The hydrostatic pressure in a liquid depends only on the depth and the pressure at the surface. Fig. 3.47 (a) shows two connected. If d1 were larger than dz• then, according to the hydrostatic pressure equation, the pressure at the bottom of the narrow tube would be higher than the pressure at the bottom of the wide tube. This pressure difference would cause the liquid to flow from right to left until the height were equal. A connected liquid in hydrostatic equilibrium rises to the same height in all open regions of the container. Fig. 3.47 (b) shows two connected tubes of different shape. The conical tube holds more liquid above the dotted line. But it is not. Both poinrs are at the same depth, thus Pj = P2• If PI were larger than p2• the pressure at the bottom of the left tube would be larger than the pressure at the bottom of the right tube. This would cause the liquid to flow until the pressure were equal.

Force acting on an inclined wall: Since pressure always get normal to the surface, therefore. the pressure at a point depends on the height of liquid column above it. [Fig. 3.49. (a)]

',,',','.'.'.'.' ..,.,',','.' :.,."~.,. /,!------- -------

h

:::::::::::::::::::::::0.::,:,::,:,::,:,::;::::::--

:::-:-:-:-:-:-:-:-:-:-:-:-: -:-:-:-:-- a

-

Fig. 3.49 (a)

The total force acting on the inclined wall is I -F = -(pgh) b 2

(h)' smS -,-

I h' =-pg~

2 sinS The force on the inclined wall may be obtained in tw"o parts: horizontal and vertical. I

i.e., Fx = -pgbh

,

2

The vertical force Fy acts due to the weight of the liquid supported hy the wall,

www.puucho.com

;_F.~ Fig. 3.49 (b)

Study Anurag Mishra with www.puucho.com

311

lOUD! AND FLUID! i. e.,

Fy

:=

1 zpgb(h)(h

cotO)

..

.. .. ..

F

=~pgbh2cot2e 2

..

I"

The magnitude of the resultant force is F:= ~F~2+ or

F

b

1

F:

= ~pgbh2COSece

Fo Fig. 3.51

Ii2

F = PIA - a) = pg(H - h)(A - a)

Pg sine =2

Since the liquid is in equilibrium,

The force exerted by liquid at the base of a container is equal to the product of pressure exerted by liquid at the base and area of the base. Since pressure at the base depends on the height of liquid, not on the weight of liquid, therefore force exerted at the base is independent of the weight of liquid inside the container. Concept: 1. The Fig. 3.50 shows two containers (a) and (b) with same base area A and eachfilled with a liquid of density p to the same height H. The container (b) is made narrower at the top so that it contains less quantity of liquid

as compared to container (a).

therefore,

Fb =Wb+F 0'

Fb =[pghA+pg(H-h)al+pg(H-h)(A-a)

or Fb = [pgHa + pgh(A - a)] + pgH(A - a) - pgh(A - a) or Fb = pgHA

HYDROSTATIC FORCE ON A SUBMERGED

CURVED SURFACE

We rake clue from example 46. We see that horizontal component of force on the inclined face, Fe sin Q, is equal to force on the vertical face, Fa' Fa = Fe sino Pae sinQ = Pee sinQ Fc sinQ = Pee slnQ = Pc (Projected area)

('I Pr~L~:t:d3r

Fig. 3.50

-+ught

!l'Yf'.... ~-

Since the height H of liquid is same in each container, therefore pressure at the base is also equal. i.e.,

=

Container (a) : Pa pgH :. Container (b) : Pb = pgH :. The weight of liquid in container Wb = pghA + pg(H

Ffl = PaA = pgHA Fb

on screen

= PbA = pgHA

(b) is

- h)a

1.1

Where (H - h) and a are the height and cross-sectional area of the narrow portion of the container. Note that the weight Wb of the liquid in container less than the force Fb exerted at the base.

(b) is

2. A liquid can exenforce greater than its weight in some situations, as shown in Fig. 3.50 (b). From the free body diagram of the liquid as drawn in figure it can be understood. Here, Fe is the force exerted by the ceiling Since liquid in the container pushes the therefore according to Newton's third law. pushes the liquid downward. The magnitude given by

of the container. ceiling upward, the ceiling also of the force F is

Fig. 3.52

Projected area can be conSidered inclined surface or a curved surface on Thus force in x-direction, Fx = Average pressure x projected x-direction. Similarly force iny~direction, F = Average pressure x projected y-direction. y

www.puucho.com

as the image of an a screen. area normal to the

area normal to the

Study Anurag Mishra with www.puucho.com

312

MECHANICS.II

Example Fig. 3E.49 shows a glass tube in the Jonn oj an equilateral triangle oj unifonn cross-section. It lies in the vertical plane, with base horizontal. The tube isfilled with equal volumes oj three immiscible liquids whose densities are in arithmetic progression. Determine the length x as shown in Fig. 3£.49.

Curved surface

A

'.

/,

Fig. 3.53

•.

,~~n'\" -,-

Force in vertical direction (z-direction), Consider a hemispherical volume of a fluid. Net force on plane surface

I

= (pgR) x rr.R2

~~ Pavg

:.~~~~m::mw~~~::~m \

B

area

In a static fluid this element is in equilibrium, i.e., forces in x-direction on plane face as well as curved surface must be equal. As we can ~ee from Fig. 3.54, plane face is projection of curved hemispherical' surface. Similarly consider a semi-cylinder, Fig.3.54 the force on plane face ABeD is same as on curved surface. •• As shown in Fig. 3.55, a hemisphere kept on a horizontal surface experiences pressure force normal to the surface. We consider a differential element on the surface of a hemisphere and set up a coordinate system with centre as origin. The resultant force on the element,

,

, -Palm r

\

d+2y

E

C

Flg.3E.49

Solution: Let the densities of liquids in DAF, FeE and EBD be d, d + y and d + 2y respectively. Pressure at a point inside a fluid depends on the venical height of the liquid above that point. We will calculate pressure at E from left ann and right arm, these values must be the same for equilibrium . PE = dg{l- x) sin 60 +{d + 2y)gxsin 60° [left arm] PE = dgxsin 60 +(d + y)g{1- xlsin 60° [right ann] Equating the two values of pressures, J3 xJ3 dg{l-x)-+-{d+2y)g 2 2 xJ3 J3 = -dg +(/ -x)-Cd+ y)g 2 2 (/ - x)-J3d + xJ3(d + 2y) = xJ3d + (/ - x)J3(d + y) (/ - x).J3d + x-J3d + 2yxJ3 0

0

= x.J3d + (/ - x)-J3d + y(/ - x)J3 y

2yx = y(l- x);

Fig. 3.55

dF = dFxl + dFyj + dFlt
a or g < a. III. Acceleration is x and y Direction A liquid ill all opell vessel is given a uniform lincar acceleration a as shown in the Fig. 3.61.

~

hi -h2 X (Ill -IJ

Jg

X=~~~~2

a Thus, pressure

r

... (2)

and

in given by

difference

y

P.4-Fa =pa(h1 -h2)g

a FA - Pa = pg(hj

-

h2)

.. (4)

t,

...... 1""

i.e., pressure along a horizontal line is not the same in case of horizontally accelerated liquid.

Concept: 1. Initially the container is completely filled with a liquid of densiry p, now the container is given a unifonn horizontal acceleration 'a'.

"' .. j

.

!

•...........................

a

Fig. 3.61

www.puucho.com

••••••.••••

x

Study Anurag Mishra with www.puucho.com

314

MECHANICS.II

Let us assign a Cartesian co-ordinate system with y-vertical

and x such that the acceleration

vector -; is in the

~ x-y plane and Z4axi.~ is normal to a and there is no acceleration

component

in that direction.

2

Therefore the pressure gradients along x, y and 'z. direction arc, iiP dx =- -pax iip

-

iiy

:.:.:.:.::. h .•..

:::::.~.:::::::::::::::::.:...

Here pressure is function

or or

Where h is the depth oj the point A below the free surface oj liquid along effective gravity and Po is the atmospheric pressure acting on free surface of the liquid.

,

iiz

Thus

PA =Po+hp"a +g2

= -p(Oy +g)

iiP =0

Therefm"c

Pressure at every point in a liquid layer parallel to the free surface, remains same for example iJ we find pressure at a point A in the accelerated container as shown in Fig. 3.62 is given as

t~~~~~~~~~~~~~~@ ..................••••••••.•.•. -•..-•............:.

x, y and z

of coordinates

.:.:.:.:.:.:.:.:.:-:-:-:.:.:.:.:.:.:.:.:.:.:.:.:.

P = I(x,y,z) iiP iip iiP dP = - dx + - dy + - dz ax r.ly dZ

J~dP = ~pJ:axdx

('J

- pf:Cay + g)dy + a

P-Po =-pa..:x-p(Oy

(bJ

Fig. 3.63

Alternatively as shown in Fig. 3.63 (b).IfII and 12are the vertical and horizonral distances of point A from the free surface of liquid then pressure at point A can aLso be given as

+g)y

P=Po-.poxx-p(Oy+g)y

PA = Po +llPg or Fa =Po +12pa

Where Po is the pressure at origin. a P. -p or y=--'-x+ 0 0y+g p(g+Oy) The above equation

..•

h"a2 II = h seeS = ~---

(1)

is equation of straight linc.

So the lines of constutlt pressure isobars have a slope.

g

Thus,

PA=Po+hpJa2+g2

Also

PA = Po + 12pa

h,la +g

",

2

" tanu=---Qy +g P = 0 in equation

(1).

IV. When liquid is in an accelerated frame as shown in Fig. 3.62. Due to acceleration of conrainer, the free surface oj liquid remaim normal to the direction of effective gravity. The inclination angle of free surface of liquid from horizontal is

e = tan-

2

12 = h coseeS = -~---

The equation of the free surface Call be obtained by putting

+g2

---"

PA =Po+hpJa2+g2

We have

A trolley containing a liquid of density P slides down a smooth inclined plane of angle 0: with the horizontal. Determine the angle of inclination 9 of the free surface with the horizontal.

,

(i)

1

B:(l

~'. !'\..

a:gsino.

?~::::::'.'.

"

a

::::::::::~::::::.:.: .•.•..•••.••••...-.. - ....

('I

~~r~fL~i:~f.tt~~:~~:.: .

.....•••....•-•.•.•...•............•.•.•.•....... .:.:.:.:.:.:.:.:.:-:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.

(bJ

Fig. 3.64

The acceleratioll oj the trolley along the inclined plane is a=gsino:

ax

Fig. 3.62

www.puucho.com

= a coso: = g sino:coso:

Study Anurag Mishra with www.puucho.com

SOLIDS AND FLUIDS

,

a.v =asina=gsin2a The inclination

of the free

surface is

tanS = ~

= g sinacosa

= tana

g(l-sin2a)

g-Oy

or

6=a

Note that the free surface becomes parallel ro the inclined plane. It is because the net effective gravity inside the trolley is perpendicular to the inclined plane. V. The Fig. 3.65 shows an L-shaped rube filled with a liquid to a height H. What .~hould be tile h01izontal acceleration

315

a of the tube

$0

thal the pressure at the point A

becomes atmosplleric ?

----------... _-----------------

:-:-:-:-:-:-:-:-:: I:--.-.-.-.-.-.-.-.-.-.._ . -----------

-----------

Fig. 3.66

gradients

The pressure directions.

dP

,

-=poo dr

dP

B

-

ay

Since therefore

along radial and vertical

r

~--jlg

P ~

I(r,y)

ap

ap

uy

dr

dP=-dy+-dr

H

jpPodP = -pgjY0 dy + poo2j'0 rdr :.:.:.:.:_:_:.:_:_:_:_:_:_ :-:-:-:-:-:-:-:-:-:-:-:-:-:-: _ A

I P-Po =-pgy+-pw

2

L

Fig. 3.65

Beginning from B we each point Gnd write pressure equation to get Pa +pgH -pal = PA Since therefore,

PB

-

PA =

Palm

a =g

(H :

"r

p = Po -pgy + .!pw2r2 2 Where Po is the pressure at origin. Now let us take the lowest "'--. point on the meniscus as origin (0.0) which is an surface as shown in figure.

,

......... I .

• r

L)

Pressure in a Rotating Frame when a liquid in a container when rotated about a vertical axis has shear stresses in the liquid and the only acceleration that occurs is directed radially inward towards the axis of rotation. . ~ . Let us consider the unit vector i in the r direction and j in the \'ertically upward diretiion with Y being the axis of rotation and z-axis (k) in tangential direction.

Then hence

Po = 0

Fig. 3.67

1 "r P = -pgy +-poo 2

To obtain the equation of free surface, Put

P=0

then Thus the surfaces of equal pressure are paraboloids of revolution.

www.puucho.com

Study Anurag Mishra with www.puucho.com 316

_ MECHANICS.II

Concept: Lengtl! of a horizontal ann of a U-tube is L and ends of both the vertical arms arc open to atmospheric pressure Po, A liquid of density p is poured in the tube sucH that liquidju.st fills the horizontal part of the tube as shown in Fig. 3.68 (a). Now one end of the opened ends is sealed and the tube is then rotated about a vertical axis passing througn the other vertical ann with angular speed {o. If length of cacn. vertical arm is a and in the sealed end liquid rises to a height y, find pressure in the sealed tube during rotation.

force must act nonnal to the surface, otherwise fluid will flow. Since fluid is static in our reference frame, ... (1) rFy = Rcos9-dmg =0 ... (2) Ux =Rsin9-dma=O

--

From eqns. (1) and (2), tane =~. g Note that if the container is given acceleration that has components ax and ay along x and y-axis respectively a

tane = --'-

The pressu.re difference across an element of width dx;

ay

which is given as

where ax and aJ' represent

dP = dxpro2x Now integrating from A to B, we get

accelerations

in x-and

y-directions respectively.

= f>ro xdx 2

FB -FA

0"

.

-'/

•

c

, d, ..-

Y

A

-

•

p+dP

P

: Y

:.

(.J

Solution: (a) As shown in the previous example, a h tan9=-=-

B

•

l

Fig.3E.51 shows a U-tube. When at rest, the liquid in both the anns is at the same level. But a difference of height occurs when the system is given an acceleration 'a' toward the right . (a) Detennine the difference in heights . (b) What can be the maximum acceleration of the system without spilliczgthe liquid?

g

(') Fig. 3.68

----

Pc = PB

-

ypg

,

and at point A. pressure is atmospheric, thus we have 2

(L2_y2)+PA_ypg _.

