Fundamentals of Acoustic Waves and Applications (Synthesis Lectures on Wave Phenomena in the Physical Sciences) [2024 ed.] 3031481992, 9783031481994

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Synthesis Lectures on Wave Phenomena in the Physical Sciences

Sanichiro Yoshida

Fundamentals of Acoustic Waves and Applications

Synthesis Lectures on Wave Phenomena in the Physical Sciences Series Editor Sanichiro Yoshida, Department of Chemistry and Physics, Southeastern Louisiana University, Hammond, LA, USA

The aim of this series is to discuss the science of various waves. An emphasis is laid on grasping the big picture of each subject without dealing formalism, and yet understanding the practical aspects of the subject. To this end, mathematical formulations are simplified as much as possible and applications to cutting edge research are included.

Sanichiro Yoshida

Fundamentals of Acoustic Waves and Applications

Sanichiro Yoshida Department of Chemistry and Physics Southeastern Louisiana University Hammond, LA, USA

ISSN 2690-2346 ISSN 2690-2354 (electronic) Synthesis Lectures on Wave Phenomena in the Physical Sciences ISBN 978-3-031-48199-4 ISBN 978-3-031-48200-7 (eBook) https://doi.org/10.1007/978-3-031-48200-7 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

This book is dedicated to the memory of my late father who introduced me to the joy of engineering, and my mother, Sonoko Yoshida, and my wife, Yuko Yoshida, for filling my life with love, joy, and happiness.

Preface

This book discusses the fundamentals of acoustic wave phenomena. I use the word “discuss” because I intend to describe each topic through discussions from various angles rather than to “provide” the information about them. I believe we can deepen our understanding of natural sciences by discussing subjects. I hope the readers apply the concept gained from this book to other fields of science and engineering. The target audience is undergraduate students majoring in physics and engineering. I wrote this book as part of Synthesis Lectures on Wave Phenomena in the Physical Sciences, which aims to describe various waves in a short book. It is impossible to cover many topics with this short-book concept. I selected a few and dug them deeper, exploring their physical meanings. Mathematical operations play significant roles in this process. Often we can sense physical phenomena intuitively by digesting the associated mathematics. I tried my best to break down mathematical operations into small pieces. This sometimes makes equations long. Please bear with me while I explain a long equation step by step. Acoustic technologies have numerous applications in a variety of engineering and scientific fields. Recent advancements in hardware and software technologies facilitate the use of devices and analyses of the outcomes. On the other hand, this convenience tends to hide the operation principles of hardware and software modules. This tendency sometimes leads to the inefficient use of a hardware device and the misinterpretation of data. A proper understanding of the science behind the scene is essential. Upon writing this book, I always bore this factor in mind. The first two chapters discuss the basics of waves, where Chap. 1 focuses on the mathematical aspect of it and Chap. 2 on the physical. Chapters 3 and 4 discuss acoustic waves propagating in air and solids, with audible frequency in air and ultrasounds in solids in mind. Chapter 5 describes the operation principles of typical acoustic transducers. Finally, I would like to express appreciation to people. I owe much to my high school and college teachers because my understanding originates from what they taught me decades ago. I am grateful to my parents for allowing me to receive such an excellent

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Preface

education. I thank my wife, Yuko Yoshida, for her continuous support. Often I spent much time on weekends for this book project. Hammond, LA, USA August 2023

Sanichiro Yoshida

Contents

1 General Discussions of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Wave as a Movement of an Oscillatory Pattern . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Terms to Represent a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Longitudinal and Transverse Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Mathematics of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Wave Function as a Periodic Function of Time and Space . . . . . . . 1.2.2 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 2 4 6 6 9 12 21

2 Wave Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Equation of Motion of Point Mass-Spring System . . . . . . . . . . . . . . 2.1.3 Oscillation to Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.4 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Acoustic Wave Equations and Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Acoustic Wave in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Longitudinal and Transverse Vibrations in Solid . . . . . . . . . . . . . . . 2.2.3 Decaying Acoustic Wave Equations and Solutions . . . . . . . . . . . . . 2.2.4 Nonlinear Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Wave as a Flow of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Acoustic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Intensity of Acoustic Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Acoustic Wave Reflection and Transmission . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Laws of Reflection and Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Behavior of Acoustic Waves Near Boundary . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 23 24 26 29 29 34 35 39 43 49 50 50 56 58 58 60 63

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Contents

3 Propagation of Acoustic Waves in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Dynamics of Air Particles and Acoustic Wave Equation in Air . . . . . . . . . 3.1.1 Particle Motions in Longitudinal Waves . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Wave-Generating Mechanism in Air . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Decaying Pressure Waves in Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Sound We Hear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Our Ear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Sounds of Speech and Music . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Transmission of Audible Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65 65 65 66 71 72 72 74 78 84

4 Propagation of Acoustic Waves in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Acoustic Wave as Solutions to the Equation of Motion . . . . . . . . . . . . . . . . 4.1.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Propagation of Acoustic Waves in Solid Under Various Conditions . . . . . 4.2.1 Plane Waves in an Elastic Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Plane Waves in a Medium with Free Surface . . . . . . . . . . . . . . . . . . 4.2.3 Surface Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Waves Propagating in Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 87 91 91 96 108 111 120

5 Electrical-Mechanical Transduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 General Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Reciprocal Transduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Electrostatic Transducer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Electrical Governing Equation of Electrostatic Transducer . . . . . . 5.2.3 Mechanical Governing Equation of Electrostatic Transducer . . . . . 5.2.4 Physics Behind Electrostatic Transducer’s Operation . . . . . . . . . . . 5.3 Anti-reciprocal Transducers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Moving-Coil Transducer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

123 123 124 125 126 127 128 132 132 139

Appendix A: Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

141

Appendix B: Gradient Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

143

Appendix C: Orthogonality of Sine and Cosine Functions . . . . . . . . . . . . . . . . . . .

149

Appendix D: Notes on Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

151

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

155

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1

General Discussions of Waves

This introductory chapter discusses wave characteristics and related terminologies in general. We first view waves as a motion of a spatial pattern (a snapshot) of an entity on the time axis and as a temporal oscillation on the spatial axis. This spatiotemporal behavior leads to the concept of phase velocity, which characterizes the medium that the wave propagates through. In the second half of the chapter, we discuss some mathematical aspects of wave dynamics. We derive wave equations from the spatiotemporal characteristics and discuss their solutions in general.

1.1

Wave as a Movement of an Oscillatory Pattern

Consider a sample wave in Fig. 1.1. This figure illustrates a wave pattern spread along the x-axis at three different times, 0, 1, and 2.µs. The wavy pattern at each time step is a snapshot of the wave. With the elapse of time (shown by the arrows in Fig. 1.1), the pattern moves to the right. The little circle put at the leftmost trough indicates that each wave pattern moves rightward in the positive .x-direction with time. The solid line connecting the little circles shows that it has an angle to the time axis. Here the dashed line is drawn parallel to the time axis. The angle made by the solid and dashed lines clarifies the rightward movement of the circle with the passage of time. We say that the wave has a rightward velocity. This type of wave velocity is referred to as phase velocity for the reason clarified shortly in this chapter.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_1

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1 General Discussions of Waves

Fig. 1.1 Schematic illustration of a wave as a function of space. Each wave pattern illustrates the snapshot taken at 0, 1, and 2 .µs

1.1.1

Terms to Represent a Wave

Wavelength; spatial oscillatory periodicity Figure 1.2 illustrates this rightward motion of the wave pattern more explicitly. Notice that the pattern has a spatial periodicity. As the dashed box indicates, the periodicity in this example is 10 mm in length. The spatial period of a wave is referred to as wavelength. We can say that this wave has a wavelength of 10 mm. From the analysis of the motion of the periodicity and how long it takes in time, we can calculate the phase velocity. In this example, the periodicity moves 0.5 mm in 1.µs, and we find that the phase velocity is 500 m/s (0.5 mm .÷ 1 µs .= 500 m/s). Period of wave; temporal oscillatory periodicity Observation of Fig. 1.2 at the same location on the .x-axis reveals the temporal oscillation of the wave. For instance, see the signal at .x = 3.8 mm indicated by a vertical line in Fig. 1.2. The signal at this location is positive at .t = 0 .µs, approximately 0 at .t = 1 .µs, and negative at .t = 2 .µs. It is clear that the medium oscillates during this duration of 2 .µs. This oscillation is called particle oscillation and the corresponding velocity is called the particle velocity. Figure 1.3 illustrates the temporal oscillation (the particle oscillation) behavior of the same wave as Fig. 1.1 in the same form of three-dimensional presentation. Here, the oscillations at three representative locations of .x = 0, 1, and 2 mm are presented. You may notice the similarity in the oscillatory pattern between Figs. 1.1 and 1.3. In fact, as Fig. 1.4 explicitly shows, the oscillatory patterns as a function of space and time are identical to each other. The temporal periodicity observed in Fig. 1.4 is referred to as period.