••••

aL

(b) Maximum difference in height that can occur is H. H a tana=-=~ L g a

Example U-tube A liquid kept in a container is subjected to a coTlstant acceleration 'a'. What angle does the surface of the liquid make with the horizontal?

""'--------.

Solution: We will solve this problem from the reference frame of the container. The forces acting on a differential element dm are as shown in Fig. 3E.SO. Here R represents the force exerted by neighbouring particles on the element under consideration. This

••

dm.

.~. ,

(Pseud force)

max.

Flg.3E.51

= Hg L

Manometers

These manometers work on Pascal's law. The pressure at a point in a vessel is obtained by writing an equation called manometric equation.

YL ,

R

P,

P,

h,

dm,

L

h=g

so

Thill pressure at point C can be given as

Pc=pw

+g

-

~

...~

•

Fig. 3.69

Flg.3E.50

To write the manometric equation, we start from the point where the pressure is to be obtained and move

www.puucho.com

Study Anurag Mishra with www.puucho.com

SOLIDS AND FLUIDS

317

towards the free end of the tube. Pressure increases in moving downwards, decreases in moving upwards and does not change in moving horizontal and finally at free end, we equate it to the atmospheric pressure to obtain absolute pressure at that point i.e.,

hmax

h

= Pz

,

"

Flg.3E.52

If the tube is accelerated

A

tan8 =g

P,

P,

i1

hm

I

In the above case, Pl + hlPg - h2pg -I sinflpg + "3pg ~ h4pg

1

,

t

FromFig.3E.S2,

hmax. =h+(~)tan8 hmin. = 1J -

(~}ane

Average pressure on the from face = pg( h~n. ) Fig. 3.70 (al

PI + h1pg -llpa

Force on the from face = pg

- hzpg -12 sinflpg

-12 cosflpa + h3pg -13pa-

h4pg = Pz .

(h;in. ) (b x hmin.)

Average pressure on the rear face = pg(

Note: The pressure at the same horizontal level in this type 'of case mayor may not be same. II should be checked"by applying manometric equation.

h;ax. )

Force on the rear face = pg( hm2ax). (b x hmax.) .

Average pressure on the side face= pg

(h

+h.)

max. mm. 4

J

. h Force on the side face = pg h maX.4+ h mLn. (

12

= p(ax

Fig. 3.70 (b)

PA + hlPlg + hzpzg -11pzQ -12,pJ,a

b xc)g

(ii) The free surface is inclined at an angle a, tan8=--

- h3pJg -h4P4g

Example

a

Force on the base is equal to weight of liquid contained in the vessel, i.e.,

--.---.~ 11

X

az +g

= PB

52;...?"

A vessel of recrangular cross-section has the dimensions a x b x c. It contains water upto a height of h. The vessel can be subject to acceleration of 'a' m/s2 in the following cases: (i) horizontally parallel to longer side (ii) vertically. De/ermine the force on the front face, rear face, side face in each case.

as ax =O,tan8=O The free surface will remain horizontaL Pressure on the base = peg + A)h Force on the base = peg + A)h x (a x b) Force on the longer faces

=

h

peg + A)-(a x h) 2

h Force on the front and rear faces = pCg+ A) - (lJ X h)

Solution: (i) The free surface is inclined to horizontal at an angle 8, given by

www.puucho.com

2

Study Anurag Mishra with www.puucho.com

318

MECHANICS-II

Liquid in a Vessel Velocity

Rotating with Constant Angular

When the vessel is rotated there is no relative motion between fluid elements. The liquid surface orients itself in permanent position relative to vessel. We will analyze a differential element in the reference frame of vessel. The forces acting on it are shown in the Fig. 3.71. R is the force exerted on the element due to neighbouring elements of fluid. Thus ... (1) R cosS = dmg R sine = dmn1l2

... (2)

Dividing eqo. (2) by ego. (1), tanS

",'r =--

... (3)

g

BN is curve AP Constant defining parabola.

subnonnal of the and is constam. subnormal is the property of a

--

r=R,

At

R2oo2 Ymax. At

=~+c r=O,

Fig. 3.72

c=Ymin. 2 2

R

Therefore

oo

Y=~+Ymin

Volume of a paraboloid of revolution is half thar of the circumscribing cYlinder. 1 Y =Z(ymax.

m

R

+Ymin.l

~Rcose

..- _.-'.---•....._~ R

.

",

:

dm Rm2

sin e

:

(Pseudo force)

,, ,

, dm,

A:,

Example ~ V-tube shown in Fig. 3£.53 rotates with angular velocity 00 about the vertical aXis AB. What is the difference in fluid level 'h' in terms of ro, the radii r1 and the fluid density p.

cbw ,

,, ,

1-",--:--, ,

,:

Fig. 3.71

At the position of the element the slope of the tangent to CUIVe is given by dy I dr. dy ro2r tan8=- =-dr

A

','

2----...j

r

.--.

8

h'

:, ----i'" .'

-..:---

.'

g

Fig.3E.53

",'r dy =-dr g

which on integration yields y = ~

Solution: Let Y min. be the lowest poim on the dotted parabola shown.

,, 2g

+c

As at r = o.y = 0 hence c = O. 0)2r2 y=--

2g

which represents a parabola. Thus liquid surface is a paraboloid of revolution. ~ The force R is exerted by air and rest of liquid is normal to surface. Force R cannO( be at an angle, otherwise the air and liquid would exert a shear, or tangential force, along the surface, which will cause the flow. -- From Fig. 3.71 PN g BN = -== constant tanS 002

Alternatively, we may between points A and B.

apply

Bernoulli's

1 2 1 2 Palm. + "2P(rioo) =Patm• + "2P(r2oo) +pgh or

www.puucho.com

h=(ri~r/)ro2

equation

Study Anurag Mishra with www.puucho.com

SOLIDS AND FLUIDS

319

Example

where A is cross-sectional area. Inirial volume was OAA m 3

54__

A cylindrical vessel of diameter 0.3 m, height 0.6 m, is filled tlvo-thirds with liquid of specific gravity 0.8. The vessel is rotated about its axis. Determine the speed of rotation:' (a) when the liquid just starts spilling,

Percentage of liquid left;=; o.3A x lOa;=; 75%

O.4A Hydrostatic Force on a Submerged Plane Surface

(b) when the base is just visible. (e) what is the percentage

left

of liquid

-..

in the vessel?

............... _.' : ,...•.... -- --''./' : '.

.

Solution: Initial height of liquid in the vessel 2 =D.6x-=OAm 3 ,,

1

.--_. :~.":-.•':_--,

"

y~

J

: .. .........•... '

-------1--..-, ,, , w2R2/4 -'.. .....•.. •. ....•.~~

!

Free SUrfa;;

w2R2/4f y

,

!

L_-'-

Hyml~,

a .....

......

Fig: 3E.54

Height of container = 0.6 m Y

max. -

w2R2 Y min. =--

...(1)

2

...(2) Fig. 3.73 2 2

Ymin.

(a)

wR

... (3)

=y---

4g

Ym~. = 0.6

From egn. (2)

Consider an elementary strip at a depth z Force on strip;=; dE ;=;FdA

, , (0.3)' w

x -

Consider a plane surface Df area A with its geometrical centreC be placed Dn an angle with respect to free surface of a liquid.

2

;=;

0.6::: 0.4 + ----4x9.81

;=;pg y sinudA

w = 18.68 rad/sec (b)

R

Net hydrostatic force

a

Y max. ::: 0.6, Y min. =

F=fdF

f

2

Y max.

=

(I}2

-2, Pliq. Here potential energy continues to fall below zero, but slowly then it would be in air. The greater the density of object (Pob.) greater is the rate of decrease of potential energy. The straight line 1 corresponds to this situation.

(ii): Pub. = Pliq In this case the potential energy remains constant, str~ight line 2 corresponds to this case Case

(iii): Pub. < Pliq. In this case the potential energy continues to increase when the object sinks in the liquid. Straight lines 3. 4, 5 corresponds to this case. The rate of increase is higher less the value of Pob. The potential energy of the object cannot exceed the initial potential energy sho\'lJl by the dotted line. This level will be attained only if all the drag forces are ignored. In this case the object will sink to a cerrain depth in the liquid, stop, and then return to the surface with same speed as it had before entering the liquid. When out of liquid, the object will rise to the height determined by initial potential energy. In real conditions the drag force of liquid will slow down the object. Case

(iii) force due to pressure of surrounding water.

Same forces act on the cork piece, The force of pressure of the surrounding water is same but the pseudo force and gravitational forces are lesser. Hence there is a net force, the difference between the force of pressure and the force of gravity and pseudo force. This force makes the cork piece to move towards the surface and at the same time towards the axis of the vessel. Similarly an object with a density greater than the density of water, in a rotating vessel filled with water, will sink and move towards the wall of the vessel.

Example

If the density of object is one half the density of liquid, I

Fig. 3£.61 shows potential energy U versus the position of the object (altitude) of an object of density Pob, dropped into a liquid of density PI from a certain height. The potential energY, of the object at the level of the liquid is taken zero and the positive direction of the Vertical axis is the one pointing upward from the liquid's surface. Determine which of the five straight lines, 1.5, corre.~ponds to an object with the highes~ density and which to an object with the lowest density. Is there a straight line among these five for which PoD. 1/2Pliq. ? The arrows on the straight lines point in the direction of motion of the object?

Pub. = ZPliq.

~

then

I PI = V(2pob.

- Pob)g = VPob.g

The difference between buoyant force and weight is equal in magnitude to weight of the object, and is directed opposite to gravity. The slope of the straight line must be the same as that of the free falling object, as represented by the straight line 4.

=

Example unifonn rod of length 2l floats partly immersed in water. The rod is supported by a string fastened to one of its ends as shown in Fig. 3£.62. If the specific gravity of the rod is 75, what is the fraction of the length of the rod that extends oilt ot the water?

j'\

...------....~~(~~!~~Q~i energy)

2 90"

, (Position)

1 60.

','

Flg.3E.61

www.puucho.com

Study Anurag Mishra with www.puucho.com 323

SOLIDS AND FLUIDS

Solution: When the tube is at rest using Pascal's law we have PA = PH Let Po be atmospheric pressure. Po + pg (~a ) = Po + (2p)gh When the tube is whirled about vertical axis such that the interface shifts to C PSI~'Vo > VI' hence the volume of the ball

submerged in mercury is decreased when water is poured.

.:.:.:.:_:_:_:_:.-.".".-."

_.-.-.-.-.-

Flg.3E.64

..

w.:.

,

Solution: As the density of iron is greater than water, some part of the chain will lie at the bottom while the other will hang vertically. If we assume that the sphere floats at the surface, the chain will hang vertically which is not possible in view of the greater density of iron. In equilibrium the force of gravity will balance the buoyant force. Let h be the depth at which it will float, then we have

(M +

m

h-D!2) 'g

::: Pw

(v

h

m

h-D!2) 'g

Piron

I

+

On solving for h, we get h=!!.+ 2

PwV-M m(l-p

I

•••/piroll)

A solid spherical ball of volume Vfloats on the interface of two immiscible liquids. The density a/upper liquid is PI and of the lower one P2 while the density of ball is P(PI < P < P2). (a) Whatfraction of the volume of the ball will be in upper liquid and what fraction in the lower one? (b) What happens if aensity of the ball is equal to that of the upper liquid and if the density of the ball is equal to that of the lower liquid? (c) Determine the density of the ball if the ballfloats with exactly half its volume in the rwo.Jiquids._. __ Solution: (a) Let VI and V2 denote the fraction of volume of the ball in the upper and lower liquids respectively. In equilibrium the buoyant force exerted by two liquids must balance the weight of the ball, i.e., (VI + Vz)pg = p] \'Ig + Pz VzK or pV=P1V1 +Pz(V-V1) where V is the total volume of the ball. Thus

VI=VPZ-P Pz -PI

Similarly,

Vz

where

Example A steel ball floats in a vessel with mercury. How will the volume of the part of the ball submerged in mercury change if. a layer of water completely covering the ball is poured above the mercu'Y. ;> Solution: Denoting the volume of the steel ball by V, volume of its part immersed in mercury by Vo before water is poured and V1 after water covers the ball completely. Before water is poured, Pst. V = Pm •.r. Vo

where Pst. and Pmer. denote densities of steel and mercury respectively. When water is poured,

(b ) If P=Pl.weget and

-

P - PI Pz - PI

(p,-p,l

VI=V---=V (pz -PI)

VZ=VPI-PI_O Pz -Pl

Similarly, if P = Pz. we get and

VI = 0 V2=V

i.e., the ball is in the lower liquid. (c) As VI = V2• we have VP2-P=VP-PI Pz -Pl P2 -PI 0'

P2-P=P-Pl P= Pz +PI 2

VI = PSt.- P... V

Pmer.

V

i.e., the ball is in the upper liquid.

Weight of ball = Buoyant force of water + Buoyant force of mercury Pst.V=Pmer.VI +P •••(V-V]) On solving for VI' we get

=

P ••.

www.puucho.com

Study Anurag Mishra with www.puucho.com

325

SOliDS AND flUIDS

Example

67

and the second case if

_

A

A

cylindrical

communicating

vessel with different cross-sectiom has mercury poured into it. An iron block with Q volume Vo i5 dropped into the broad vessel and as a result the level a/the mercury in it rises. The water is pourcd into the ves.~eluntil the mercury reaches the previous level. Find the height of the water column h

vessel is AJ

if the cross-section

of the

narrow

•

Solution: When the block is dropped into the broad vessel the mercury level in both the vessels will rise by the amount x to a level shown in Fig. 3£.67 by AB. Height of the water column can he determined by equalising pressure at level CD. A

Example

,

>PI(PO-PZ)V23 PZ(PI -pz)

68

0

_

How far does a wooden (spherical) ball of specific gravity 0.4 and radius 2 feee sink in water?

Solution: By '~rchimedes' principle" we know that the ball will sink until it displaces a weight of water equal to the entire weight of the ball: (weight of ball) ""(weight of displaced water) We know that (weight of ball) "" (volume) (specific gravity)(density of water) where the density of water w is measured in pounds per cubic foot. We must first calculate the volume of the part of the sphere immersed in water when the sphere sinks to a depth of h, and then solve for h from the given information [see Fig. 3£.68 (a)]. y

~,.y ~_R)2=R2

Fig.3E.67

(y + x)p\g

:: hP2g

... (1)

(x+ y)A\

"" V2

..•

(2)

where V2 is the volume of the mercury displaced by the block after the water is poured in. If the water covers the block entirely, from Archimedes principle, we have VoPog = VZPlg + (Vo - V2)P2g

... (3)

On solving eqns. 0), (2) and (3), we get II"" Pl(PO -P2)VO P2(P\ -P2)A1

(b)

(.)

where Pmer.and p", denote the densities of mercury and water. As the volume of the mercury is constant, we have

Finding this \'Dlume is equivalent to finding the volume generated by rotating the area bet\veen the curves x2 + (y ~ R)2 and y ""h, about the .v-axis [see Fig. 3E.68(b)].

We note that the general formula for a circle is + (y _ k)2 == r2, where (h, k) are the coordinates of its centre and r is the length of its radius. In this case the coordinates of the centre are (0, R) and r "" R. Recall that [see Fig. 3£.68 (c)]: dV"" nx2 dy "" It [R2 _(y _R)2}2dy

(x _ h)1

If the block is partially submerged, we have

= It(2Ry - y2)dy

VoPog = V2Plg + p2Ahg

where A "" V023 is the cross-section area of the block. In this case the required height is h ""

where dV is the differential cylindrical volume element, x being its radius and dy its height. Finally, we integrate dV benveen y =

PoVo

a and y

"" h.

V=nJo

pz(Al + VOZ3)

, (2Ry-y')dy

=n[2R~_y3]'

The first case is valid if Al :::;Pl(PO -PI) PZ(Pl -Po)

(c)

Fig.3E.68

V Z3 O

2

www.puucho.com

3

0

Study Anurag Mishra with www.puucho.com

326

MECHANICS.II (M + m)::: )M7tR2aL :::~rrR2Lp.1tR2aL

Hence, from the equation for the weight of the ball: (Weight of ball) = (volume of submerged portion) (density of water) ~

(~7tR3jc(4)W

-i)w h'( 2-%)

In:::

LJCiP -rrR2Lp

7tR2

2

= Tth ( R

~(2)3(0.4)=

Example A conical cup of height b, semiverrical angle a rests open end down on a flat surface as shown in Fig. 3£.70 (a). The cup is filled to height H with liquid of density p. What is the upward lifting force on the cup ?

h3-6h2+12.8=O

We find by synthetic division that h ", 1.75 feet.

Example A wooden stick of length L, radius R and density p ha.~a small

metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float verticallx in equilibrium in a liquid of densitx u. Solution: The stick will be vertical, that is, in rotational equilibrium if centre of gravity lies below the centre of buoyancy. For minimum m, the two points will coincide. Let h be the length of immersed portion. For translatOl)' equilibrium, combined weight of stick and mass attached is equal to brought force,