1.1 Wave as a Movement of an Oscillatory Pattern

3

Fig. 1.2 Schematic illustration of a wave as a function of space emphasizing the motion of the spatial periodicity

Fig. 1.3 Schematic illustration of a wave as a function of time. Each wave pattern illustrates the particle oscillation at a fixed .x location

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1 General Discussions of Waves

Fig. 1.4 Schematic illustration of a wave as a function of time. Each wavy pattern is a snapshot of the wave at the corresponding time

1.1.2

Longitudinal and Transverse Waves

Longitudinal waves Figures 1.1, 1.2, 1.3 and 1.4 illustrate the oscillatory behavior of a wave as a function of space and time. However, they do not indicate the direction of particle oscillations. It can be parallel or perpendicular to the direction of the wave propagation. We call a wave a longitudinal wave when the particles oscillate parallel to the direction of the wave propagation. A wave traveling through a sling is an example of a longitudinal wave. A sound wave traveling through air is also a longitudinal wave. (See Sects. 2.2.1 and 3.1.) Figure 1.5 illustrates the propagation of a sample longitudinal wave. The regions where the dots are dense (sparse) represent the portion of the medium where the density is high (low). The illustrations at each row present the pattern at a given time, where the time elapses from the top towards the bottom row with an increment of 0.25 .µs. Notice that the illustrations on the left column indicate that the dens-sparse pattern moves to the right, whereas those on the right column represent that it moves to the left. Since the direction of the wave motion is parallel to the particle motions, these waves are longitudinal.

1.1 Wave as a Movement of an Oscillatory Pattern

5

Fig. 1.5 Propagation of longitudinal waves. The wave moves parallel to the dens-sparse patterns of the particles’ motions

Transverse waves If the particle oscillation is perpendicular to the wave propagation, we call the wave a transverse wave. A wave formed by fans in a soccer stadium is a transverse wave. Figure 1.6 illustrates the propagation of a transverse wave in the same fashion as Fig. 1.5. The dots represent particles. The rows represent the progression in time, whereas the left and right columns illustrate the waves moving rightward and leftward, respectively. In this case, the oscillatory pattern does not exhibit a change in the density. Instead, it indicates a swinging motion of the particles in the perpendicular direction to the wave motion. A sound wave can travel through a solid as a longitudinal or transverse wave. (See Chap. 4.) An ocean wave is a mixture of longitudinal and transverse waves [1]. The particle oscillation (the motion of the water molecules) has longitudinal and transverse components to the propagation.

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Fig.1.6 Propagation of transverse waves. The waves move horizontally whereas the particles oscillate vertically

1.2

Mathematics of Waves

1.2.1

Wave Function as a Periodic Function of Time and Space

The fact that a wave has an identical pattern in its spatial and temporal characteristics allows us to put a wave function in the following form. .

f (θ ) = f (ωt ± kx)

(1.1)

Here, .t is the time coordinate variable, .x is the spatial coordinate variable, and θ ≡ ωt ± kx

.

(1.2)

is called the phase of the wave. For the physical meaning of .ω and .k, see the following paragraphs. Since . f (θ ) is a periodic function, we obtain the following relation. .

f (θ + Θ) = f (θ ± N Θ) = f (θ )

(1.3)

1.2

Mathematics of Waves

7

Here .Θ is the period of the function . f (θ ) and . N is an integer. Equation (1.3) literally indicates the fact that every time the phase changes by .±Θ the function takes the same value. We can always find the same change in .θ by fixing .t and varying .x or fixing .x and varying .t. This fact stipulates the above observation that a wave has identical temporal and spatial patterns. We can interpret the quantities .ω and .k as the temporal and spatial frequencies of the wave by making the following argument. Find the total derivative of .θ by differentiating (1.2). ∂θ ∂θ .dθ = (1.4) dt ± d x = ωdt ± kd x ∂t ∂x Consider the spatial periodicity using (1.4) in association with Fig. 1.2. Since the spatial wave patterns in this figure are the snapshots taken at the respective time, we can put .dt = 0 in (1.4). Under this condition, express (1.4) for one period. We can put the left-hand side as the phase period .dθ = Θ and the right-hand side as the wavelength .d x = λ. Thus, Θ = kλ

.

(1.5)

Since period .Θ is a constant for a given wave, (1.5) indicates that .k is proportional to the reciprocal of .λ. Here the wavelength .λ represents the periodicity in length (m in the SI unit). The reciprocal of .λ measures the number of wavelengths in the unit length, i.e., how frequently the wavelength appears as we move over the unit length on the spatial axis. Thus, it is called the spatial frequency of the wave. In the case of sinusoidal waves (as we will discuss in more detail in later sections), the periodicity in phase is.2π . In this case, the spatial frequency becomes as follows. 2π .k = (1.6) λ We can repeat the same type of argument on (1.4) in association with Fig. 1.4. In this case, .d x = 0 and we obtain the following equation as the temporal version of (1.5). Θ = ωτ

.

(1.7)

Here .τ is the period of the wave. Similarly to (1.6), we can define the temporal frequency .ω for a sinusoidal wave. ω=

.

2π τ

(1.8)

Repeating the same argument as for .k, we can say that .ω represents how frequently the period appears as we move along the time axis. To distinguish from the terms the spatial and temporal frequency in the unit of 1/m or 1/s, we call .k and .ω the angular spatial frequency and angular temporal frequency. They are in rad/m and rad/s, respectively.

8

1 General Discussions of Waves

Traveling wave and phase velocity As indicated by (1.3), the phase determines the value of a wave function. The motion of a wave is the trajectory of a constant phase value. In other words, the wave velocity we discussed above is the velocity of a phase value; that is why we call the velocity phase velocity. A surfer stays at a crest of an ocean wave where the phase makes the vertical position of the water molecule at the highest. So, we can say that the surfer moves at the phase velocity of the wave he is on. Here we consider the concept of phase velocity quantitatively. When we move along with a constant phase, the phase does not change over time. Mathematically, we can express this situation by saying .dθ/dt = 0. We can discuss it using (1.4) by differentiating the phase .θ with respect to time and setting it to zero. .

dθ dx =ω±k =0 dt dt

(1.9)

From (1.9), we obtain the following equation. .

ω dx =∓ k dt

(1.10)

Consider the meaning of .d x and .dt that appear on the right-hand side of (1.10). These are the changes in the coordinate variable .x and .t necessary to keep the phase .θ = ωt ± kx constant. In other words, if the time elapses by .dt, we need to change the location on the .x coordinate by .d x to keep the phase unchanged. From this viewpoint, we can interpret that .d x/dt is the velocity of the constant phase location on the . x-axis. Thus, we can say that the quantity expressed by the left-hand side of (1.10) represents the phase velocity of the wave .v p . ω dx (1.11) .v p = =∓ k dt Note that the negative (positive) sign on the right-hand side of this equation corresponds to the positive (negative) sign on the right-hand side of the phase expression (1.2). This means that . f (ωt + kx) corresponds to .ω = −d x/dt, i.e., a negative phase velocity for positive .d x and .dt, indicating that the wave travels in the negative . x direction, and . f (ωt − kx) represents a wave traveling in the positive .x direction. In other words, to keep the phase .θ = ωt − kx constant when the time elapses by .dt (i.e., .t increases), we need to increase . x by .d x = (ω/k)dt. A wave expressed in the form of (1.1) is referred to as a traveling wave. As clear from the above discussion, when the sign between the time and space terms of the phase is negative (. f (ωt − kx)), the wave travels in the positive .x-direction. If the sign is negative, the wave travels in the negative .x-direction. According to (1.10) and (1.11), we can distinguish the direction of the wave motion by the sign of the phase velocity .v p .

1.2

Mathematics of Waves

1.2.2

9

Wave Equations

Consider the property of the wave expression (1.1) further by differentiating the wave function . f with respect to time and space. .

∂2 f = ω2 f '' ∂t 2 ∂2 f = k 2 f '' . ∂x2

(1.12) (1.13)

Here . f '' denotes the second-order derivative of the function . f (t, x), e.g., if . f = cos(ωt − kx), . f '' = − cos(ωt − kx). Equating . f '' appearing in (1.12) and (1.13), we obtain the following wave equation. ( ω )2 ∂ 2 f ∂2 f = (1.14) . ∂t 2 k ∂x2 Note that .ω/k appearing on the right-hand side is the phase velocity (1.11). Expressing the unit of the quantity represented by . f with .[ f ], we find that the unit of the left-hand side of (1.14) is .[ f ]/s.2 and the unit of .∂ 2 f /∂ x 2 is .[ f ]/m.2 , indicating that .(ω/k) is in m/s. Indeed, it has the dimension of velocity. We can express the wave equation using (1.11) in (1.14). .

2 ∂2 f 2∂ f = v p ∂t 2 ∂x2

(1.15)

Extension to three dimensions It is clear that the differential equation (1.14) yields the wave solution . f (ωt ± kx). We call this equation a wave equation. As discussed in the preceding section, (1.14) yields a wave solution that travels along the .x-axis. Since its traveling direction is limited in one axis, this type of wave is known as a one-dimensional wave. What if the traveling direction is not in line with the .x-axis, e.g., it has some angle to the . x-axis? Even if the traveling behavior is physically the same, we cannot express the phase term in the form of .ωt − kx. With the passage of time, the change in the spatial coordinate account for the same phase cannot be expressed only by the change in .x. In this situation, we say that the traveling wave is three-dimensional, and call the spatial frequency .k the propagation constant. Here, the propagation constant has a directionality, i.e., we need to move along an axis that has an angle to the .x-axis to define the spatial frequency. In other words, the propagation constant .k is a vector referred to as the propagation vector. Figure 1.7 illustrates the situation where the same physical wave travels through space represented by two different coordinate systems. In Fig. 1.7a, the .x-axis is in line with the direction of the wave propagation, hence the propagation vector .k has one component. In Fig. 1.7b the direction of propagation has angles of .θx , .θ y , and .θz with the .x, . y and .z-axis, respectively. The cosines of these angles are called direction cosines [2]. The propagation

10

1 General Discussions of Waves (a)

(b)

Fig. 1.7 The same wave propagates through space represented by two different coordinate systems. a The propagation is in line with the .x-axis. b The propagation has angles of .θx , .θ y , and .θz with the . x, . y and .z-axis. . xˆ is the unit vector along the . x-axis. .lˆ is the unit vector along the wave’s propagation

vector’s .x, . y and .z-components are the vector’s magnitude multiplied by the corresponding directional cosine, e.g., .k x = k cos θx . The conversion from a one-dimensional to a three-dimensional wave expression leads to the following change in the phase term of wave function . f (ωt ± kx) defined in (1.1). .

.

f (θ ) = f (ωt ± k · r)

(1.16)

Here .r is the vector that represents the spatial coordinates; the three-dimensional version of x in (1.1). ˆ .r = x iˆ + y jˆ + z k (1.17)

Here .iˆ etc. are the unit vectors for the corresponding directions. The scalar product in the phase term of (1.16) represents the spatial phase change in the direction of propagation vector .k, i.e., along the propagation path of the wave. ( ) k · r = k x x + k y y + k z z = k cos θx x + cos θ y y + cos θz z

.

(1.18)

Here .k is the magnitude of .k, i.e., .k = |k|. Defining vector .lˆ as the unit vector in the direction of .k, we can express (2.89) in a compact form as follows. ( ) k · r = k lˆ · r = k l x x + l y y + l z z

.

(1.19)

ˆ Here .l x etc. are the components of .l. Comparison of (2.89) and (1.19) reveals that .lˆ has direction cosines as its components. lˆ = cos θx iˆ + cos θ y jˆ + cos θz kˆ

.

In Appendix A we prove that .lˆ is a unit vector.

(1.20)

1.2

Mathematics of Waves

11

Using (1.19) or (1.20), we can express wave function (1.16) f (ωt ± k lˆ · r) = f (ωt ± k(l x x + l y y + l z z)) = f (ωt ± k(cos θx x + cos θ y y + cos θz z)) (1.21) The extension to three dimensions requires one more procedural change regarding differentiation. Differentiating a function is the process to find the change in the value of the function associated with an infinitesimal change of independent variables. When the wave function has one spatial independent variable, like the above case of . f (t, x), the change in the value of function . f is definitely determined by the function’s dependence on .x, i.e., . f ' = d f /d x. However, if the function has three independent spatial variables like . f (t, x, y, z), the spatial derivatives become a vector quantity. We need to replace the differential operation .d f /d x with the gradient operation .∇ f [3]. In Cartesian coordinates, it is given as follows. ( ) ∂ ˆ ∂ ˆ ∂ ˆ ∂fˆ ∂f ˆ ∂f ˆ (1.22) .∇ f = i+ j+ k f = i+ j+ k ∂x ∂y ∂z ∂x ∂y ∂z .

Appendix B discusses some more description of the gradient operation. By repeating the gradient operation twice, we can derive the following expression as the three-dimensional version of the second-order spatial differentiation .∂ 2 /∂ x 2 . ( ) ( ) ∂ ˆ ∂ ˆ ∂ ˆ ∂ ˆ ∂ ˆ ∂ ˆ 2 .∇ f = ∇ · ∇ f = i+ j+ k · i+ j+ k f ∂x ∂y ∂z ∂x ∂y ∂z =

∂2 f ∂2 f ∂2 f + + ∂x2 ∂ y2 ∂z 2

(1.23)

The operator .∇ 2 is called the Laplace operator or Laplacian. Using (1.21) in (1.23) we find the following expression as the three-dimensional version of (1.13). 2 2 2 2 2 '' .∇ f = k (l x + l y + l x ) f = k 2 f '' (1.24) Here .lˆ = l x iˆ + l y jˆ + l z kˆ is the unit vector in the direction of the propagation vector, and we used that .(l x2 + l y2 + l x2 ) = 1 on the right-hand side of (1.24). From (1.12) and (1.24), we derive the following equation. .

( ω )2 ∂2 f = ∇2 f ∂t 2 k

(1.25)

Wave equation (1.25) is the three-dimensional version of wave equation (1.14). It should be noted that above we derived wave equations based on the temporal and spatial periodicities. As will be discussed in the following section, this type of wave equation yields solutions that continue unlimitedly spatially and temporally. It is because the periodic function we started with has such a property. In reality, waves decay due to various causes such as viscosity. We will discuss this topic in the next chapter by considering the physics behind wave dynamics.

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1 General Discussions of Waves

1.2.3

Wave Solutions

Consider solving the wave equations derived above. Since we derived the wave equations by considering a periodic function in the form of (1.1), the solution should take this form. As a simple example, we know that sine and cosine functions are periodic. Let’s take a moment and confirm that a simple cosine function in the following form satisfies the one-dimensional wave equation (1.14). . f (t, x) = A cos(at ± bx) (1.26) Here . A, .a, and .b are constant. Differentiating this function with respect to time and space twice, we obtain the following set of equations. ∂2 f = −a 2 cos(at ± bx) ∂t 2 ∂2 f . = −b2 cos(at ± bx) ∂x2 .

(1.27) (1.28)

We find that with the following conditions, (1.26) satisfies wave equation (1.14). a=ω

(1.29)

b=k

(1.30)

.

.

Adding a constant phase term .φ does not affect the temporal or spatial differentiation. Thus, we find the following function is a solution to the wave equation (1.14). .

f (t, x) = A cos(ωt ± kx + φ)

(1.31)

Of course, we can extend solution (1.31) into three dimensions. .

f (t, x) = A cos(ωt ± k · r + φ)

(1.32)

The freedom to add a phase .φ indicates that a sine function of the same temporal and spatial frequency can also be a solution. Above, we casually used sinusoidal functions as examples. However, this discussion does not lose generality if we remember Fourier’s theorem that state “a periodic function can be expanded into a series of cosine and sine functions.” In the next section, we briefly discuss this topic. Fourier series According to Fourier’s theorem [4, 5], the temporal periodicity of a wave function . f (t) can be expanded into a series of cosine and sine functions as follows.

1.2

Mathematics of Waves .

a0 + a1 cos(ω0 t) + a2 cos(2ω0 t) + · · · + an cos(nω0 t) 2 .+ b1 sin(ω0 t) + b2 sin(2ω0 t) + · · · + bn sin(nω0 t)

f (t) =

13

(1.33) (1.34)

Here .ω0 is the fundamental frequency, .2ω0 , .3ω0 , .. . . . N ω0 are the second, third, .. . . . Nth harmonics. For a given function . f (t) we can find the coefficients .a0 .· · · .bn using the orthogonality of the sine and cosine functions. See Appendix C. Consider these harmonics in the context of wave solution (1.31). We discussed that a cosine sine function in the form of (1.31) satisfies the wave equation (1.15). This wave equation indicates that the secondary temporal differentiation of a wave function is proportional to the secondary spatial differentiation, where the constant of proportionality is the square of the phase velocity. As will be discussed later, the phase velocity is a material constant. Each material has its unique value of acoustic phase velocity. We also learned that phase velocity is the ratio of temporal frequency over spatial frequency (see (1.11)). These facts lead to the following discussion. When an acoustic wave travels through a material at the frequency determined by the source, the phase velocity determines the spatial frequency, hence the wavelength. The combination of .ω and .k in solution (1.31) satisfies this relation, i.e., its ratio is the material’s unique phase velocity. The above situation indicates that if an acoustic wave at the fundamental frequency .ω0 in the Fourier series (1.33) satisfies the wave equation, a harmonic of frequency . N ω0 is also a solution to the same wave equation where the spatial frequency is .(N ω0 )/v p , i.e., . N times greater than the fundamental wave. Since the wavelength of the fundamental frequency is .λ0 = v p /ω0 , the wavelength of this harmonic is . N times shorter. This situation is the same for any . N . The above discussion indicates that the same physical system (the acoustic source and the medium through which the acoustic wave travels) can generate harmonics. Shortly, we will discuss the phenomenon known as resonance, which is an example where the same physical system generates harmonics (higher resonator modes). Standing wave In Sect. 1.2.1, we discussed that a wave function. f (ωt − kx) travels in the positive.x direction and . f (ωt + kx) travels in the negative .x direction. Call the former a forward-going wave and the latter a backward-going wave. Since these two waves have the same phase velocity, .v p = ω/k, the same physical system can generate both waves at the same time, raising the following questions: “Can a pair of forward-going and backward-going waves coexist in a medium? If so, do they add each other, and if that is the case, what is the direction of the combined wave?” The answer is “Yes, they can coexist and add to each other. The combined wave can be stationary. This type of stationary wave is called a standing wave.” I say “can be stationary” because, under some conditions, the combined wave can travel. We will discuss such conditions in Appendix D. Here, we focus our discussions on standing waves.