Fig. 3E.70 (a)

~~~~~~f' .I~~~~~

.....-..... .... ----_._-

..._---_ .

_---- ..

:::::: :: ::::::

:::::: _: m :::::

Flg.3E.69

... (1)

M:::7tR2Lp

The height of centre of mass from bottom (M)L/2 + m x 0

=

m+M

= 2(m +M) ..

h = 2

h=

m, this

F =

J:

x2 = 0.39845

or

x= Q.631yl4

Fig.3E.97

www.puucho.com

Study Anurag Mishra with www.puucho.com 343

SOLIDS AND FLUIDS

ExamJ:>le

(b) Rate at which liquid is discharged,

."

99~

-=adx -=Qt'=a

_dV

dr

A syringe, as shown. in Fig. 3E.99, contains an ideal fluid of density p. The cross-section area of plunger is A and the small orifice at the end of the cylinder has area a.

or

(aJ Determine the velocity of outflow of a liquid from the syringe if (he plunger moves with a constant velocity under

or

the action afforce F. (b) If the volume a/fluid is V, how much work must you do to empty the cylinder in time t ? F

dr

-J'v dV=a \tpA ~J'

0

V=a

~ IpA

dr

~t

\ pA pAV2 F=-2a2t2

or

Work done by force

V A

W =F-

pAV2 V =-_x_

2a2t2

A

3

-= pV

2a2t2 Fig.3E.99

Example

Solution: Distance covered by plunger in time t is ut, the work performed by force F is W = Ful. The mass of liquid flowing out during rime t is pAut. If outflow velocity is v, from equation of continuity we have Au = UP. The change in kinetic energy of the liquid during the time t is j,KE = pAllt

lV' ,,' J l J 2" - 2

100

_

A test tube is filled with water and closed with a loose fitting stopper that allows air co pass freely. There is a small hole in the base of the test tube as shown in Fig. 3E. 100. The test tube is rotated with angular speed w. The distance of small hole from the axis of rotation is R As the tube spins, water sprays out of the hole. Determine the speed with which the water flows through the hole.

From work~energy theorem, we have Fut =pAut On eliminating

u, we get 02

-=

2F x

pA If a

V' "' 2-2

1 l_a2/A2

Flg.3E.100

« A, then

@f

V=

Alternatively

\ pA

we may apply Bernoulli's

1 (Po +P)+-pu

2

2

+pgh

where Po is atmospheric

,,=

theorem.

or

1 2 -= Po +-pv +pgh

pressure

F

2

-= _pAw2

J

rdr

r'

and P is pressure

to external force F. As a « A, velocity nearly zero, we have F 1 , p -= - -= ~pv A 2

or

Solution: We consider a differential element dr at a distance r from the axis of rotation. The centripetal force on this element is dF -= _pA(I)2rdr

inside

cylinder

due is At pressure.

=_pAW2~+C 2 P, r=R-DF=~ where 'A Hence

(3£

\'pA

www.puucho.com

p" "

is

the

atmospheric

Study Anurag Mishra with www.puucho.com

MECHANICS.II Thus

Po pAw2 -+-+--(R-D) 2 A 2

2,2

F=-pAw

2

Force just inside the hole can be determined substituting, = R; we get 2 2 2 F} = _pAoo R + Po + pAw (R_D)2 2

A

by

pv' -=P1-P2 2

A

+ poo2 (2RD _D2) 2

••• (1)

From Bernoulli's theorem, we have 1 2 1 2 ~+-~1=~+-~2

2

=pgH

... (2)

Since the fluid flow velocity is constant, the pressure varies according to the law

2

Pressure at this point is PI = F1 = Po

where lim is a small element of fluid, !ih is height of fluid element and lim = pAlih is the mass of the element. Hence

P = Po - pg(h - x)

similar to a liquid at rest in a vessel. Here Po is atmospheric pressure and x is the distance from the upper end of the pipe.

... (2)

2

As cross-section of tube is large, we have

vI

... (3)

p

,= 0 and

v2 = v. From eqns. (1) and (2), we get I

-pv 2

,

Pl:PO +pg (H-h) ---Po ••••

=P1-Po

-- ••••••••

-

P2:Po-pgh

H-h

H

Fig. JE.101 (b)

or

The Fig. 3E.101 (b) shows the change in pressure with the height.

Example Fig. 3E.101 (a) shows a broad vessel connected with a narrow tube. How are the pressure and velocity of the liquid ill the vessel and in the pipe distributed along a vertical when the liquid flows out?

::~::~::::::~~::::::::~~:

1

•••••_._.- ---.-.-.."...•••••-.---_.

----:::::

H

U

Example Two holes are drilled in the wall of a vesselfilled with water. The distances of the holes from the level of water are hand h + H. Find the distances x and y as shown in Fig. 3E.102 where the streams flowing our of the holes imersect. Assume that the level of water is maintained in the vessel by regulated supply of water. A -.-.-.:.-.-.-.-.-.----

.-.". •••••

---.-

------

-

.:-:.:-: c .. _---:-:.:-:.:-: .._ ....

-:-:-:-:-:-:-:-:-:-:0

FIg. 3E.101 (a)

Solution: Taking cross-section area of vessel and tube A and a respectively, with A » 0, we see that the velocity of liquid in the pipe is constant along the entire cross-section. Velocity of efflux is v = ~2gH. There must be a pressure difference PI - P2 at the junction of vessel and pipe that gives kinetic energy to the fluid. From conservation of energy, we have I , - .1.mv = (PI - P2)alih 2

h

-:-:-:-:-:-:-:-:-:-: a----_

... (1)

p

Fig. JE.102

Solution: At the point p, point of intersection of streams, the range for the two streams are same. As the range of liquid issuing from a depth h from the surface and from the base range is same, hence CD=AB=h

or y=H+h+h=H+2h The time taken by water coming out of C to reach P,

www.puucho.com

Study Anurag Mishra with www.puucho.com

345

SOLIDS AND flUIDS t=

V

(3E

P==Po+-dg+-2dg+-dx

i.e.,

3 P==Po+-Hdg

2

g

Velocity of water coming out of C, t' =

H H

i.e.,

!'t

\ g

.ExampJe

i.e.,

103 ~

of large uniform cross-section area. A resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities d and 2d each of height (H/2) as shown in Fig. 3E.103.

A container

f

1 +-Ldg

5

1 = Po+-(6H

2

g(3H - 4h) = 2",2

or

or v = ..,j(g/2)(3H - 4h) (ij) Liquid issues out horizontally from the hDle, SDtime taken by it to reach the ground, ~_

d

-'-~W.----1". ::::=:::::::: 2cf:::::::~-~ -,.

so that

,._-....-.-.-.-.-.-.-.-.-.' -----_._ ..----

Hf2

+L)dg

Po+Hdg+(H-hJ2dg=Po+~(2d)V2 2 2

T = \1(2h/g_) __

Hf2

1

xgxA

1 2 1 Z PI +-Pt'l = Pz +-pl'2 2 2

.. ./2g(H + 2h) = 2.Jh{H + Zh)

= ~x

-xL

4

2 4 4 (b) (i) Applying Bernoulli's theorem for a point just inside and outside the hole,

,l2gCH + 2h)

Hence required value of x =

5 (A J

2

-----_ .... ---- __ I :-:-:c-:-:-:-:-:-:-:-:-:-:.-------------

x=

". ,,' ..

l't

= ,/!(3H-4h) 2

x ~

, g

= Jh(3H - 4h)

..

'," , '

(iii) FDrx to be maximum x2 must be maximum, i.e., ~(X2)=O

dh

Fig.3E.103

The lower density liquid is open to the amlOsphere having pressure Po- (a) A homogeneous solid cylinder of length L(L < Hj2), cross-sectional area (A/5) is immersed such that it floats with its axis vertical (u the liquid-liquid interface with length (L/4) in the denser liquid. Determine (i) rhe density of solid and (ii) the total pressure at the bottom afthe container. (b) the cylinder is removed and the original arrangement is restored. A tiny hole of area s(s < A) i.o; punched on the vertical side of the container at a height h(h < H/2). Determine (i) the initial speed of efflux of the liquid at the hole, (ii) the horizontal distance x travelled by the liquid initially, and (iii) the height hm at which the hole should be punched so that the liquid travels the maximum distance x", initially. Also calculate xm.

Solution: (a) (i) The cylinder flDats in equilibrium hence weight of the cylinder is balanced by brought forces exerted by the two liquids. Weight cylinder = brought force of liquid 1 + brought force of liquid 2. Vpg = Vj djg + V2d2g Ot

i.e.,

or

~(3Hh-4h2)=O dh

or

i.e.,

and Example

2

Ii == (3/8)H X

= max-,S

13H(3H-~HJ=~H 24

104

_

A heavy rod of length I is released from a height h above the water surface as shown in Fig. 3£.104. The rodfall.~ in air and then in water, always remaining in vertical position. Find tlte velocity of the rod when it reaches the bottom of the rank. Assume the water level remains uTlcllQngcd due to fall of the rod. The relative density of the rod L~2.

L(~Jp=(%LJ(~Jd+(~LJ(~J2d 3

3H-8h=O

fm h

________ L f

...... ..... __ _ .......••• _ . ....... _ _-_. -__

-_

-_ .. .

::::::::::::::::::::::0:::::::::

5

-----------, ........... , =J2g(h-%)+3;L =J2gh+g~

When the rod is completely inside the water: Resultam force = LAag - LApg

g[2(H +h)_l]_PW

2

g{ZH -I] =

l'2•

2

where Pm = density

2

U"'=~g{H+2h)-g~

of metal.

Jg[2lH + hl -

= 2gh +gL +gH -gL 2

= glH + 2hl_gL

Pm t' =

= u2 + 2g (H -L)

2

-I]

1 ='2PmAl.l'

2

Let u'" be the final velocity of the rod when it reaches the bottom of the tank.

l' .•.2

1 ='2ml'

=!

lAo

l' .•.2

-~]-PwAlg[~+H

21.

Acceleration, d = !AcrE - !Apg

PWAl'g +PwAIg(H-l)

222

or

-x)

l,dv=~(21.-x)dx 2L "

= PwA12g

2

mg[ H +h

a = LAag - xApg = ~(21. lAo 2L dv g v-=-(21.-x) dx 2L

"

of rod and P••.is density of water. Work done by rod against average buoyant force is

mg [ H+h--

So acceleration,

n -~[2H -I]

SURFACE TENSION

V=NlH+2hl-~] Method 2. Let area of cross-section of the rod be A. density of rod be 0' and density of water is p. Velodty of rod when its lower end touches the water surface,

-----------il----- fl


R=(¥}

Putting this value in eqn. 0), we get h

Z/Tcosa (1- h)'pg

=

hl- h2 = 2/Tcosa rpg

That is, the fluid immediately adjacent to plate 1 is essentially at rest; the fluid immediately adjacent to plate 2 moves with velocity Vo (and is thus at rest with respect to plat 2); and the fluid velocity varies smoothly and uniformly between the t\vo plates, being everyv.rhere parallel to the velocity of plate 2. Due to smooth variation of velocity, the fluid is being divided into an infinite number of parallel layers of infinitesimal thickness dy. Each of these layers is called a lamina (plural laminae). This comparatively simple type of flow, in which cach lamina slips past its neighbours with relative speed dl', is called laminar flow. A force F must be applied to plate 2, as shown in Fig. 3.96 to maintain the constant velocity Vo in the presence of the viscous drag exened on the plate by the fluid. At the

L

Which is a quadratic in h, therefore,

i

1~_2ITcosa 2-\4,pg

h: +

==:C-------~-.

dy dy __

If the tube be only slightly conical then cos a = l.

= 1'0

£x D

v v+ dv -'dv

~

~

p",,~

FLOW

when you move your hand through water and thus set the water into motion, you feel frictional resistance. This resistance, often called drag, always opposes any forces that tend to make a fluid move. Like kinetic friction, fluid drag always tends to remove kinetic energy from the system and bring it to rest. The magnitude of the viscous drag force depends on the "thickness," of the fluid. Fluids such as molasses and grease have large viscosities. Fluids such as water and mercury have smaller viscosities. The viscosity of air is very much smaller still. and the ideal fluid has zero viscosity by definition. 1\'10 very large parallel plates. each of surface area A. are separated by a distance D. the space between them is filled with a fluid. Plate 1 is stationary; an external mechanism (not shown) drives plate 2 with constant velocity Po relative to plate 1. [n Fig. 3.96 the direction of motion is the x direction. If the fluid has non-zero viscosity, the moving plate drags it along. Measurement shows that. if 1'0 is not too large, tile moliol! of rhe fluid is everywhere ill rhe x direction. The velocity of the fluid varies linearly with distance y from the stationary plate according to the simple rule. l{y)

LZ:::.._-_-_-_-_-_-_-_-_-_-_-_-_-_-,_~v

yD_F/

h=i:tI~_2IT 2 'i 4 "pg

VISCOUS FLOW; lAMINAR

p,,,.,;;f

... (1)

Fig. 3.96

same time, a force -F must be applied to plate 1 to keep it in place. The two forces constitute a couple; thus there is a shear stress a :. FjA applied to the fluid between the plates. But a fluid can respond to a shear stress only by moving. Consequently, plate 2 moves through an ever-increasing distance X = vor relative to plate 1. That is, the constam shear stress a results in a shear strain P = KID that increases without limit. In solids the stress is directly proponional to the strain and the proportionality constant is the shear modulus G defined by Egn., G = criB. But, laminar flow, the stress is directly proportional to the time derivative of the .~train. We call the proportionality constant 'l and define

a aD ~ = d~dr = dlCdr or

aD

~=-

...(2)

"0

Coefficient of the quantity T\ is a property of the fluid and is called the viscosity. It is the quamitative measure of the opposition of the fluid that. Because the 51 unit of stress is the pascal, it follows from egn. (2) that the 51 unit of viscosity is the pascal-second (Pa-s).

www.puucho.com

- -

Study Anurag Mishra with www.puucho.com

354

MECHANICS.II Scale

The torque t = F'l exened by the fluid on the inner cylinder and the wetted area A Qf the cylinder. The shear stress exerted by the fluid on the inner cylinder (Qr by the inner cylirider Qn the fluid) is:

~::::~

F 'ti'l G=-=--=-_

A

2xr, h

t

2xrr2 h

From the knQwn value Qfthe angular speed Qfthe outer cylinder tQ calculate the linear speed vo. Vo

,

...

, ..

.

= oor2

So equation (2) becomes t('2 -'I) ~=~~~ 2XO>T'I2r2h

-

... (3)

A Microscopic View of Viscosity

FIg. 3.97 Rotatrng

cylinder

vlscomete~

Eqn. (2) properly describes the behaviour of fluids only up to some critical value of va at which the flow ceases to be laminar. The critical value of Vo depends on the fluid, the temperature and pressure, and the shape and size of the system, Above that value, turbulence, become dominant. Thrbulent flow dissipates mechanical energy more effectively than laminar flow. Viscosity is by measured device the rotating-cylinder

viscometer shown schematically in Fig. 3.97. The outer cylinder, which has a bottom so as to contain the liquid, is mounted on a variable-speed turnable. An inner cylinder hangs from a torsion balance calibrated so that its scale directly reads the [Qrque exerted on the inner cylinders. The two cylinders are coaxial. The difference between their radii, '2 - 'I' is much smaller than either radius. Thus the annular (ring-shaped) space between the twQ cylinders is a reaSQnable approximation Qf the planar space between the twQ parallel plates of Fig. 3.97. The space between the cylinders is filled tQ a depth h with the fluid whose viscosity is to be measured. The outer cylinder is made tQ rQtate, and it pulls the fluid alQng with it. As a result of the visCQUSdrag Qfthe fluid, the inner cylinder experiences a torque t. The distance D between the plates is just the radius difference; D = '2 - 'I'

CQnsider twQ adjacent laminae, such as those labeled b and c in"Fig. 3.97. The velocity of the fluid in a lamina is equal to the mean velocity of the molecules contained in that lamina. Thus the mean velocity of the molecules in lamina b is v, while the molecules in lamina c have the slightly." greater mean velocity v + dv. Consequently, moleCules are continually migrating in large numbers between the two laminae. On the average, molecules passing from lamina c to lamina b will be moving too fast for their new lamina by an amount dv and will slow down as a result of collisions with the molecules in lamina b. The result is a transfer of momentum from faster-moving laminae to their neighbouring slower-moving laminae and thus eventually to plate 1. Because the original source of this transferred momentum is plate 2, the overall result is a transfer of momentum from plate 2 to plate 1. If no external forces were applied, this momentum transfer would soon diminish to zero the speed of pate 2 with respect to plate 1. The externally applied couple required to maintain constant v consists of the forces required to overcome viscous drag. The reduction in the velocity of the molecules in the direction of laminar flow is because their rebound directions after collision tend to be random. This randomization process, amounts to an increase in the thermal energy of the fluid at the expense of its macroscopic kinetic energy and process is dissipative, or "frictional." Also in liquids there is an additional, stronger interaction between molecules in adjacent laminae, owing to the intermolecular forces that distinguish liquids from gases. Poiseuille's Law The flow rate Q is in the direction from high pressure to low pressure. The greater the pressure difference between two points, the greater the flow rate. This is precisely stated as Q= P}-P2 R

www.puucho.com

Study Anurag Mishra with www.puucho.com

SOLIDSAND FWIDS

355

(a)

(b)

Fig. 3.98 Fig.3E.105

where Pj and Pz are the pressures at two points, such as either end of a tube, and R is the resiSTance to flow. The resistance R includes all the factors that affect flow, except pressure. For example, R is greater for a long tube than a short one. The greater the viscosity of a fluid, the greater is R. Thrhulence greatly increases R, whereas increasing the diameter of a tube decreases R. Poiseuille derived his result by taking the fluid in !:J.E evaporation. i.e., if

PV = J1RT.i.e.,~ = RPVT]

pL

Since this radius is of the same order of magnitude as the size of one water molecule, a drop of water with this radius cannot exist. Therefore, there is no water drop that can evaporate without absorbing heat, or losing internal energy.

Exam",le .'_._- ~ .

--b

Water in a clean aquariumfonns Fig. 3£.133 (a).

The conditions are isothermal, so temperature is constant, i.e., Ta = Tb = T" so the above expression reduces

-----.--_

to

PaVa +PbVb = PcV, From eqns, (1) and (2), we get

2 _

2T 7 x 10-"m. R T2 (c) T\ = T2

< Tz (d) data insufficient 83. A capillary tube of radius r is placed in a liquid. If the angle of contact is B, the radius of curvature R of the meniscus in the capillary is : (a) r (b) rsine

(b) ~ '1t(r24 - ,,4) (0 - p)g

3

(e) ~ n(rt - '13)(0 - p)g

3

Cd) ~ 3

,

.---- ._.A. .._ ._____ _

"

(a) The body will move up (relative to liquid) (b) The body will move down (relative to liquid) (c) The body will remain stationary (relative co liquid) (d) The body will move up for some inclination Band will move down for another inclination B 82; The diagram shows the change in length & of a thin uniform wire caused by the application of stress cr, at two different temperature T\ and Tz. The variations shown suggest that :

4 TC(r2 '2

4 '1

+ rl

lca -

p)g

,

BO. An ideal liquid is flowing in two pipes, one is inclined

and second is horizontal. Both the pipes are connected by two vertical tubes of length hI and hz as shown in the figure. The flow is streamline in both the pipes. If velocity of liquid at A, Band C are 2m/50 4m/s and 4m/s respectively, the velocity at D will be :

(a) 4m/s

(b).JJ4

m/s

(c) J28 m / s Cd) none of these 81. A body B is capable of remaining stationary inside a liquid at the position shown in Fig-] if the whole system is gently placed on smooth inclined plane (Fig. II) and is allowed to slide down, then (0JM (b) ~ rrpH \ 2rrpH

h

2 the

of a flux at these two holes, then (a) -

Co)

Cb) (iil (c) (iii)

(e)

h

liz

1

(a)

I~

h

from 2 surface of lighter liquid. If 1'1 and 3h

170. The three water filled tanks shown have the same volume and height. If small identical holes are punched near this bottom. which one will be the first to get empty ?

Cb) 9.8 kW (d) 4.9 kW

)jlM pH

(d) None of these

172. Which of the following is not an assumption for an ideal fluid flow for which Bernoulli's principle is valid? (a) Steady flow (b) Incompressible (c) Viscous (d) Irrotutional 173. A body of density p' is dropped from rest at a height Il into a lake of density p, where p > p'. Neglecting all dissipative forces, calculate the maximum depth to which the body sinks before rerurning to float on the surface? (a)

h

p-p (e)

l!L p-p

Cb) hI"

P (d) ~ P-P'

174. A Newtonian fluid fills the clearance hetween a shaft and a sleeve. When a force of 800 N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.S em/sec. If a force of 2.4 kN is applied instead, the shaft: would move with a speed of : (a) 1.S cm/see (b) 13.5 em/sec (e) 4.5 em/sec (d) none 175. A solid metallic sphere of radius r is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity, then the tenninal velocity of the sphere is proportional to \'lhich of the following?

www.puucho.com

Study Anurag Mishra with www.puucho.com '"

394

MECHANICS-II

m-m' (b) r (b) m+m' , (a) Cd) r1/2 •..•. .'(c) (m+m') 176. Two drops of same radius are falling through air with (d) (m - m' )r2 steady velocity of vernis. If the two drops coalesce, what would be the terminal velocity? 181. Which of the following is the incorrect graph for a (al 4v (bl (4)v, v sphere falling In a viscous liquid? (Given at t = 0, . , " velocity v = 0 and displacement x = (e) 2v (d) 64v ., 177. A cubical block of side 'a' and density 'p' slides over a fixed inclined plane with constant (a) ~ (b) velocity 'v', There is a thin film of viscous fluid of thickness 't' between the plane and the block. Then the coefficient of viscosity of the thin film will be : (a) ,2

,

(e) r312

,

"

°)

l

~:

(al 3pagt Sv (e) pagt

'L

=....:

(b) 4pagt Sv

Cd) none of these

"

182. The displacement of a ball falling from rest in a viscous medium is plotted against time. Choose a possible option.

178. Which of the following graphs best. represents motion of a raindrop?

(alL

(a)

.......

(b)

(d)!1

(e)

/

~ 183. 179. A spherical ball of density p and radius 0.003 m is dropped into a Oem •. tube containing a viscous fluid :~:~:~:~:~:i".~' filled up to the a em mark as .~.::.~ ..::.:;.:: ..:. 1ocm: shown in the figure. Viscosity of ::::~:~::::::::~ the fluid = 1.260 N_m-2 and its 20cm '-.:':.:'-.:';': ..: density PL = £ = 1260kg. m -3. -';. ..;.:--':-:.:..: 2 184. Assume the ball reaches a terminal speed by the 10 em,. ~ mark, The time taken by the ball to traverse the" distance between the 10 cm and 20 cm mark is: (a) 500 IlS (b) SO ms (c)0.5s (d)5s (g = acceleration due to gravity = 10 ms-2)

:~:E~:E~

180. A ball of mass m and radius r is gently released in a .. viscous liquid. The mass of the liquid displaced by it is' m' such that m > m'. The terminal velocity is proponional to :

There is a 1 mm thick layer of glycerine between a flat plate of area 10 cm2 and a big fixed plate. If the coefficient of viscosity of glycerine is 1.0 kgim.s then how much force is required to move the plate with a velocity of 7 cmls? (a) 3.5 N

(b) 0.7 N

(c) 1.4 N (d) None of these There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then what will be the tension in the thread? ltR' (al (b) ltR'T T (e) 2ltRT (d) 2RT 185. A container, whose bottom has round holes with diameter 0.1 mm is filled with water. The maximum

www.puucho.com

•

Study Anurag Mishra with www.puucho.com

SOUDS AND flUIDS height in em upto which water leakage will be what? (Surface tension (a) 20 em ee) 30 em

=

can be filled without (a)

75 x 10-3 N/m and g = 10 m I $2) (b) 40 em Cd) 60 em

(e)

186. If two soap bubbles of different radii are connected by a tube: (a) air flows from the bigger bubble to the smaller bubble till the sizes become equal (b) air flows from bigger bubble to the smaller bubble till the sizes are interchanged (e) air flows from the smaller bubble to the bigger Cd) there is no flow of air 187. A liquid is filled in a spherical container of radius R till a height h. At this positions the liquid surface at the edges is also horizontal. The contact angle is:

395

,

(~)j ,

lz:r

(d) none of these

189. A long capillary

tube of radius 'r' is initially just vertically completely immerged inside a liquid of angle of contact 0°. If the tube is slowly raised then relation between radius of curvature of meniscus inside the capillary tube and displacement (h) of tube can be represented by : R (a)

(b) h

R

L.... r ••.•••••.•

(e)

(d)

R

0 Cc) cos-t(h;R) (a)

1(R;h)

Cd)Sin-I(R;h)

188. A soap bubble has radius R and thickness

h

h

Cblcos-

d( v" =PO-pg(h1 =VB

=

=VE

+hz) = J2gh\

+vB

SVA

4. n drops of a liquid each with surface energy £ join to form a single drop. Then: (a) Some energy will be released in the process (b) Some energy will be absorbed in the process (c) The energy released or absorbed will be £(0_n2,/3)

(d) The

energy

n£(22/3

www.puucho.com

_ 1)

released

or

absorbed

will be

Study Anurag Mishra with www.puucho.com 401

SOLIDS AND FLUIDS

5. A cylindrical vessel of 90 em height is kept filled upto the brim. It has four holes 1, 2, 3, 4 which are respectively at heights of 20 em, 30 em, 40 em and SO em from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from:

,,~~~~~~~~,;

4

:::::::::::.

3

::::::::::::

2

..-----_---_ ..-----_ ..--------_ ...

..---_ ..----

p

Q

(a) hole number 4 (b) hole number 1 (e) hole number 2 Cd) hole number 3 6. Siphon is a device to transfer liquid from a higher level to a lower level. The condition of working of a siphon is :

(c) Ball hits the top of container at end Q after a time

t=Hi (d) Ball hits the top of container at end Q after a time t

=

r¥

\ 3g

8. A cylinder is floating in rvvoliquids ~: as shown in figure. Choose the correct options : (a) net force on cylinder by liquid 1 is zero (b) net force on cylinder by liquid 1 is non-zero (c) net force on cylinder by liquid 2 is equal to the upthrust (d) net force on cylinder by liquid 2 is more than the upthrust 9. Three different liquids are filled in a U-tube as shown in figure. Their densities are Pl ,P2 and P3 respectively. From the figure we may conclude that:

h2~lp, IJ-p'I'

.0

~j

.:'~,:, '.,. (a)h2>hj (b) h2 =' 2h1

(e) hI should be less than the height of corresponding liquid barometer Cd) h] should be greater than the height of corresponding liquid barometer 7. A small solid b(ll1of density P is held inside at point A in a closed cubical container of side L, filled with an ideal liquid of density 4p as shown in the figure. Now, if the container starts moving with constant acceleratioll a horizontally and the ball is released from point A simultaneously, then:

(a) P2 > PI

(b) PI > P2

(c) P3 = 2(P2 - PI)

(d) P3 = P2 +PI

~)~>w

~)P>P

2 A block is floating in a liquid as 10. shown in figure. Support w = weight of block, P = pressure at bottom of block and F = upthrust on the block. Now suppose container starts moving upward with some positive acceleration. The new values are suppose w' ,P and F. Then: (c) F' > F

(d) F -w' = 0 There are rvvo holes on a water tank as shown in 11. figure. Area of hole 2 is rvvo times that of area of hole 1. Suppose v is the speed of liquid coming out and Q the volume of liquid flowing per second. Then:

Q

p

U21 ~m~mmmm A

HI

U2111!!!mmmm R •

(a) (b)