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1 General Discussions of Waves

Before starting the discussions here, I would like to note some physical significance of standing waves. As will be discussed later, a wave carries energy. Therefore, if a wave stays at the same location, the energy carried by the wave does not flow and thereby the local energy grows in time. Indeed this phenomenon occurs under the condition of resonance. We can construct an instrument to embody such a resonance phenomenon, and the instrument is called a resonator [6]. Most musical instruments [7] (see Sect. 3.2.2) are resonators [8] to generate standing waves [9]. Consider the formation of a standing wave by adding two oppositely traveling waves. Assume the two waves have the same amplitude . A. .

A sin(ωt − kx) + A sin(ωt + kx) = 2 A cos(kx) sin(ωt)

(1.35)

Equation (1.35) indicates that when the forward-going and backward-going waves have the same amplitude, the superposed wave oscillates without traveling in either direction (forward or backward). The time function part (.sin ωt) of the right-hand side of (1.35) tells us every time the time .t makes .ωt = N π (. N is an integer), the wave function becomes zero over the entire.x. This situation is referred to as complete destructive interference. The forward-going and backward-going waves (call these waves component waves) interfere with each other so that the superposed wave becomes zero at all .x locations. When time .t satisfies .ωt = π/2 + N π , on the other hand, .sin ωt takes .±1 and thereby the component waves interfere with each other constructively, maximizing the amplitude of the superposed wave. The “2” in front of . A on the right-hand side of (1.35) indicates that under the condition of complete constructive interference, the resultant amplitude is doubled as compared with the component waves. Note that destructive and constructive interference each repeats at a period of .π (an increase of 1 in the integer . N changes the phase by .π ). This period is half of that of the component waves. Figure 1.8 is an example of a wave expressed in the form of Eq. (1.35). Here the left column illustrates oppositely traveling component waves and the right column the standing wave resulting from the superposition of the component waves. Each row represents time at every one-eighth of the period .τ . It is seen that the first complete constructive interference occurs at .t = 2τ/8, i.e., a quarter period after .t = 0 when complete destructive interference is seen. At .t = 4τ/8, i.e., half a period later the second complete destructive interference occurs. We can easily imagine that the second complete constructive interference occurs a quarter period later, i.e., at .t = 4τ/8 + 2τ/8 = 6τ/8. As mentioned in the preceding paragraph, the completely destructive and constructive interference each repeats every half period or .π in phase. The left column illustrates that the component waves are completely out of phase under complete destructive interference. Under complete constructive interference, they are completely in phase. The complete destructive and constructive interferences alternate at every quarter period (.t = 2τ/8), meaning that destructive (constructive) interference occurs in the middle of consecutive constructive (destructive) interferences.

1.2

Mathematics of Waves

15

Fig. 1.8 An example of a standing wave

In the above sections, we assumed that the amplitude . A is a constant. This assumption means that the wave has the same strength over the plane perpendicular to the .x-axis (the axis along which the component waves travel). Since every wave carries energy, this in turn indicates that the energy is infinite. Such a wave is unrealistic, but we often use this type of solution as it exhibits important characteristics of waves. We call a wave with a constant amplitude a plane wave. The standing wave solution (1.35) has the form of the product of a temporal function and spatial function. This reminds us of the variable separation method. Let’s solve the wave equation (1.15) using the variable separation method [10]. Put the solution as the product of time function .T (t) and space function . X (x) as follows. .

f (t, x) = T (t)X (x)

(1.36)

Substitution of (1.36) it into the wave equation (1.15) yields the following set of equations. ) ∂2 ( f¨ = 2 T (t)X (x) = T¨ X ∂t ) ∂2 ( '' .f = 2 T (t)X (x) = T X '' ∂x .

(1.37) (1.38)

16

1 General Discussions of Waves .

T¨ X = v 2p T X ''

Divide (1.39) by .T X . .

X '' T¨ = v 2p T X

(1.39)

(1.40)

Here.T¨ /T is a function of time independent of.x, and.v 2p X /X '' is a function of.x independent of .t. (1.39) says these two terms are equal to each other. This means that both terms must be a constant. Let .−c2 represent that constant and put (1.40) in the following form. T¨ = −c2 T '' 2 X = −c2 .v p X .

(1.41) (1.42)

Put .T (t) in the following form. .

T (t) = Tc cos(ct) + Ts sin(ct)

(1.43)

Differentiate .T (t) twice with respect to time. .

) ( T¨ = −c2 Tc cos(ct) + T s sin(ct) = −c2 T (t)

From (1.44), we can easily find that (1.43) satisfies (1.41). Similarly, we find the following form of . X (x) satisfies (1.42). ( ) ( ) c c . X (x) = X c cos x + X s sin x vp vp

(1.44)

(1.45)

Thus, by substituting (1.43) and (1.45) into (1.36), we find that the wave function takes the following form. [ ( ) ( )] [ ] c c . f (t, x) = Tc cos(ct) + Ts sin(ct) X c cos x + X s sin x (1.46) vp vp We now in a position to determine amplitudes .Tc , .Ts , . X c , and . X s , and the constant .c. The values of these quantities depend on the initial and boundary conditions. Initial condition In Fig. 1.8, we set the time origin .t = 0 when the component waves interfere completely destructively. It is clear that we can set the time origin freely by adjusting constant phase .φ in (1.31). Here we use the same initial condition as Fig. 1.8, i.e., at .t = 0, the standing

1.2

Mathematics of Waves

17

wave is null. Substituting .t = 0 into (1.46) and setting the value of the function zero, we find .Tc = 0. So, now the wave function takes the following form. [ ( ) ( )] c c . f (t, x) = Ts sin(ct) X c cos x + X s sin x (1.47) vp vp Boundary condition 1 Use the following boundary conditions for . X (x). This condition represents the situation where the standing wave is null at .x = 0 and .x = L. .

X (0) = 0, X (L) = 0

(1.48)

Substituting the first condition of (1.48) into the spatial function in (1.47), we obtain the following equation. .

X (0) = X c cos 0 + X s sin 0 = 0

(1.49)

Thus, we find. X c = 0. The second condition of (1.48) and (1.57) yield the following equation. ( ) c . X (L) = X s sin L =0 (1.50) vp It follows that the following condition holds. .

c L = nπ vp

(1.51)

where .n is an integer. Since the constant .c is in the form of .sin(ct) in the time function .T (t), we can interpret this constant as the angular frequency of the wave. Call it the eigen frequency .ωe . From (1.51) it follows that .ωe satisfies the following condition. nπ , n = 1, 2, 3, . . . .ωe = c = v p (1.52) L Note that the eigen frequency .ωe is derived from the boundary condition (1.48), which indicates that the standing wave has a spatial periodicity whose integer multiple is equal to . L. The physical meaning of this statement becomes clear shortly. Now substitute (1.52) into the spatial function . X (x) using the above-found condition . X c = 0. ( ) ( nπ ) ωe x . X (x) = X s sin x = X s sin (1.53) vp L Here (1.52) is used in going through the last equal sign. So in this case the standing wave function takes the following form. ( nπ ) ( nπ ) v p t sin x . f (t, x) = A sin L L Here . A = Ts X s .

(1.54)

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1 General Discussions of Waves

Fig. 1.9 Phase condition for a passive resonator. In the case of an active resonator, a source is placed inside the resonator

Figure 1.9a illustrates standing waves under Boundary condition 1 in a scenario where a pair of reflectors generate the forward-going and backward-going waves that form the standing waves. The reflection at each of the reflectors is referred to as fixed-end reflection. Boundary condition 2 Now consider the following boundary condition instead of (1.48) using the same initial condition .T (0) = 0. This condition represents the situation where the standing wave has crests or troughs at .x = 0 and .x = L. .

X (0) = A2 , X (L) = ±A2

(1.55)

Substituting the first condition of (1.55) into the spatial function in (1.47), we obtain the following equation. . X (0) = X c cos 0 + X s sin 0 = A 2 (1.56) Thus, we find . X s = 0. The second condition of (1.55) and (1.56) yields the following equation. ( ) c . X (L) = X c cos L = ±A2 (1.57) vp It follows that the following conditions hold. .

c L = nπ, X c = A2 vp

(1.58)

where .n is an integer. Using (1.58) , we find the spatial function . X (x) in this case as follows. ( ) ( nπ ) ωe x . X (x) = A 2 cos x = A2 cos (1.59) vp L

1.2

Mathematics of Waves

19

Here whether the standing wave is at a crest or trough in the boundary condition (1.55) depends on the integer .n; when it is odd, .x(L) = −A2 , and when it is even .x(L) = A2 . Repeating the same procedure as above using the first condition of (1.58), we find the same eigenfrequency .ωe as Boundary condition 1. Thus, in this case, the standing wave function takes the following form. ( nπ ) ( nπ ) v p t cos x . f (t, x) = A sin (1.60) L L Here . A = Ts X c = Ts A2 . Figure 1.9b illustrates standing waves under Boundary condition 2 in a scenario where a pair of reflectors generate the forward-going and backward-going waves that form the standing waves. The reflection at each of the reflectors is referred to as open-end reflection. Boundary condition 3 The last boundary condition considered here represents the situation where the standing wave is null at .x = 0 and a crest or trough at .x = L. .

X (0) = 0, X (L) = ±A3

(1.61)

Here . A3 represents the peak amplitude when the time function .sin(ct) = 1. Substituting the first condition of (1.61) into the spatial function term . X (x) in (1.47) and obtain the following equation. ( ) ( ) c c . X c cos x + X s sin x = X c cos (0) + X s sin (0) = 0 (1.62) vp vp It follows that . X c = 0. Substituting this condition and the second condition of (1.61), we obtain the following equation. ( ) c . X s sin L = ±A3 (1.63) vp It follows that .

π c L = nπ − , X s = A3 vp 2

(1.64)

Condition (1.64) leads to the following expressions of the wave under Boundary condition 3. ) ( (2n − 1)π x . f (t, x) = A 3 sin (ωe t) sin (1.65) 2L where the eigen frequency is as follows ωe =

.

(2n − 1)π vp 2L

(1.66)

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1 General Discussions of Waves

Figure 1.9c illustrates standing waves under Boundary condition 3 in a scenario where a pair of reflectors generate the forward-going and backward-going waves that form the standing waves. The reflection at the left reflector is open-end reflection and that at the right reflector is fixed-end reflection. Resonance of wave The situation depicted in Fig. 1.9 is associated with a phenomenon called Resonance [8]. Imagine that a longitudinal wave traveling in air is incident to a medium, e.g., a tube of length . L. Consider that the frequency of the incident wave is equal to one of the eigen frequencies discussed in the preceding section. We can easily imagine that one of the situations depicted in Fig. 1.9 is established. Due to the difference in acoustic impedance (see Sect. 2.3.1), the incident wave experiences partial reflection at the left end of the tube. Similarly, reflection occurs at the right end of the tube. The wave reflected at the right end travels back toward the left end, and there it experiences reflection. In this fashion, reflected waves go back and forth in the tube. It is instructive to consider the resonant condition for each of the three boundary conditions discussed in the preceding section. First, rewrite (1.47) using the wavenumber or wavelength. Since.c represents the angular frequency.ωc and the phase velocity is given as.v p = ωc k, we can express .c/v p with the wavelength.λ and thereby rewrite the spatial function term as follows. .

f (t, x) = Ts sin(ωe t) [X c cos (kx) + X s sin (kx)] [ ( ( ) )] 2π 2π . = Ts sin(ωe t) X c cos x + X s sin x λ λ

(1.67) (1.68)

Using .v p = λ(ωe /(2π )) and (1.51) we find the following equality under Boundary conditions 1 and 2. c ωe 2π L= L= (1.69) . L = nπ vp vp λ Solving (1.69) for .λ or . L we find the following equation. λ=

.

2L nλ , or, L = n = 1, 2, 3, . . . n 2

Similarly from (1.64) under Boundary condition 3 we find as follows. ( nπ π ) .ωe = − vp L 2L ) ( n 2L 1 λ n = 1, 2, 3, . . . , or, L = .λ = ( − ) 2 4 n − 21

(1.70)

(1.71) (1.72)

Using (1.68), the initial condition . f (0, x) = 0, and other conditions used in the above section we find the following expressions of standing waves under the three Boundary conditions.

References

21

Boundary condition 1

.

( nπ ) x L nλ nπ 2L vp, λ = , L= , n = 1, 2, 3, . . . ωe = L n 2

f (t, x) = B sin(ωe t) sin

(1.73)

Boundary condition 2

.

( nπ ) x L nλ nπ 2L vp, λ = , L= n = 1, 2, 3, . . . ωe = L n 2

f (t, x) = B sin(ωe t) cos

(1.74)

Boundary condition 3 ) (2n − 1) π x . f (t, x) = B sin(ωe t) sin 2L 2n − 1 (2n − 1)π 4L ωe = vp, λ = , L= λ, n = 1, 2, 3, . . . 2L 2n − 1 4 (

(1.75)

The integer .n represents the order of harmonics. Sometimes this number is called the resonator mode number. As will be discussed in Sect. 1.2.2, the phase velocity .v p is determined by the medium. The eigen frequency expressions (1.71)–(1.75) indicate that for a given resonator length . L and a medium inside it, the eigen frequency increases with the mode number .n. The frequency determines the oscillatory cycle of particles. We can easily imagine that the higher the frequency the greater the kinetic energy of the particles as they move faster. Generation of higher-order modes usually requires higher input energy [11]. For example, if you apply an acoustic transducer to a steel structure and monitor the resultant frequency with an acoustic sensor, higher-order modes are observed as the transducer’s energy is increased.

References 1. Komen GJ, Cavaleri L, Donelan M, Hasselmann K, Hasselmann S, Janssen PAEM (1994) Dynamics and modeling of ocean waves, Ch. 1. Cambridge University Press, pp 1–68 2. Roithmayr CM, Hodges DH (2016) Dynamics: theory and application of Kane’s method, Append. I. Cambridge University Press 3. Harper PG, Weaire DL (2009) Introduction to physical mathematics, Ch. 28. Cambridge University Press 4. Prof. Brad Osgood (2014) Lecture notes for EE 261 the Fourier transform and its applications. CreateSpace Independent Publishing Platform

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5. Bracewell RN (1999) The Fourier transform and its applications, 3rd edn. McGraw-Hill, Boston, New York 6. Rienstra SW, Hirschberg A (2023) An introduction to acoustics. https://www.win.tue.nl/~sjoerdr/ papers/boek.pdf (accessed on July 17, 2023) 7. Fletcher NH, Rossing TD (1998) The physics of musical instruments, 2nd edn. Springer Link 8. Standing waves and resonance in university physics vol. 1 (British Columbia/Yukon open authoring platform) https://pressbooks.bccampus.ca/universityphysicssandboxbook1/chapter/ standing-waves-and-resonance/ (accessed on July 17, 2023) 9. King GC (2009) Vibrations and waves. Wiley, Chichester, UK, pp 137–158 10. Borden B, Luscombe J (2017) Essential mathematics for the physical sciences, vol. 1 homogeneous boundary value problems, Fourier methods, and special functions, Ch. 2 Separation of variables. Morgan and Claypool, San Rafael, CA, USA 11. https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/16-4-energy-and-powerof-a-wave/

2

Wave Dynamics

In this chapter, we focus on the physics behind wave dynamics. After reviewing the harmonic oscillation of point mass-spring systems, we discuss how oscillatory motions transition into a wave. We derive wave equations from the equation of motion for oscillatory dynamics in air and solids. Considering some solutions to the wave equation, we discuss the propagation of waves as a flow of energy, along with related topics such as acoustic impedance and acoustic wave intensity. Finally, after briefly reviewing the laws of reflection and refraction, we discuss the propagation of acoustic waves through the boundary of media having different elastic properties.

2.1

Oscillations and Waves

A vibrating object generates sounds. A tuning fork [1] (used by a musician) generates sound because it vibrates at a specific frequency. The physical mechanism underlying this vibration is the elasticity of the material of the tuning fork. If you hit a tuning fork made of steel against something rigid (like an edge of a table), the steel atoms near the contact with the table edge shift from their equilibrium positions. Due to the elasticity of steel, these shifted atoms return to the equilibrium locations after they reach the farthest point. Due to inertia, they pass the equilibrium location and swing in the opposite direction from the initial shifts. This motion pushes the atoms of the neighboring lattice, shifting them from their equilibrium locations. In this fashion, the vibratory motion transfers through lattices reaching the other surface of the tuning fork. Due to the difference in acoustic impedance from the air, the vibratory motion reflects toward the initial surface hit by the table edge. This process is an example of the resonant phenomenon associated with the reflection of the open-open ends discussed in the last chapter. The size (corresponding to . L in Fig. 1.9b) and the phase velocity of steel (see (1.74)) determine the frequency (the eigen frequency) of the fork.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_2

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2 Wave Dynamics

The sound you hear in this fashion has the following properties. If you hit the fork more strongly you will hear the sound louder. However, the pitch of the sound will most likely be the same. Here, I say “most likely” because depending on how strongly you hit the fork, harmonics can be heard [2]. If you hit another tuning fork of a different dimension, you will hear a sound of a different pitch from the first tuning fork. It is because the second fork is different in material and/or size from the first fork. In other words, eigen frequency is different. This is how you tune a musical instrument using tuning forks. In this section, we explore the physics underlying these phenomena.