~~~~mmH I

~::. H/4 :::::::: 2 ~~~~~~~~~~~~~~~~~l-------,. -------_ ----------... L

(n) 1'1 =

.s

(c)

For ball to hit the top of container at end Q, (/ 0:= 3g For b L12. Solution: We choose a mass element dm at a distance x, as shown in Fig. 4E.9. All -V2 such elements produce a dm gravitational field at point P, that points towards negative x-axis. The resultant field can be Flg.4E.9 calculated by integrating the magnitude of the field produced by dm from x = -L/2 to

-'-

Solution: We choose an element that is an arc of length ds ::0 R de, its mass is dm ::0 AR dO. For every element at. + x there is an equivalent element at -x whose contribution is symmetrically opposite. From this symmetry we can see that the field strength has no x-component at the centre. The y-component of the field is dgy ::0 dg sinS G dmsinS =

The resultant field strength is

f

g = dg, = -GAf" smS d S

R

x = +L/2 N;

dg = Gdm

The Gravitational Field of a Spherical shell By Integration

M

dm= -dx L

and

r::o

o -

X

0

2GA =-R

,2

where

R2

X

First we will detennine the gravitational field on the axis of a ring of unifonn mass density. We then apply our result to a spherical shell which can be considered to be a set of coaxial rings.

www.puucho.com

-

Study Anurag Mishra with www.puucho.com

G~VITATION

427

Fig. 4.13(a) shows a ring of mass m and radius a and a point P on the axis of the ring. The field at P due to the element dm is toward the element and has the magnitude given by

dg:.Gdm

= M sinSdS 2 Now the eqn. (2) becomes G dM GM sinSdS dg = ---cosa = -----cosa s2

r

"

We have three variables s,8 and a. We will convert all the variables into s which varies from s = r - R at e = 0 to s = r + Rat e = 180°. We use the cosine law

RdO

$2

,

p a

... (3)

2052

On differentiation,

P

= r2 +R2

-2rRcos8

we get

2.~ds=2rRsinSdS

dgcos a

or

. BdB = sdo< Sin

By applying

cosine law to the same triangle,

R

(.J

(bJ

Fig. 4.13

coso: =

A symmetrical element dm on the other side has the same field. But symmetry shows that perpendicular component due to directly opposite element of the ring will be cancelled. The net field will therefore be in the negative x-direction. The x.component of the field due lO the element dm is

dg r :. -dg casa Cdm

dg =_GMsdss2+r2_R2 r

f " = -fe dm cosa

,

= - ~~

GM f"R (1 +-~"-R'Jd $2 r~R

by

,

2

= _ ~~R[$-

on the ring, they Thus,

r

~R2]"R ,.R

GM = - r2

dm

co,,, [f

dm =

M](1)

Now we consider a spherical shell of mass M and radius R. We choose circular strips of mass dM as our elements [see Fig. 4.13 (b)]. The field due to this scrip is radial and given by GdM dg = ---coso: ... (2)

,

~~R(I+r2:2R2)ds

gr =--4r2R

" for all points

gx ::.-2cosaf

2~r

rR

2052

=_

g" = dgx

Since s and a are the same are constants for the integral concerned. G

2.~r

these results in eqn. (3), we get

The field due to the entire shell can be obtained integrating from s = r-R(S = 0) to $ = r+R(e = 180°).

= ---cosa

Now,

Substituting

we have

S2+r2_R2

"

The radius of the strip is R sinS, so the circumference is 2n:R sine. The width is R de. If M is the total mass of the shell, and A = 4n:R 2 is its total area, the mass of the strip of area dA is dA dM=M A M(21tR sinS)(RdS)

=------4n:R

If the point P is inside the shell the limits of R - r to R + r, thus

$

are now

=0

GRAVITATIONAL

POTENTIAL

The change in potential energy when a mass m is moved from initial position ri to final position rj, near a mass M,is given by ~PE=

GMm[! - ~]r rj

j

Th\ gravitational potential difference betwcen initial and final pomts due to mass M is defined a.~(he change in potential energy per unit mass. Thus,

2

www.puucho.com

Study Anurag Mishra with www.puucho.com

MECHANICS,II

il28 V(rt)-V(rj)=

[-~~]

-

Fi~a1

•.. To find the gravitational potential at a given point due to a collection of n particles we scalarly add potential due to each individual particle.

[-~] initial potential

Thus we say that the potential due to mass M at a distance r from it is given by V(r)=_GM r

Alternatively: Change in potential energy = negative of work done by field

m

--

m

Wexternal agent

Gravitar10nal potential difference between two points is also defined as the work done by an external agent in moving a unit rnassfrorn initial position rj to final position rf along any path. •.. Relation between gravitational field intensity and gravitational potential Work done by an external agent in moving a unit mass through differential displacement ~ d r is ....•

. Consider a differential element dm on the ring. The potential at P due to mass dm is dV=Gdm (x2+R2)12

G~_r~ ~

r -.....

dr I+-

Fig. 4.14

Fig. 4.15

The total potential is ....•

v=-f

....•

dW = F. dr = g.dr By definition, this is potential difference, hence

fVv, Q

dV =

_fr+dr '

In general,

-

f

VQ = - r

=-

g. d;

r+dr.....

Vp

Gdm

(x2 +R2)l'2 G (xz +RZ)l'Z (xz +R2)~'Z

g.dr

The gravitational field intensity at P can be determined

....• dV" g = --(r)

from

dr

~

d

g = - dx C':')

g=gxi+gyj+g;k =-~[-(xZ~~2l2]

dr = dxi+ d.r.J+ d;k we have

dV=-[gxdx+gydy+g;dz]

where

gx = - ax ,gy = - Cly ,gz = - Clz

av

~g=-

av

iJy

=-

av

[av, av, av,] -l+-J+-k ax

Idm

GM

=

....•

In terms of cartesian coordinate system,

and

ri

Potential Due to a Uniform Ring On its Axis

m

...•

i"l

r

VCr!) _ VCr ) = ~PE = _ Wgrav. i

i"l

m.

-'

The total potential is obtained by integrating the function dV over the entire object. Since the potential V is a scalar function of position, it is ~ often preferable to calculate the field strength gdue to a continuous object by first calculating V and then differentiating the potential function.

= work done by external agent

= Wenemalagent

n

•.. To find the gravitational field at a point due to a continuous object, we find the potential dV due to a differential mass element dm, which contributes dV=_Gdm --

6PE = -Wgmv.

n

v=:LV;=-G:L

GMx (x2

+Rz)3.Z

Potential Due to a Disc on its Axis Surface mass density of disc is

a,

M" 0=--

7tR'

www.puucho.com

Study Anurag Mishra with www.puucho.com

GRAVITATION Now

429 we

consider

-.

a

differential ring of radius rand thickness dr.

"'"....~ -'d'......... ,.... ..\"

Gravitational

..

field on the axis,

.'

E = _ 2GM [1 _

-.'x-.-.:: ..........••. p

dm == (2T1:rdr)a

Mass,

2. Uniform disc of mass M and radius R

2

R

ZMrdr

=---

Gravitational

R'

The

pmenrial

differential

due

on the axis,

V = _ 2GM [x_)R2

to

x ] + x2

+x2J

R'

ring dV=-

3. Thin spherical shell of mass M and radius R

Gdm

Gravitational

.Jx2+r2 2iJMrdr

=

field at a distance

r from centre:

R2,/x2+r2

Total potential

can be obtained

r

limits r= Oto r= R. v=_2GM

°.Jx

2

R T h.e

potential

Fig. 4.16

,IR2

mtegra I I

=;

by integrating

dV in the

rdr +r2

......

2

f ~~~== ,d, ../x2

can

bid e so ve

:r, = R :

by

1

go2

Fig. 4.18

(i) Inside the shell,

GM

V(r riowork done is negative, as gravitational force is opposite to displacement. If rj > rl' work done is positive u.s gravitational force is in same direction to displacement.

ciA =JGMaO-e') dt T=

""~ JGMa(l-e')

•.

Ia'

As defined before, the potential energy is negative of work done by a conservative force. Wgrav. =-(PEf

1-;;"'-

= '" \GM

We compare this expression correlate the terms.

or

GRAVITATIONAL

Concept: We can choose any arbitrary path between initial and final poines, and break it up into radial and circular segments u.s shown in Fig. 4.27. Thus the work done by the gravitarionalforce is path independent. It depends only on .initial and final positions. Gravitational force is a conseMlative force.

POTENTIAL ENERGY

W sm.

A particle of mass m moves from position A (a distance ri from mass M) {Q point B (a distance rj from mass M) as

rj (-

=-GmM(_~+l) rf

rj

'/

"

where r is a unit vector in radial direction. We break up the path into circular and radial segments. Work done by gravitational force along circular segments is zero because ~ displacement d r is perpendicular to force. Only work done is along radial segments.

f"

earlier, and

GmM PEf =---

F=_GMmt

W '"

obtained

=-[(-G~H-G~)]

shown in Fig. 4.27. Origin is located at mass M. The gravitational force of mass M on m is

Thus

-PEi)

, fJ. dri

GmM PEi=---

"

Thus we say that the mass m has a potential energy given by =.:.GMm

PE l(l'llV.

G~M

r

due to its position in gravitational field of M.

www.puucho.com

Study Anurag Mishra with www.puucho.com

437

GRAVITATION ~

If g is the magnitude of local acceleration due to gravity, GM GM

If masses m and M move with velocities!' and V relative to an inertial frame and are subject to their gravitational interaction, their total energy is E = KE + PEgra,", 1212GMm =-MV --/Ill'

2

---

2

,

When M is much larger than In (M »m), e.g.,sun and planets, earth and moon, etc., we assume that M is at rest in all inenial systems. Thus total energy is E =~nw2

•.

R2 = ---

R

+ mgh

Since we are interested only in potential energy change when a body is moved a distance 1J above the surface of the GMm earth, the constant term --cancels out. R

,

When a particle of mass m moves in a circular orbit, we have

mv2

GMm

Rf:

GMm

PEgra,'.

~

_ GMm

2

g=->-

-;T=-,-

The total energy of the system is negative. All elliptical (or bound) orbits have a total energy (E < 0) negative, when the gravitational potential energy is chosen a zero value at infinite separation. The energy for elliptical orbits is given by E = _ GMm

2a Therefore

kinetic energy is

1 2

-ml'

2

GMm

•.

=--

2,

The total energy of the circular orbit reduces to _ GMm GMm E -----2, ,

GMm

=--2,-

Potential of a Body on Earth's Gravitational Surface •. If a body is at a height II above the surface of earth with IJ «RE,

where a is semi-major axis. The potential energy of mass m is taken to be zero when it is so far from the mass M that the gravitational influence of mass M vanishes. i.e., at ri = Xl . Thus.

U = _ GMm

,

As the mass m approaches M the potential energy then becomes negative. The work needed to move the mass from a distance ri to an infinite distance r1 ="1) is W =GMm

"

ri

GMm

=--R+h

GMm

=

= _

G~m[l

+

~rl

Now we use the binomial theorem to expand the expression in brackets.

(l~*rl =1-*+(*)\... However, h «R;

Concepts: The negative potential energy indicates that an amount of energy W YO i.~ required to remove the mass m from the influence of mass M. W,,(rj) is called the binding energy of m when it i.5 at rest on the surface of M. In general the binding energy of a system at any location i.5 the additional energy the system requires to reach an infinite distance away with final speed zero. (1) A particle whose total energy is positive has a velocity l' at a distance r such that E > O. The particle can then reach infinity with some kinetic energy. At r = r.cJ, 1 2 GMm E=-mv --2 •

we can neglect higher powers of

0;

(h/R), Thus

(l+~r >(1-~) =-:G~m(I_*) PEgmv.

=_GMm

R

+ GM mh

R'

v

00

=(~r2

(2) The escape speed i.5 defined as the speed needed for the mass m to leave the influence of the gravitational attraction of another mass M (such as planet, asteroid or star) so that m never returns. In order to reach infinity the total energy of the particle must be zero or positive, therefore minimum velocity will correspond to zero total energy. Therefore we have

www.puucho.com

Study Anurag Mishra with www.puucho.com

438

MECHANICS.II 1

2

GMm

2

'

R

-mv ---=0 or

t'r;:;

Binding energy ==IK +U1= GMm R

-R(2GMJ'"

= 1.12)( 104 m/s

(for earth)

escape velocity is independent of the mass of tlie body, however thrust required to accelerate a body until it reaches escape velocity depends on the mass of t.he body, heavier satellites need powerful boosters.

(3) Total mechanical energy E On solving for speed 1

we

=..!

m1l2

+ PE

2

have

=[~

(E-PE)]

"

It is the minimum energy is supplied to the particle in any (orrp. the panicle no longer remains bound to the eanh. Escape Velocity This is the minimum velocity that should be given to a point mass so that it can escape from the gravitational field in which it was present. If a point mass is placed on the surface of a fixed sphere then the binding energy can be calculated from the equation given below. 1 2

-mv

For the speed of the particle to be real E ~ PF, during the motion at all times. Fig. 4.28 shows a graph of potential

,

No turning point for unbound particle

g =9.Bm/s2

Unbound

Turning point

, ,=

o

Prohibited t values for bound particle

.. ,

PE E 1 ,

..

'

er

Hyperbola e > 1;

v ••".'

L'

Parabola e

PE=----

rnder

2,

l'

> ,./2GM/R

to

=.fiGMiR

v ~2GM --

Artificial satellites are launched from the earth for telecommunication. The path of these satellites are elliptical with the centre of earth at focus. The planets closer to the sun move in nearly circular orbits where as the farther planets move in elliptical orbits. Motion of satellites: Here we discuss the motion of artificial earth satellites which moves in circular orbits.

E > GMm R

R

--

R

E=GMm

v~~2~M

Hyperbolic

GMm

-- m,R, = m,R,

(ii)R=Rt+R2

b=a,h-e2

m,R

Rt =---ml + m2

Now from equation we have the total area of ellipse traced by the planet is given as L A=-T

R2 =

2m or

T = 2m A = 2m1tQb = 2mrrab L L mUir}

or

T =-

2m1U+~j

R

mlm2

W 3

System of Bodies

.

\", ..,

L=(It =

.\

.

'\

C.M..

+12)ro=(mlRl2+m2R;)w

(TnIRI2 + m2:~Rf)w

, ,--

m,

.. "

',:

'.-.-'

m,

:

=(

,

'

"

R,

"-_.-

.'

,

=-(ml +m2)R1w

"

.•~--~.~.~~ .. ",

R 3 Kepler's third law

",

:•

,

ex;

G(ml + m2)

Angular momentum of the system about centre of mass.

....- ..

,.- ."..

:m,:

T2

=>

Figure shows two particles moving due to mutually attractive gravitational force about centre of mass. Since there is no external force eM of system remains fixed and time period of revolution must be same .

.'

R

T=21t

GM,

,.,-

R3

" "1"2. TIme ta ken to camp 1ete one clrc e IS-.

T2=~a3

, ,

2

G(ml + m2)

w=

_

m[G:, (: ::)}a(l-dJ

(9) Gravitational

ml +m2

=----w ml +m2

,

or

m,R

m,m, +

Tnl

m2

)R'W

Moment of inertia of the system

R,

TnlTn2 M =-= -~--

Fig. 4.44

ml

Both bodies have comparable mass and both are moving in circular orbit about centre of mass as shown in diagram

Kinetic energy

=

mlm2 + m2

ml

)R'

reduced mass

.![w2 2

=.!( 2

www.puucho.com

=-

+ m2

1=(

m1m2 + Tn2

m)

)R2W2 =.!MR2oi 2

Study Anurag Mishra with www.puucho.com 4

C

MECHANICS.II

448

---------

(10) Multiple Gravitational Bodies Figure shows multiple point objects rotating' about common centre of mass, due to mutual attraction we wish to determine time period of revolution.

where

F

, Fig. 4.46

Resultant gravitational force on any mass is 2F cos 3D'" = mrro2 F.J2+F) :mroi 'R' is the distance of the masses irom the centroid of equilateral triangle Gm2.J2+ a2

2

-mn,l

Gm

(a.J2)2

Xm' __ ,_ cos300= a

mr())2

with

0

as a focus (v) if v = v~ escape from the G.E and path is a parabola (vi) if v = l'~ again the path is a hyperbola

mtw2 (Pseudo force) F

mrr02(pseudo force)

(iv) if va < v < v~ ellipse

Vo

escape

but

o ~.qq.

Fig. 4.49

= ~G~• orbital velocity and v

escape velocity -- In case (i) to (iv) total energy is negative. Hence these are closed orbits. In case (v) total energy is m1) assume that the mass distributions are spherically symmetric and therefore each can be considered as a particle in calculating the motion of their centres. Furthermore assume that the system is isolated, implying that we can choose an inertial frame of reference which translaus with the centre of mass. Show that the motion ofm, relative to m2 is the same as ifm2 were fixed and m) were replaced by a reduced mass ~:=

m)m2 m) +

,the force flZ being unchanged.

mz

Solution. Let r1 and rz be the position vectors ofm)and m2' respectively, relative to the centre of mass. The mutual

force between m, and mz is

Region that can receive communication signals

Unsh: 10-11 )(5.98>: 1024 )(470) _ 2

" _c

2a

x[4.23~ 107 A 470 kg communication satellite is released from a space shuttle at a height of 280 km above the surface of the earth. From [hi.s height a rocket engine boosts it into a geosynchronous orbit. What is the energy spent by the engine in transfer ring the orbit? Solution: The orbit period of a geosynchronous satellite is one day, i.e., the satellite travels once around the earth in the same time that the earth spins once on its axis. We can apply Kepler's third law to find the tadius of the orbit.

r2

.r

M< 103

www.puucho.com

Study Anurag Mishra with www.puucho.com

GRAVITATION

457

Example

35

Solution: From the equation mn,}:! == GmMjr2 for a body m to orbit around a fixed body M under gravitation, we

_

Distance between the centres of two stars is lOa. The masses of these stars are M and 16 M and their radii a and 2a, respectively. A body of mass m is fired straight from the surface of the larger star towards the smaller star. Wllat should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of G. M and a:

,'. ,

10a.----

, ,

•

r2(02 =GM. Then if M~, M m are the masses and reI rm are the radii of the earth and moon respectively, and the periods of revolution of the earth and moon satellites are the same, we have

,

", = M, " M Me

where Veand Vm are the volumes of the earth and moon respectively. It follows that the earth and moon have the same density.

Flg.4E.35

Solution: The distance (from the smaller planet) where the gravitational pulls of the two planets balance each other will be given by -GMm -G(l6M)m (loa-x)2

i.e., x = 2a So the body will reach the smaller planet due to the planet's gravitational field if it has sufficient energy to cross the point B (x = 2a), i.e., 1

2

but

Velocity of a point on the equator of a rotating spherical planet is V. The a.ngular velocity af the planet is such that the value af g at the equator is half of g at the pole. Detennine the escape velocity for a polar particle on the planet as a function

ofv. Solution: Value of from

>m(VB -Vs) 2a

GM

]

At the equator 2

Sa

VB=_[16GM+GM] 8a

GMm ---mg ,mV =-R' R 2a

or

20GM

=---Sa

!mv2 2

i.e.,

vmin

[g"

I f gatt h eequator ] lsvaueo

mg mV2 mg--=-2 R

or

> m[65GM _ 20GM] Sa Sa =

GM =gR2

giving

(l0a-2a)

65GM

so

at the pale can be determined

GMm mg =--

=---and

g

R'

V __ [16GM+ ., -

' m

M

_IlW2

m

-V:-=V

or

~=

Mill

m

,.B

"

find

2V' g=T

...(1)

Let u be the required escape velocity at the pole.Then total energy of the panicle is

~)5~M

E =.!mv2 2

_ GMm R

For the particle to escape its total energy must be at least equal to the minimum energy of a body at infinity, i.e., zero. Given that an earth satellite near the earth's surface takes about 90 min per revolution and that a moon satellite (of our moon, i.e., a space ship orbiting our moon) takes aLsoabout 90 min per revolution, what interesting statement can you derive about the moon's composition?

1 2 GMm -ml' ---=0 2 R or

www.puucho.com

v2

=

'2l1M = 2gR = 4V2 R

v

=

2V

Study Anurag Mishra with www.puucho.com

MECHANICS-II

458

Example Two small particles of mass m lie on either side of earth as shown in Fig. 4£.38. The particles are located along joining

the line

the centres of the earth and the moon. The difference

between the force exerted by the moon on the particles is called tidal force. (a) Determine

the magnitude

of the tidal

force assuming'" » R. (b) Now consider an object consisting of two uniform spheres of mass In and radius a. Determine the Roche limit, r, for which the spheres will fall apart due to cidal forces. (c) Using the result of part (a) show that a satellite of

densityps can exist near a planet of densicypo and radius R if Roche limit is

the moon when the ocean is closest compared to when it is farthest is much greater than the corresponding differential fo~ce exerted by the sun. This differential force is respo!,!sible for tides, hence the name tidal force. , (b) The astronomical objects are held together by force of gra~ty. If the tidal force on such an object is greater than the. gravitational forces holding the object togemer, the object will fly apart. The tidal force exerted by a planet of mass M on a satellite varies as M/ r 3 , so there is a minimum

distance rm at which a satellite can exist. This minimum distance is called Roche limit. The gravitational attraction force between the two spheres, each of mass m and radius a is Gmm/C2£t)2. Substituting R :: a and taking r:: rm, the Roche limit, we have

Earth

Moo"

F('':)---

R

2

= 2~

2

\'2R

4

=>

/2 19 19/2

e = cos

=>

Example

\!g~e= 5656 m/s

r

va

2../2

IGM

\ 4R,

\!g:e

==

4000 mls

Let the velocity of the satellite at perihelion of elliptical orbit be increased to L' p Now, semi.major axis, a = 3R~, distance of focus, e = Re. Using conservation of angular momentum, velocity at the aphelion is

( /-;::;;;-IGM

cosO=-x-=--=--

IGm

,2R,

Orbital velocity for the second orbit,

19GMm 12 R

RpcosB = 2",\

\iG~where m is the mass of

(a-e) (a+e) =up

=Up

[3R,-R,]

3R +R e e

Vp

or

V

, =-2

19 Applying energy conservation perihelion, we have 1 2 GMm 1 2 GMm -mv --=-nw --a 2P2R 2 4R , ,

_1(2/2} -19

~3_~

or

of

R

e v 22GM -v g =--=-

a

p

,

2R

A space vehicle of mass 300 kg is in a circular orbit of radius 2R£ about the earth. It is desired to transfer the vehicle to a circular orbit a/radius 4Rt•

2

Fig. 4E.43 (b)

Since,

Fig. 4E.43 (a)

(a) What is the minimum energy required for the transfer? (b) If the transfer is accomplished through an elliptical orbit as ShOWIl in thejigure, what initial and final velocity changes

are required?

mls

2 ar the earth's surface and R~ = 6400 km Take g = 10 (radilLS of earth).

"nd Initial:

l'p lip

=21'" = 6532m/s -Vj

= 6532-5656

= 876 mls Final:

t'2-L'"

www.puucho.com

=4000-3266=734

m/s

aphelion

and

Study Anurag Mishra with www.puucho.com

MECHANICS.II

462 Minimum energy required is 12 121212--mul E =-mu +-mv2 --mug p 2 2 2 2 =.!(300)[v~ -t'~+ v~ -v~]

2 = 2,4,. 109 J mu2

12121222

-mv

2

=-mvA

2

+-mw

2

(R -r)

where the second term on right hand side of the equation corresponds to work done by the gravitational force. From eqn. (4), we get

GMm

v=~~(2h+R)R-r2

T=Ji2'

[asg=~~]

I.e.,

vo=~=~

i.e..

Vo =~(9.8x 6.4)( 106) ",,8 kmls

or

or

and for escaping from close to the surface of earth, GMm 1 2

--=-mv R 2 ('

J:

v,=J~M =J~R

i.e.,

vt =

J2 x Vo

'"

[asg=~~]

-dr

=

~(2h+R)R-r2

=-~~

j' 0

IIdr

\R

sin-l[J(2h~R)RI =JIr

or

ie.

... (4)

t=~sin-l[2h:Rr2

,Let c' be time taken by the particle to reach the surface of the earth.

1.41)( 8 kmls t'=~

= 11.2 krnIs

So additional velocity to be imparted to the orbiting satellite for escaping

.

The time period of oscillation = 4(t +t')

"

=11.2-8=3.2kmjs

= ~{JRsin-'(JR:2h

+'/2h)}

Example. Example. Consider an imaginary tunnel dug along the diameter of the earth. A particle is dropped from a height h (h «R) above the earth's surface. Determine the tim"eperiod of motion. Solution: Considering earth as a solid sphere of mass M and radius R, the gravitational field intensity at point P is GM I = -r ...(1) R'

A body is projected from the earth's surface ar an angle of projection a with the horizontal. Show that the semi-major axis of the elliptical orbit of the body is independent of the angle of projection.

Force experienced by particle at paine P GMm mg F = --r = --y R' R

... (2)

B Fig.4E.44

The acceleration of the particle, GM a =--r

Flg.4E.45

... (3)

R' Comparing eqn. (3) with expression for S.H.M., we get 00= ~~~

=Jf

Solution: Let Vo be the velocity of projection and r be the maximum and minimum separations from earth. From conservation of angular momentum, we have m(vo cosa)R

Velocity of particle at A = ~2gh Let v be the velocity of the particle at P then

= mvr

From conservation of energy, we have 1 2GMm 1.2GMm -mvo --=-mv --2 R 2 r

www.puucho.com

... (1)

...(2)

Study Anurag Mishra with www.puucho.com

GRAVITATION

463

Now we substitute the expression for

l'

from eqn. 0) in

v=~

(2) to get

R

v~_[t'OC~SaRr ~2GMC~R) or

[asg=~~.J

From law of conservation of energy, we have GMm 1 ---+-mV R 2

v~r2-t'~R2cos2u=2,gRr(r_R)

The roots of this quadratic equation are rmaxand rmin. The semi-major axis a

=

r

max

2

GMm 1 2 =---+~mv r 2

On eliminating V,we get '2CM 2102, 2 2 -'2CMR+lI r =0 ( -,--I-'

!'

+ r. .

mm.

2

Sum of roots of quadratic equation "------~---~--2

Therefore

~M2g~~:,;]

R k -~--

or

r

2-k

gR' Example

2gR - v~ So we can see that the semi-major

axis is independent

of

angle of projection a.

47l--""""

Two identical spheres each of mass m and radius r are at rest with their centres 4r apart on a smooth horizontal surface. When they are released they move towards each other due to gravitational attraction. How long after will they collide?