2.1.1

Harmonic Oscillations

In Chap. 1 we discussed that a wave is a spatiotemporal oscillation. The temporal oscillation discussed in Sect. 1.2 indicates that a recovery force acts on the particle, making it oscillate about an equilibrium position. We can call this type of recovery force an elastic force and the resultant oscillation is a harmonic oscillation [3]. A harmonic oscillation occurs generally in two modes. In the first mode, the system oscillates after the external disturbance is removed. The above example of a tuning fork falls in this category. This type of oscillation is referred to as unforced oscillation [4, 5]. In the second mode, the external agent keeps applying an oscillatory force and the system oscillates in response to the external force. This type of oscillation is referred to as forced oscillation [4, 6]. Unforced and forced oscillations occur when the system exerts an elastic force. (Forced oscillation can occur without an elastic force but in that case, the system simply follows the oscillatory motion provided by the external agent. This type of oscillation is not a harmonic oscillation and will not be discussed.) The elastic force is defined as the force whose magnitude is proportional to the displacement from the neutral (equilibrium) position and the direction is opposite to the displacement. Naturally, the neutral position is where the elastic force is null. When the oscillating entity (called the particle) comes back to the neutral position, it passes there due to inertia. Consequently, the particle goes to the other side of the neutral position. As soon as the particle passes the neutral position, the elastic force changes its direction because it is always toward the neutral position. We can say that the acceleration of the particle is always toward the neutral position (according to Newton’s second law). Thus, the acceleration is opposite to the displacement of the particle. Point mass-spring system Spring-mass systems [7–9] are simple yet informative to understand harmonic oscillations. Consider the above dynamics using a system of a point mass connected to a horizontally placed spring. In this case, the spring exerts the elastic force that causes the oscillation of point mass and the point mass is what we called the particle in the above paragraph.

2.1

Oscillations and Waves

25

Fig. 2.1 Spring oscillation. The mass (particle) is on the right side of the neutral point (top), at the neutral point (middle) and on the left side of the neutral point (bottom). The thicker arrow represents the spring’s elastic force and the thinner arrow represents the displacement from the neutral position

Figure 2.1 illustrates a sample spring-mass system. Here the object connected to the right end of the spring is considered to be a point mass. The middle illustration represents the moment when the point mass passes through the neutral point from the right. The top and bottom illustrations represent the situation before and after the mass passes through the neutral point. The arrows represent the elastic force exerted by the spring. Consider the situation after the mass passes through the neutral point. Since the magnitude of the elastic force is proportional to the displacement, the acceleration toward the neutral position keeps increasing. This causes the velocity of the mass to decrease. Eventually, the velocity becomes null. At that point, the acceleration is still toward the neutral point. Therefore, the mass starts to move back towards the neutral point with zero initial velocity (after a momentary stop). From this nature, the point where the mass stops momentarily is called the turning point. After some time, the mass comes back to the neutral point from the opposite side from the previous time. The mass passes the neutral point again due to the inertia and moves toward the other turning point. There, by the same mechanism as at the first turning point, it switches the direction of motion after a momentary stop. In this fashion, the mass keeps oscillating back and forth around the neutral position. It is instructive to analyze the above dynamics qualitatively. The questions I would like to ask ourselves are (a) How long does it take the mass to come back to the neutral position? and (b) How far is the turning point away from the neutral position? The quantity asked by the first question is called the oscillation period and the one by the second question is the oscillation amplitude. Let’s consider what determines the oscillation period. Above I said that the mass moves toward the neutral position due to the elastic force and that it passes the neutral position due to the inertia. It is easily expected that the stronger the elastic force the sooner the mass returns to the neutral position and that the greater the inertia the slower the oscillatory motion of the mass. Thus, it is speculated that the stronger the elastic force the shorter the oscillation period, and the greater the mass the longer the oscillation period becomes. Indeed, as we will discuss in the next section quantitatively, the oscillation period is an increasing function of the elastic constant divided by the mass of the particle. Here the elastic constant measures the strength of the elastic force for a given displacement.

26

2 Wave Dynamics

Energy of oscillatory systems As for the oscillation amplitude, neither the elastic constant nor the weight of the mass plays a role in determining it. It is determined by the initial position of the mass. In fact, the amplitude is similar to the distance between the initial and neutral positions of the mass. (I said “similar” for a reason, which will become clear shortly). How can we rationalize it? The best way is to consider energy conservation. Above, I said, “The mass keeps oscillating back and forth around the neutral position”. This statement indicates that the oscillation continues forever with the same period and amplitude. You can easily imagine that such a motion is unrealistic. Indeed, it is impossible for any system to oscillate forever. Eventually, the oscillation stops, and the mass becomes stationary at the neutral position. This is because there is always some mechanism that dissipates the oscillatory energy. Many oscillatory motions are dissipated by the mechanism known as velocity damping [4], which will be discussed quantitatively later in this chapter. The velocity damping force is proportional to the velocity of the mass in magnitude and opposite in direction. In a system in which the damping effect is low, the rate that the oscillation energy is dissipated is low. The oscillation energy consists of the elastic potential energy and the kinetic energy of the mass. At the initial point where the mass starts the oscillatory motion, its velocity is null. All the oscillatory energy is in the form of the elastic potential energy which is determined by the elastic constant and the initial elastic force. Since the initial elastic force is proportional to the initial displacement, the initial mass position determines the maximum elastic energy. At the neutral point where the elastic potential energy is null, the oscillation energy is in the form of the kinetic energy. As the system loses the oscillation energy due to the damping mechanism, both the elastic potential energy and kinetic energy decrease. Consequently, the displacement at the turning point decreases and the velocity at the neutral point decreases. This in turn means that the oscillation amplitude keeps decreasing as the oscillation repeats. If the system is a low damping system, the oscillatory energy diminishes slowly, and therefore the amplitude decrease slowly. When the damping effect is significant, the amplitude decreases fast.

2.1.2

Equation of Motion of Point Mass-Spring System

Unforced oscillation A system comprising of a spring and a point mass with a velocity-damping mechanism is a simple yet very illustrative physical system to understand the decaying oscillatory dynamics. Let .ξ(t) be the displacement from the equilibrium (neutral point). Being proportional to the displacement in magnitude and opposite in direction, the elastic force . f sp can be put in the

2.1

Oscillations and Waves

27

following form. (See Fig. 2.1 and note that the displacement vector and spring force vector are opposite and that the spring force is the only external force acting on the mass.) .

f sp = −ksp ξ(t)

(2.1)

Here the constant of proportionality.ksp is called the spring constant or stiffness. It represents the strength of the elasticity of the spring. From Newton’s second law, the external force on the point mass, . f ex , is proportional to the acceleration where the mass is the constant of proportionality. This enables us to write the following equation. .

f ex = m

d 2 ξ(t) dt 2

(2.2)

If the spring force is the only component of the external force, . f ex = f sp and (2.1), and (2.2) lead to the following equation of motion. .

ksp d 2 ξ(t) ξ(t) =− 2 dt m

(2.3)

Equation (2.3) indicates that the solution .ξ(t) is a function of time that has the following property; if differentiated twice the result is proportional to the original function with a negative proportionality constant. We know that sine and cosine functions have this property. Thus, it is clear that the motion resulting from a spring force is oscillatory, and can be represented by a sine or cosine function. The oscillation is called the harmonic oscillation. When the mass undergoes velocity damping (viscous) force, we need an additional term on the right-hand side of the equation of motion. The resultant equation of motion looks as follows. d 2 ξ(t) dξ(t) .m = −ksp ξ(t) − b (2.4) 2 dt dt Here .b is the damping coefficient. For simplicity, introduce the following parameters .ω0 and .β. √ ksp .ω0 = (2.5) m b .β = (2.6) 2m Here .ω0 is referred to as the natural (angular) frequency and .β as the decay constant. It will become clear in the next section why these parameters are called in these ways. Dividing both-hand sides by.m and using (2.5) and (2.6), rewrite Eq. (2.4) in the following form. dξ(t) d 2 ξ(t) + ω02 ξ(t) = 0 . + 2β (2.7) 2 dt dt

28

2 Wave Dynamics

Equation (2.7) is a linear differential equation. The displacement of the point mass is given as a solution to this linear differential equation. Note that this differential equation has no source term, i.e., the right-hand side of the equation is zero. Hence, (2.4) is classified as a linear homogeneous differential equation. Physically, this corresponds to unforced oscillation. In other words, there is no driving force in the system. Being represented by a linear differential equation, the system is called a linear system. Forced oscillation When an additional force (the driving force) is acting on the mass, the solution.ξ(t) represents forced oscillation. In this case, the differential equation has a source term as below. .

d 2 ξ(t) dξ(t) f dr + 2β + ω02 ξ(t) = dt 2 dt m

(2.8)

Here . f dr is the driving force and .m is the mass. It is possible to view the driving force as an input to the linear system. In this view, the solution .ξ(t) is the output of the linear system. As long as the coefficient .β and .ω0 are constant, the output can be viewed as the linear response to the input. The linear system associated with elastic force can be argued from the viewpoint of elastic (potential) energy. As a conservative force, the elastic force can be characterized as the first-order spatial differentiation of the potential energy. When the potential energy has quadratic dependence, the force has linear dependence. In an actual physical system, it is possible that the potential energy has a dependence on the displacement with a third-order or higher polynomial function. In this case, the system is no more linear. Such a system can be interpreted as the case where the stiffness .ksp is a function of displacement .ξ . A typical example of such a nonlinear system is an inter-atomic potential. The famous Lennard-Jones potential curve [10] is approximately quadratic near the equilibrium (near the potential well). This means that as long as the displacement from the equilibrium is small, the inter-atomic force can be approximated as a spring-like force. When the displacement is so large that it exceeds the quadratic part of the potential well, the dynamics becomes nonlinear. A good example is the acoustic response of atoms in a solid material having residual stress. Residual stress can lock the atom in a range where the potential curve is not quadratic. Figure 2.2 illustrates the situation schematically. If the specimen is free of residual stress, the atom is located near the bottom of the well (the equilibrium position .ξ0 in Fig. 2.2) where the potential curve has a quadratic dependence on .ξ − ξ0 . In this case, the inter-atomic force .dU (ξ )/dξ is a linear function of .ξ − ξ0 , and therefore the stiffness is constant. (Here .ξ is not defined as the displacement from the equilibrium. .ξ = 0 is the zero interatomic distance. Therefore, .ξ here is different from .ξ(t) used in the differential equations (2.3), (2.7) and other equations associated with these two equations.) If residual stress locks the atom far from the equilibrium position, and therefore the potential energy cannot be approximated as a quadratic function of .ξ − ξ0 , the stiffness

2.1

Oscillations and Waves

29

Fig. 2.2 Stress energy potential with residual stress. The slope of the potential curve represents inter-atomic force. Potential curve is approximately quadratic near equilibrium .ξ0 . On .ξ < ξ0 side potential is steeper than quadratic and on .ξ > ξ0 side it is less steeper than quadratic

is not a constant anymore; it depends on .ξ − ξ0 . In this case, the atom shows nonlinear behavior when excited by an acoustic wave. Analysis making use of this non-linearity to probe residual stress is known as acoustoelasticity [11].

2.1.3

Oscillation to Waves

In the preceding section, we discussed the temporal oscillatory behavior of a wave as a harmonic oscillation of a point mass. In Chap. 1 we discussed a wave is a spatiotemporal oscillation of particles. We can view the phase velocity of a wave as the motion of a spatial oscillatory pattern in time, or the motion of a temporal oscillatory pattern through space. In other words, if the harmonic oscillation experienced by a portion of a medium has a time lag from the neighboring portion, the harmonic oscillation travels through the medium forming a wave. In this section, we look into the physics behind this mechanism.

2.1.4

Wave Equations

One dimensional wave We first discuss how a harmonic motion of a point mass propagates as a wave in one dimension. Consider the series of point masses connected with springs in Fig. 2.3. Imagine you pull the rightmost point mass to the right from its equilibrium point and release it. When you pull this point mass, the spring connected to it stretches, pulling the next (second from the right) point mass. In this fashion, the point masses to the left of these two will be pulled to the right one after another. Now the question is “Are these all point masses pulled simultaneously?”. Intuitively, we know that the answer is “No”. The ones located to the left will be pulled after a certain delay, and the more to the left, the longer delay time they experience.

30

2 Wave Dynamics

Fig. 2.3 Series of point masses. a1 unstretched; a2 rightmost point mass pulled; a3 rightmost point mass released. b elastic force on .n.th point mass. c elastic force acting on .Δx portion of elastic continuum

When you release the rightmost point mass, a similar delayed behavior is observed in other point masses. At the moment you release the rightmost point mass, the next point mass to the left is still moving to the right. Shortly after, this second point mass changes its direction being pushed by the spring on its right. Thus, all the point masses change their direction with a certain delay time. This pattern continues where each point mass oscillates around its equilibrium point. We can view this oscillatory pattern with the delay time as a longitudinal wave that propagates the harmonic oscillation of the point masses. Then, the next question is what is the delay time? Again, intuitively we know that the weaker the springs the longer the delay time. However, it is not straightforward to express this delay quantitatively. In fact, this delay time represents the wave’s phase velocity. Below, we analyze this phenomenon quantitatively. Equation of motion Consider the dynamics of the .n.th point mass. This mass receives spring force from the two springs, one that connects it to the .(n − 1)th point mass and the other that connects it to the th .(n + 1) point mass. The equation of motion that governs this motion is as follows. m

.

d 2 ξn = ksp dξn+1 − ksp dξn−1 = ksp (dξn+1 − dξn−1 ) dt 2

(2.9)

Here .m is the mass of the point mass, .ksp is the spring constant of the two springs, and .dξn+1 and.dξn−1 are the differential displacement of the.n.th point mass relative to the.(n + 1)th and th point masses, respectively. These differential displacements represent the stretch .(n − 1)

2.1

Oscillations and Waves

31

of the respective springs, and therefore their products with the spring constant represent the spring force. Now we extend the above argument into a limit case where the springs are infinitesimally small so that the entire series of the point masses can be considered as a continuous elastic medium. In this limit, we can write the equation of motion (2.9) in the following form. ( ) d 2 ξ(x) dξ(x + d x) − dξ(x − d x) = k sp dt 2

μΔx

.

(2.10)

Here .μ (kg/m) is the linear density of the medium and .Δx is the length of the medium along the .x-axis that we apply the equation of motion (corresponding to the .n th point mass in (2.9)). On the right-hand side of (2.10), .dξ(x + d x) and .dξ(x − d x) represent, respectively, the stretch of the section of the elastic medium neighboring to the section represented by .Δx on the negative and positive sides. In the infinitesimal limit, we can express the differential displacement (the stretch of the elastic medium) .dξ(x) as follows. dξ(x) =

.

∂ξ(x) δx ∂x

(2.11)

Here, the quantity .δx represents the length of the section we consider the stretch. Using (2.11) on the right-hand side, we can rewrite (2.10) as follows. .μΔx

d 2 ξ(x) = ksp dt 2

(

∂ξ(x − d x) ∂ξ(x + d x) δx − δx ∂x ∂x

)

( = ksp δx

∂ξ(x − d x) ∂ξ(x + d x) − ∂x ∂x

)

(2.12)

Expressing the spatial derivative of the displacement .ξ(x) as .

∂ξ(x) = ξ ' (x) ∂x

we can write the quantity inside the parentheses on the right-hand side of (2.12) as follows. ∂ 2ξ ∂ξ ' ∂ξ(x + d x) ∂ξ(x − d x) − = ξ ' (x + d x) − ξ ' (x − d x) = dξ ' = Δx = 2 Δx ∂x ∂x ∂x ∂x (2.13) Here, we used the following expression on the rightmost-hand side of (2.13). .

dξ ' =

.

∂ ∂ξ ∂ 2ξ ∂ξ ' Δx = Δx = 2 Δx ∂x ∂x ∂x ∂x

(2.14)

The above mathematical process lets us rewrite (2.10) as follows. μ

.

d 2 ξ(x) ∂ 2ξ = k δx sp dt 2 ∂x2

Note that in deriving (2.15) we canceled .Δx from both sides of the equation.

(2.15)

32

2 Wave Dynamics

Spring constant and elastic constant Let’s take a moment to consider the physical meaning of the spring constant and elastic constant. These two constants are similar concepts representing the elasticity of a medium. However, a spring constant is not a material constant, whereas an elastic constant is a material constant. Each medium has its unique elastic constant, which characterizes the propagation of acoustic waves. On the other hand, a spring constant depends on the dimension of the medium, and thereby it is not a material constant. Below, we consider the difference between these two constants. According to Hooke’s law [12], the elastic force at .x is the product of the spring constant and the local differential displacement .dξ . .

f = ksp dξ = ksp (ξ(x + δx) − ξ(x)) = ksp

∂ξ ∂ξ δx = (ksp δx) ∂x ∂x

(2.16)

Here, .dξ represents the elongation between .x and .x + δx. In Fig. 2.3c, the two arrows depict the elastic force at the two ends of the material having the length of .Δx. Here, the elastic force is due to the elongation over the segment of thickness .δx. In (2.16) the quantity .∂ξ/∂ x represents the stretch (the elongation per unit length or .δx = 1). We call this quantity the normal strain .∈ = ∂ξ/∂ x. Figure 2.4 illustrates the situation where the same normal force . f acts on a segment of the same material with various thicknesses and cross-sectional areas causing the same normal strain. In Fig. 2.4a, the segment has a thickness of .δx and a cross-sectional area of . A; in (b) the thickness is .2δ and the cross-sectional area is . A; in (c) the thickness is unity (.δx = 1) and the cross-sectional area is . A; and in (d) the thickness is unity and cross-sectional area is unity (. A = 1). (a)

(c) (d)

(b)

Fig. 2.4 Spring constant and Young’s modulus

2.1

Oscillations and Waves

33

' inserted in the figure According to (2.16), the spring constant for (b) is half of (a), as .ksp indicates. This is because the segment in (b) is twice as thick as (a), and therefore, the elongation is double when the same force causes the same strain. This comparison indicates that the spring constant depends on the thickness of the medium. The medium thickness of (c) is unity, i.e., .δx = 1. In this case, the quantity corresponding to the elongation in (a) and (b) becomes .(∂ξ/∂ x)(δx) = (∈)(1) = ∈, i.e. is the strain .∈. According to continuum mechanics [13], Hooke’s law for a unity thickness is given as follows. .σ = E∈ (2.17)

Here .σ is the stress defined as the force per unit area and . E is Young’s modulus. In the present case, it is the normal stress as the force is normal to the area. Figure 2.4d illustrates Hooke’s law represented by (2.17). .f = σA (2.18) Since . E is defined by the force per unit area and the elongation per unit length, it does not depend on the dimension of the medium, hence it is a material constant. From (2.16), (2.17), and (2.18), we obtain the following set of equations. .

f = ∈(ksp δx)

(2.19)

.

f = σ A = E∈ A

(2.20)