A small satellite revolves round a heavy planet in a circular orbit. At a point on its orbit an impulse acts suddenly and instantaneously increases its kinetic energy k times without change in its direction of motion. Show that in irs subsequent motion the ratio 0jits maximum and minimum distance from planet is (_k_

2-k

Solution: From conservation of energy, when the separation between centres is x < 4r, we have 2

Solution: Let 1.'0 be the velocity of the satellite at A when it is in a circular orbit. From Newton's second law,

x

, {4'-XJ

or

_.!.dx

or

f

/Gm(4r-X)

=11=

2dt

or M R

=_Gm 4r

v =Gn-x-

assuming the mass of the satellite is

negligibly small compared to that of the planet.

2

2

2('!'mIl2J_Grn

4rx

\

J4c

d, ~-

2.JGm

.Jc

c = ,/Gm

f"'" VI(4r x- x) dx x=4r

f" \I~ ~

dx

2r

.Jc

P

= r;:-[r(;c+2)] ,,IGm

Fig.4E.46

GMm

~ (rr

,

DC

, ,.

, GM

vii

Example ... (1)

Due to impulse the velocity va is increased to v such that A' of the satellite from P be R. From conservation of angular momentum, we have mVR = mvr 1.'2

2),'-Ir' '=-Gm

mvii

--~--

"

+

= kv~. Let the maximum distance at

48~

Given a thin homogeneolLS dL~cof radius a and mass m A panicle of mass m2 is placed ar a distance lfrom the disc" on its axis of symmetry (Fig. 4E.48). Initially both are motionless in free space, but they ultimately collide, becalLSe of the gravitational attraction. Find the relative velocity at the time of collisions. ~sume a « I.

www.puucho.com

Study Anurag Mishra with www.puucho.com

MECHANICS.II

'164

v, m,[

2G

(~ _

ml + rn2

a

=

m

~)]V' I

So the relative velocity at impact is

m,

:(1+ ::,t =[2Gim,+m,lG-t)]"

vreJ :vl-t'2

Flg.4E.4B

Solution: The potential on the axis of symmetry at a distance I from the disc, 2Gm

V=_

(Ja2+12

l

a'

.

Vm:z,

Va =- 2Gmlm2 eJa2 +12_l)

a' The final potential energy VI' occurring just before impact, is found by setting I = 0 in the foregoing expression, since, I from symmetry considerations, the impact must occur at the centre of the disc. U

I

Example

-1)

So we find that the initial potential energy is or

Note that this relative velocity does not depend upon the magnitude of the individual masses, but only on their sum.

Delhi and Mexico are connected by a straight subway tunnel [see Fig. 4E.49(a)). A train travels between the wo cities powered only by the gravitational force of the earth. Calculate the maximum speed of the train and the time taken to travel from Delhi to Mexico. The distance between the two cities is 300 km and the radius of the earth is 6400 km. Neglect friction.

. .=

~

=_2Gmtm2

, ,, .'

a

~•• R

We assumed that a « I and therefore we can expand the square root in powers of a2 2 to obtain the approximation [

Hence the change in potential energy is Vo -Vf =Gmlm2(~ -~)

I ,I,+'2

m2v:z =Gm\m2

Zm1l'1

(2-;:;-1"I)

+ m2v2 :

or

V2

2

Solution: Define x, h, r as in Fig. 4E,49 (b) and assume the earth to be a stational)' homogeneous sphere of radius R. Taking the surface of the earth as reference level, the gravitational potential energy of the train at x is

2

2

+ GmM(r

2

m, m,

for

V2

a

2R3

)

2

R

r2

2

= h2

is the acceleration of gravity at the

+ (150-X)2

= (R2

=R2_300x+x2 gx(300-x)

v =-----

1

R

Forvm"",

or

www.puucho.com

=0,

v =~---,

or

into the energy

2

_R

2R' 2 g(R _r2)

where g =GM/R2 earth's surface. As

=--v,

R3

where m, M are the masses of the train and the earth respectively. Conservation of mechanical energy gives, as the train starts from rest at the earth's surface,

+~):Gmlm2(~ -~) rn2

GmMrdr:GmM(r2_R2), R

0

Substituting this expression equation, we obtain

..!..mtvr(l

Flg.4E.49

mv

Also, since [here are 09 external forces on the system, the total linear momentum is conserved. Therefore, mtVl

(b)

V=f'

From the principle of con~ervation of energy, we see that the total kinetic energy just before impact is equal to Vo -VJ, i.e.,

6400 km

(.)

11

= Gmlm2 U 0--

,

d

-[x(300-x)]=0 dx

300-2x=O

_1502) + (150 _X)2

Study Anurag Mishra with www.puucho.com 465

GRAVITATION t'

6xl03o ( = 4rrx1.3xl014

is maximum when x '" 150km. 1~9~.8~-xl~5~O~x~1-5-0--xl-0-0-0

=

l'

m,,~ \,

6400

= 1.5x 105m = 150km

185.601/s

=

(c) The nuclear density is given by mp

The time from Delhi to Mexico is

J:

OO

T =

Pnudear

d..,

j'

0

= n:

Example

iR

de

~g,.iC(l-f)

3 '

4rrRoi3

where mp is the mass of a proton and is approximately equal to the mass mH of a hydrogen atom. This can be estimated as follows:

dx

" fi ,!x(300-x)

=

'"

c'

~ 1:00 rR

2 x 10-3

. min.

mp '" mH =

{!f = 42. 3 min.

=

\g

With

2 x 6,02 x 1023

1.7 x 10-27 kg

Ro '" 1.5 x 10-15 m,

We obtain

50:---

17

3

Pnud.'ar '" 1.2x 10

(aJ A spherical object rotates with angular frequency (Ill! the only force preventing centrifugal disintegration of the object is gravity, what i.s the minimum deru;itythe object must have? (b) If the mass of the pulsar is about 1 solar mass (_ 2 x 1030 kg or - 3 x 105 M rarth)' what is the maximum (c) In/act the density is closer to that of nuclear matter. What then is the radius ? Solution: (a) Consider the limiting case that the pulsar is just about to disintegrate. Then the centripetal force on a test body at the equator of the pulsar is just smaller than the gravitational force: 2

= mReo2 < GmM , ~ R2

r

or

M ell -,'-, R G

where m and M are the masses of the test body and the object respectively, R is the radius of the pulsar, II is the speed of the test body, and G is the gravitational constant. Hence the minimum density of the pulsar is M 3CJl

r~--'-4 rrR 3

4rrG

3 3{2It x 30)2

z,

(b) As

30

R,

6x 10 -----

( 4rrx 1.2x 1017

mr2

where m and M are the masses of Mariner 4 and the sun 2 respectively, G is the gravitational constant, and 11 = r 0 is a constant. At the perihelion and aphelion of the elliptical orbit, r = 0, r = RE and r = RM respectively. Then . £=~GmM+mh2 RM 2R,~1 ---

4:tR' 3M

"

1

mh2

GmM

E ~-----+--, 2 2 r 2r

4rrx 6.7 x 10-11 1.3x 1014 kg/m3,

4JtPmin

",17km.

Solution: As the gravitational force on Mariner 4, which is a central force, is conservative, we have

-GmM

(

J"

Mariner 4 was designed to travel from earth to Mars in an elliptical orbit with its perihelion at earth and its aphelion at Mars. Assume that tile orbits of earth and Mars are circular with radii RE and RM respectively. Neglect the gravitational effects of the planets on Mariner 4. (a) With what velocity, relative to earth, does Mariner 4 have to leave earth, and in what direcrion ? (b) How long will it take to reach Mars? (c) With what velocity, relative to Mars, will it reach the orbit of Mars?

3M

$

•

If P = PnucleaT' the pulsar would have a radius

--32':Pmin'

R

kg/m

.Exp~pJe

possible radius of the pulsar?

mll

JI3

Rf

giving

www.puucho.com

mh2

+--

o '

2Ri

h = /2CMRMRE \ RM + Rf

Study Anurag Mishra with www.puucho.com

466

MECHANICS.II

At the perihelion we obtain its velocity relative to the sun is V

h 2GMRM ~ .1 RE Re(RM +RE)

,;-c""-";;---:

=-

Solution: Both the angular momentum and mechanical energy of the rocket are conserved under the action of gravity, a central force. Considering the initial state and the final state when the rocket achieves maximum height, we have

Suppose Mariner 4 is launched in a direction parallel to the earth's revolution around the sun. The velocity relative to the earth with which Mariner 4 is to leave the earth is [hen r =v-v£

V

r-C;'2f]7'M"R7'M-- JGM =

Re(RM +RE)

-

R ' E

RM(RM

JGM

+RE)

-

R ' M

Applying Kepler's third law we have for the period T of revolution of Mariner 4 around the sun 2

T

=r1(RE

~RM

r = ~ = ..!.(RE + RM 2 2 2R£

)"

A long-range rocket is fired from the surface of the earth ~ (radius R) with velocity u = (up vo} Neglecting air friction

and the rotation of the earth (but using the exact gravitational field), obtain an equation to determine the maximum height H achieved by the trajectory. Solve it to lowest order in (WR) and verify that it gives a familiar result ~

1 2

( -- R )2 v,2 __ GMm R+H R+H

,

1 , ,

-m(v 2'

1(

GMm ""-m +ve)--R2

H)

2H)u, ,--GMm( 1-1-R

R

R

and hence

YRi3,

where T£ = period of revolution of the earth = 1 year. Hence the time taken for Mariner 4 to reach Mars in years is

for the case that v is vertical.

1 2 2 GMm -m(ve+v,)---=-m 2 R

which gives the maximum heighr H. Considering only terms of first order in H/R, we have

M

2GMRE =

2

where the prime refers to the final state at which the radial component of its velocity vanishes, m and M are the masses of the rocket and the eanh respectively. Combining the above two equations, we obtain

where VE is the velocity of revolution of the eanh. Similarly at the aphelion the velocity, relative to Mars,which Mariner 4 must have is v', =v'-v

mRl'e = meR + H)ve , GMm 1 ,2 GMm )---=-mpe --r R 2 R +H'

1 2 -m(ve+v 2

xample A meteorite of mass 1.6x 103 kg moves abour the earth in a circular orbit at an altitude of 4.2 x 106 m above the surface. 1t suddenly makes a head.on collision with another meteorite that is much lighter, and loses 2.0% of its kinetic energy without changing its direction of motion or its total mass. (a) Describe the shape of the meteorite's orbit after the collision. (b) Find the meteorite's distance of closest approach to the earth after the collision. Solution: The laws of conservation of mechanical energy and conservation of angular momentum apply to the motion of the heavy meteorite after its collision. (a) For the initial circular motion, E < 0, so after the collision we still have E < 0. After it loses 2.0% of its kinetic energy, the heavy meteorite will move in an elliptic orbit. mv2 GmM (b) From

--~--,-, , ,

we obtain the meteorite's kinetic energy before collision: 1 -mv 2

2

GmM =-_

2r 2

Flg.4E.52

mgR ~-2r

www.puucho.com

Study Anurag Mishra with www.puucho.com

467

GRAVITATION III

= =:

x

9.8 x 103 X 64002 2(6400 + 4200)

1.89 x 107

III

GmlW

,

=-mll =

2

-3.78

X

107 III joule.

During the collision, the heavy meteorite's potential energy remains constant, while its kinetic energy is suddenly reduced to 1.89 x 107 III x 98% = 1.85 x 107 IIIjoule.

Hence the total mechanical energy of the meteorite after the collision is E = (1.85-3.78) x 107 III = -1. 93)( 107 From

E

III joule.

away from the sun with just sufficient velocity (0 escapefrom the sun's gravitationaljield. It is timed so that it will intercept Jupiter's orbit a di.~tance b behind Jupiter, interact with Jupiter's gravitational field and be deflected by 90", i.e., its velocity after the collision is tangetltial to Jupiter's orbit. How much energy did the satellite gain in the collision? Ignore the sun's gravitational field during the collision and assume that the duration of the collision is small compared with Jupiter's period. Solution: Let r represent the distance from Jupiter to the sun. 1'i the velocity of the satellite with respect to the sun at the time it intercepts Jupiter's orbit a distance b behind it and before any interaction with it, and m and M J the masses of the satellite and the sun respectively. As the satellite just escapes the sun's gravitational field, we have

= _-_G_mM_ = _-_mR_'_g_ 2n

54~

A satellite is launched from the eartll on a radial trajectory

joule,

where III is the mass of the meteorite in kg.. ~h.e potential energy of the meteorite before collision is ---

Example

ml'T

we obtain the major axis of the ellipse as R2g 2n=---~-

giving

Vi=~

1.93)(107 (6400)(

103)2 x

GmMJ

--=--2 ,

2n

2x 4.01 x 1014 x 3.33x 105 V 7.78x 1011 4 =1.8Sxl0 mjs=18.Skmjs =

9.8

7

1.93>< 10 = 208>< 107 m

= 2.08)( 104 km

As after the collision, the velocity of the heavy meteorite is still perpendicular to the radius vector from centre of the earth, the meteorite is at the apogee of the elliptic orbit. Then the distance of the apogee from the centre of the earth is 6400 + 4200 = 10600 kIn and the distance of the perigee from the centre of the eanh is r min = 20800 -10600 = 10200 km. Thus the meteorite's distance of closest approach to the earth after the collision is 10200 - 6400 = 3800 km. From the above calculations, we see that whatever the mass of the meteorite, the answer is the same as long as the conditions remain unchanged.

where we have used M. = 3.33)< 105M, (M, is the earth's mass), GM. = gR Z (R is the radius of the earth) =4.01x1014 m3/s2,r=7.78xl0Ilm. The velocity v J of Jupiter with respect to the sun is given by

v;

GM

,

,2

VJ

= -,-

s

-=--, I.e.,

JGM

h

v. = ..•

:=

13.1 kmls.

When the satellite just enters the gravitational field of Jupiter, its velocity in the Jupiter frame is

or

www.puucho.com

Vr

13.12

=

.,has2)(

=

22.67 kmls.

Study Anurag Mishra with www.puucho.com

468

MECHANICS-II

If b does not change during the encounter, conservation of the angular momentum of the satellite in the Jupiter frame orbit shows that this is also the speed of the satellite in the Jupiter frame when it leaves the gravitational field of Jupiter. After the encounter, the satellite SUN leaves the gravitational field of Flg.4E.54 Jupiter with a velocity in the sun's frame tangential to Jupiter's orbit. Thus the speed of the satellite with respect to the sun is

" j~~i~~;';-' If a planet

~.

~vr +V}

VI

2267 + 13.1 = 35.77 km/s The energy gained by unit mass of the satellite in the collision is therefore =:

(35.77)2 _(1&5)2 2

=

Solution. If the mass of sun is M and radius of the planet's orbit is r, then as 1.'0 == .J(GMjr), T == 2n-r == 2nr

from

4n2r3

Now if the planet [when stopped in the orbit] has velocity v when it is at a distance x from the sun, by conservation of mechanical energy, 2

earth with a velocity sufficient to carry it co infinity. Calculate the time taken by it to reach height h. Solution: If at a distance r from the centre of the earth the body has velocity v, by conservation of mechanical energy,

+(_ GMm) r

2

+(_ GMm) ~ R

==v~ +

~M[7-1]

==..!.mv2

or

1.'2

But as

u,=.j2gR

and

...(1)

GM

..!.m1.'2

the surface of the

r GM

T2 = --

i.e.,

+(_

GMm) = 0- GMm

x

(-~r=~M[r:xJ

or

A body is projected vertically upwards

J

1.'0

468.6x 106 J/kg.

t::Xc:lIT\Rle

..!.mv2 2

was suddenly stopped in its orbit supposed to be circular show that it would/all onto the sun in a time (nis) times the period 0/ the planet's revolution.

J: 1027 kg, and the distance between Jupiter and Ganymede is 1.071 x 109 m. Be sure to include the gravitational effect due to Jupiter, but you may ignore the motions

Ganymede

Jupiter

t'min =

2330 mls ]

36. A planet has mean density p. Show that the period. of a minimum-orbit satellite about the planet is independent of the mass M and radius r of the planet and that it depends only on 37. A quite large proportion of all stars are members of double star systems. Consider such a system in which two stars of roughly comparable mass orbit about their common centre of mass with period •. Consider a pair of such stars separated by a distance R and having total mass M. Show that Kepler's third law takes the remarkably simple form

1/.Jr.

[Aos. 15.6 kmls ]

earth satellite of mass m from a circular orbit of radius 2RE to one of radius 3R£. [Ans. 2GMm]

12R, 32. 1\"'0 hypothetical planets of masses ml and n12 and radii r} and r2' respectively, are nearly at rest when

they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. When their centre-to-centre separation is d. find expressions for the speed of each planet and their relative velocity. [Hint: Both energy and momentum are consented] (a) VI = m2(2G/dj-'2(ml + m2rl.'2,

GM

7 = 41t2'

Fig. 4.30

31. Derive an expression for the work required to move an

(b)

[Ans.

R3

of Jupiter and Ganymede as they revolve about their centre of mass (Fig. 4.30)

(Ans.

33. Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet. 34. A particle of mass m is located inside a uniform solid sphere of radius R and mass M, at a distance r from its centre. (a) Show that the gravitational potential energy of the system is U = (GmM/2R3)r2 - 3GmM/2R. (b) Write an expression for the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its centre. 35. Small, glassy button like objects called tektites are found strewn over areas in various locations over the earth. One view of the origin of tektites is that they were ejected from the moon by the impact of meteorites on the lunar surface. At what minimum speed must a body be ejected from the lunar surface if it is to reach the earth ? [Hint: A minimumspeed tektite must have negligiblespeed when it passes through a certain point on its moon-earth trajeetory.]

38. If an object has just enough energy to escape from the earth, it will not escape from the solar system because of the attraction of the sun. Use equarion for escape velocity lit = ~GM £ /R£ with M s replacing M £ and the distance to the sun TS replacing RE to calculate the speed VeS needed to escape from [he sun's gravitational field for an object at the surface of the earth. Neglect the attraction of the earth. Show that if v. is the speed needed to escape from the earth, neglecting the sun. then the speed of an object at the earth's surface needed to escape from the solar system "b'glVen y ve,solar = lie"dll IS + Vts, an ca cu ate lIe,rolar . V.s == 42.2km/s; lie, ,ola, == 43.7 km/s ] 39. Show that the escape speed from a planet is related to the speed of a circular orbit just above the surface of [he planet by v. = J2v;, where v{ is the speed of an object in the circular orbit. [Ans.

t'2=m1(2G/d)]

www.puucho.com

Study Anurag Mishra with www.puucho.com

GRAVITATION

477

40. A s g2' (3) toward m; (e) towards m (d) 8 = 0; (e) 0) m /m2 ;- rf /rJ, j;

45.

]

The assumption of uniform mass density is rather unrealistic. For most galaxies, the mass density increases greatly toward the centre of the galaxy. Using a surface mass de~sity of the form nCr) '" Clr,

the 'work is given

GMm[.!.-.!.)l Ij

(2) gl

Fig. 4.44

J

x

=g2'

j

(3) 8 =-0]

48. A non-unifoml stick of length L lies on the x-axis with one end at the origin. hs mass density J. (mass per unit length) varies as A = ex, where C is a constant. (Thus, an element of the stick has mass dm = I.dx.) (a) What is the total mass of the stick ? (b) Find the gravitational field due to the stick at a point Xo >L. [Ans. (a)

Ce; 2

www.puucho.com

(b) 2G~ L-

[In[2LJ-[-L-)i] x(J-L

x(l-L

I

Study Anurag Mishra with www.puucho.com

MECHANICS.II

478 49. A uniform rod of mass M and length L lies along the x-axis with its centre at the origin. Consider an element of length dx at a distance x from the origin. (a) Show that this element produces a gravitational field at a point Xo on the x-axis (xo dgx =

Gm L(xo

-xl

>.! L) given 2

dx

(GMl(~+ M2l; a 2

4.84

y

.'

(c) 0]

1.21a2'

51. A sphere of radius R has its

y

centre at the origin. It has a uniform mass density Po. except that there is a spherical , In , It , 0f ra d'IUS r =lR cavny 2 centered at x

,

,-----

(b) GmMl.

9

A uniform sphere of mass M is located near a thin, . ,uniform rod of mass m and length L as in Fig. 4.52. Find the gravitational force of attraction exerted by the sphere on the rod.

by

(b) Integrate this result over the length of the rod to find the total gravitational field at the point Xo due to the rod. ee) What is the force on an object of mass rna at xo? (b) Show that for Xo »L, the field is approximately equal to that of a point mass M. concentric uniform 50. l\vo spherical shells have masses M 1 and M 2 and radii a and 2a as in Fig. 4.50. What is the magnitude of the gravitational force on a point mass m located (a) a distance 3a from the centre of the shells? (b) a Fig. 4.50 distance 1.9 a from the centre of the shells? (c) a distance 0.9 a from the centre of the shells ? The inner spherical shell is shifted such that its centre is now at x = Q.Sa.The points 3a, 1.9a and 0.9a lie along the same radial line from the centre of the larger spherical shell. (a) What is the force on m at x = 3a? (b) What is the force on m at x = 1.9a? (c) What is the force on m at x = 0.9a ? [Ans. (a)

52.

,

=.!. R

as in Fig. 2 4.51. (a) Find the gravitational Fig. 4.51 field at points on the x-axis for Ixl > R. [Hint: The cavity may be thought of as a sphere of mass 4 , m = 3" Itr Po plus a sphere of mass -m.] (b) Show that the gravitational field inside the cavity is uniform, and find its magnitude and direction.

[l\ns.G(4ITP;R3)[> ( \ 8 x--R 2

r]l

o

FIg. 4.52 [Ans. Gm2Aa(a+

53.

2)]]

Consider two identical uniform rods of length Land mass m lying along the same line and having their closest points separated by a distance d (Fig. 4.53). L

L

m

m

Fig. 4.53

Show that the mutual gravitational these rods has a magnitude F =

Gm'l{

. L2

54.

(L+d)') d(2L+d)

A uniform rod of mass M is in the shape of a semicircle of radius R (Fig. 4.54). Calculate the force on a point mass m placed at the centre of the semicircle. . GmM'h [An2 S. --,-

stralg t up

force between

'h' In

M

Fig. 4.54

t e picture 1

'"

55. A uniform solid sphere of m, mass ml and radius Rj is inside and concentric with a spherical shell of mass m2 and radius R2 (Fig. 4.55). Find the gravitational force exerted by the spheres on a particle of mass m located at (a) r = a, (b) Fig. 4.55 r = b, and (c) r = c, where T is measured from the centre of the spheres. [Ans. (a) Gm]~a; (b) Gmim; (e) G(ml \m2)m] R]

b

c

56. A sphere of mass M and radius R has a non-uniform density that varies with T, the distance from its centre, according to the expression p = Ar, for 0'::; r ::;; R.

www.puucho.com

Study Anurag Mishra with www.puucho.com

r.:G::RA::V1;;T:;;AT;;IO~N;-------------------------------;:==:;4;7;;;9 (a) What is the constant A in terms of M and R ? (b) Determine

an expression

for the force exerted on a

particle of mass m placed outside the sphere. (e) Determine an expression for the force exerted on the particle if it is inside the sphere. M

. [Ans. (a) A = -4;

""

(e) 57.

GMm F = -y--

,

4"G[1-PaR 2 2 --PayR 7

Po =-3 [Ans. (a)

toward the centre J

2

aCr) == ~

12

3 +-y 3 2R 4] .

16

;tGpor-GIt"f,2]

3

toward the centre;

An interstellar explorer discovers a remarkable planet made entirely of a uniform incompressible fluid. The radius of the planet is R, and the acceleration of gravity at its surface is g. What is the pressure at the centre of the planet? [Ans. P

58.

GMmr2 F '" --.R

(b)

Show that the pressure at the centre of the earth can be written

59. 1\vo identical spherical hollows are made in a lead sphere of radius R. The hollows have a radius R/2. They touch the outside surface of the sphere and its centre as in Fig. 4.59.

1

= - pgR ]

2

Even though the material of which the earth is composed is "incompressible" under the conditions usually econountered on its surface, the pressures within the earth are so large that density varies substantially with depth. As an improved. approximation. assume a linear variation of the form. per) = Po -yr,

where Po is the density of the matter at the centre of the earth, y is the rate of change of density, and p(r) is the density at a distance r from the centre of the earth. (a) Express the acceleration of gravity oCr) in terms of Po, 1, r, and the universal gravitational constant G. (b)

m

,

FIg. 4.59

The mass of the lead sphere before hollowing was M. (a) Find the force of attraction of a small sphere of mass m to the lead sphere at position shown in Fig. 4.59. (b) What is the attractive force if m is located right at the surface of the lead sphere? [Ans. (a) -(GMm/d2)[

(b)

_(d3/4)/{d2

-O.821(GMmjR2)1]

www.puucho.com

+ R2/4)32]

I

Study Anurag Mishra with www.puucho.com

MECHANICS-II

ePevel

CD---_._--------

Only One Alternative is Correct

1. The density of the core of a planer is PI and that of the outer shell is p 2' The radii of the core and that of the planet are Rand 2R respectively. The acceleration due to gravity at the surface of the planet is P, same as at a depth R. The ratio of density PI I P2 will be : (a) 7/3 (b) 5/3 (e) 8/3 (d) 1/3 2. An exploratory rocket of mass m is in orbit about the sun at a radius of RES /10 (one tenth of the radius of the earth's orbit about the sun). To exit this orbit, it fires its engine over a short period oftime. This quickly doubles the velocity of the rocket while halving its mass (due to fuel consumption). Immediately after the bum, what is the kinetic energy of the rocket ? Take mass of sun as Ms : (a) GM sm

(b) lOGM

2RES (e) 20GM sm RES

snl

RES

Cd) SCM sm RES

3. A shell is fired vertically from the earth with speed tJe$cIN. where N is some number greater than one and V",C is escape speed for the earth. Neglecting the rotation of the earth and air resistance. the maximum altitude attained by the shell will be (RE is radius of the earth) : (a)

-..!!:.L 2 N

(e)

-1

(b) R,

N' (d) N2Rf.

4. A smooth tunnel is dug along the radius of the earth that ends at the centre. A ball is released from the surface of earth along the tunnel if the coefficient of restitution is 0.2 between the surface and ball then the distance travelled by the ball before second collision at the centre (a)

!'R

(b):!:R

2R

(d)~R

5

(e)

5

5 2 5. A planet of mass m is in an elliptical orbit about the sun (m «M .un) with an orbit period T. If A be the area of orbit. then its angular momentum would be ; (a) mA /

zr

(e) 2mAT

(b) mAT

(d) 2mA / T

6. A satellite of mass ms revolving in a circular orbit of radius rs round the earth of mass M has a tOtal energy E. Then its angular momentum will be : (a) (ZEmsr/)l12

(b) (2Emsrs)

(c) (2Emsrs2)

(d) (2Emsrs )112

7. A satellite is launched in the equatorial plane in such a way that it can transmit signals upto 60' latitude on the earth. The angular velocity of the satellite is : (a) ~GM

2R3 (e) ~GM 3

4R

(b) ~GM 8R3 (d) \

13J3GM 3 8R

8. A planet of small mass m moves around the sun of mass M along an elliptical orbit such that its minimum and maximum distance from sun are rand R respectively. Its period of revolution will be :

N2 -1

www.puucho.com

Study Anurag Mishra with www.puucho.com 481

GRAVITATION (r+R)3

(a)

2"',IO~",--

(e)

1"1:

6GM

~(r+R)3

b)

(

2)"[

Cd) 2,

"\

2GM

icr+R)3

(a)

/Cr+R)3

(e)

1

3GM GM

(a)

8r2

10. A satellite revolving around the planet in a circular orbit is to be raised to a bigger circular orbit. The required energy can be supplied to the satellite for achieving the bigger orbit : (a) in one stage (b) in minimum two stages ee) in minimum four stages Cd) in minimum three stages 11. Three identical stars, each of mass M, form an equilateral triangle (stars are positioned at the corners) that rotates around the centre of the triangle. The system is isolated and edge length of the triangle is 1. The amount of work done, that is required to dismantle the svstem is :

z

2.

GM (a) --

(b) --3GM

2L

2L

41.

I.

(b) 2J2E (d) .,/2 E

(a) 4E (c) 2£

13. l'w"omasses Mland M z at an infinite distance apart are initially at rest. They start interacting gravitationally. Find their velocity of approach when they are separated by a distance s : /G(M1 +Mz)

(b) ~GM1Mz .