From (2.19) and (2.20), we find the following relationship between the spring constant and Young’s modulus. ksp δx A , or ksp = E .E = (2.21) A δx The first expression of (2.21) indicates that Young’s modulus is in N/m.2 , and the second expression tells us the greater the cross-sectional area and the less the thickness, the stronger the spring constant (stiffness) of the elastic medium becomes. This situation is similar to the relationship between the electric conductivity .σe and conductance, .G e ; .G e = σe (A/l) where . A is the cross-sectional area of the conductive material and .l is the length. This relationship tells us the greater the cross-sectional area and the shorter the length, the same material exhibits higher conductance. Phase velocity as a material constant The above discussion lets us state that phase velocity is a material constant. Consider the wave equation (2.15). d 2 ξ(x) ∂ 2ξ .μ = ksp δx 2 (2.15) 2 dt ∂x

34

2 Wave Dynamics

The quantity .μ on the left-hand side of this equation is linear density. Expressing the crosssectional area of the medium with . A, we can relate the linear density and volume density .ρ as follows. .μ = ρ A (2.22) Using (2.22) on the left-hand side of (2.15) and the first expression of (2.21), we can rewrite the wave equation in the following form. .

ksp δx ∂ 2 ξ E ∂ 2ξ d 2 ξ(x) = = dt 2 ρ A ∂x2 ρ ∂x2

(2.23)

A comparison with (1.14). we can identify the quantity . E/ρ as the square of the phase velocity .ω/k. ( ω )2 ∂ 2 f ∂2 f . = (1.14) ∂t 2 k ∂x2 Thus, we find the phase velocity of the longitudinal displacement wave as follows. √ E long .v p = (2.24) ρ This discussion tells us that the phase velocity is a material constant. Each elastic material possesses a unique phase velocity. This causes reflection at a boundary between elastic media having different phase velocities. Here (2.25) represents the phase velocity of a longitudinal wave because the underlying equation of motion represents the normal force acting on the segment of the material and the resultant elongation (and strain) is parallel to the force. We can repeat the same argument for shear force and resultant shear strain to derive the phase velocity of the shear (transverse) wave in the following form. √ v shear = p

.

G ρ

(2.25)

Here .G is the shear modulus.

2.2

Acoustic Wave Equations and Solutions

In Sect. 2.1.4 we discussed that an equation of motion yields a mechanical wave equation. We also discussed that an elastic force is necessary for the oscillatory behavior that generates wave dynamics. These statements indicate that an expression representing the elastic force and an equation of motion associated with the elastic force expression are essential to derive an acoustic wave equation. Elasticity is intrinsic to a medium and therefore the expression of the elastic force including the expression of the elastic constant depends on the medium. In this section, we consider wave equations in air and solids.

2.2

Acoustic Wave Equations and Solutions

2.2.1

35

Acoustic Wave in Air

Hooke’s law in air We can express Hooke’s law in the following form. .

P = −B

ΔV V' − V = −B V V

(2.26)

Here, . P is the pressure that causes the volume change .ΔV , .V is the initial volume, and V ' is the volume after compression. . B (N/m.2 ) is the bulk modulus, which is the elastic constant corresponding to Young’s modulus or shear modulus for solids. Note that, unlike Hooke’s law for solids (see (2.17)), we need to use a negative sign. This is because in solids conventionally positive stress is expansion whereas in air pressure is compression. Referring to Fig. 2.5 consider the change in volume due to pressure . P.

.

.V

'

= (Δx + dξx )(Δy + dξ y )(Δz + dξz ) = ΔxΔyΔz + (ΔyΔzdξx + ΔzΔxdξ y + ΔxΔydξz ) + (dξ y dξz Δx + dξz dξx Δy + dξx dξ y Δz) + dξx dξ y dξz ∼ ΔxΔyΔz + (ΔyΔzdξx + ΔzΔxdξ y + ΔxΔydξz ) = ( ) dξ y dξx dξz = ΔxΔyΔz + ΔxΔyΔz + + Δx Δy Δz

(2.27)

Here, .ΔxΔyΔz = V . Therefore, )) ( ) ( ( dξ y dξ y dξx dξx dξz dξz ' −V =V .ΔV = V − V = V +V + + + + Δx Δy Δz Δx Δy Δz (2.28) In the infinitesimal limit the fractions .Δξx = ∂ξx /∂ x, etc. can be replaced with the corresponding partial derivatives. Thus (2.26) becomes as follows. ) ( ∂ξ y ΔV ∂ξx ∂ξz = −B∇ · ξ (2.29) . P = −B = −B + + V ∂x ∂y ∂z Fig. 2.5 Change in volume due to pressure at a single point

36

2 Wave Dynamics

Fig. 2.6 Gradient of volume expansion .∇ · ξ due to pressure gradient

The quantity .∇ · ξ represents the divergence of particles from a single point due to the local displacement field. In other words, .∇ · ξ , hence the pressure . P is defined at this single point. The fact that the dimension of .∇ · ξ is m/m, i.e., unity indicates that this quantity is defined at a single point. We can view .∇ · ξ as the volume expansion at the point. Now consider that the pressure . P has a gradient, meaning that it is not uniform. From (2.29) we put Hooke’s law as follows. ∇ P = −B∇(∇ · ξ )

.

(2.30)

Consider the meaning of .∇(∇ · ξ ) in Fig. 2.6 that illustrates a cubic unit volume of air. Due to the pressure gradient, .∇ · ξ has .x, . y, and .z dependence. In other words, at the corners of the cube, the air experiences different levels of volume expansion illustrated in Fig. 2.5. The left-hand side of (2.30) represents the forces that apply on this unit volume experiencing .∇(∇ · ξ ). By breaking .∇ P into the .x, . y, and .z components, we can visualize that the differential pressure causes the differential volume expansion .∇ · ξ . ) ) ) ) (( ( ( ∂P ˆ ∂P ˆ ∂P ˆ ∂(∇ · ξ ) ˆ ∂(∇ · ξ ) ˆ ∂(∇ · ξ ) ˆ . (2.31) i+ j+ k = −B i+ j+ k ∂x ∂y ∂z ∂x ∂y ∂z Equation of motion We can view the left-hand side of (2.31) represents the net force acting on the unit volume characterized by mass .ρ. Thus, we can write the equation of motion for this unit volume as follows. ) ) ) ( ( ( (( 2 ) ( 2 ) ( 2 ) ) ∂ ξy ∂P ˆ ∂P ˆ ∂P ˆ ∂ ξx ˆ ∂ ξz ˆ ˆ (2.32) . i+ j+ k = −ρ i+ j+ k ∂x ∂y ∂z ∂t 2 ∂t 2 ∂t 2

2.2

Acoustic Wave Equations and Solutions

37

Using (2.30) we can also write this equation of motion expressing with the temporal and spatial derivatives of the displacement vector .ξ and bulk modulus. ρ

.

∂ 2ξ = −∇ P = B∇(∇ · ξ ) ∂t 2

(2.33)

Equation (2.33) explicitly represents that the pressure gradient causes the acceleration of the unit volume. Thus, we can view it as the equation of motion governing the unit volume. Note that we need a negative sign on the right-hand side of (2.32) because the inward pressure is defined to be positive; if the pressure at .x + d x is greater than at .x, the net force is in the direction of negative.x (the pressure at .x + d x pushes the unit volume more strongly inward than at .x). Taking the divergence of both-hand sides of (2.32), we obtain the following equation. .

∂ ∂x

(

∂P ∂x

) +

∂ ∂y

(

∂P ∂y

) +

∂ ∂z

(

∂P ∂z

)

(

) ) )) ( ( ( ∂ ∂ 2 ξx ∂ ∂ 2ξy ∂ ∂ 2 ξz + + ∂x ∂ y ∂t 2 ∂ x ∂t 2 ∂t 2 ) ( ∂ξ y ∂ξz ∂ 2 ∂ξx + + = −ρ 2 ∂x ∂y ∂z ∂t = −ρ

We can write this equation more compactly as follows. ∇ 2 P = −ρ

.

∂ 2 (∇ · ξ ) ∂t 2

(2.34)

We can view (2.34) as the equation of motion governing .∇ · ξ where the right-hand side represents the acceleration (times the mass) and the left-hand side the net force. Volume expansion (compression) wave equation Express the equation of motion (2.32) in the following compact form. ∇ P = −ρ

.

∂ 2ξ ∂t 2

(2.35)

Hooke’s law is as follows. ∇ P = −B∇(∇ · ξ )

.

(2.30)

Equate the right-hand sides of (2.35) and (2.30), and take the divergence of the resultant equation using the identity .∇ · ∇ = ∇ 2 . Then we obtain the following equation. .

∂ 2 (∇ · ξ ) B = ∇ 2 (∇ · ξ ) ∂t 2 ρ

(2.36)

38

2 Wave Dynamics

We can view (2.36) as an equation that represents a volume expansion .(∇ · ξ ) wave that √ travels at the phase velocity . B/ρ. Pressure wave equation Using Hooke’s law (2.29), we can replace .∇ · ξ on both sides of (2.36) with .−P/B and obtain the following equation. B ∂2 P . = ∇2 P (2.37) 2 ∂t ρ Equation (2.37) is a wave equation that represents that pressure . P travels at the phase veloc√ ity of . B/ρ, the same velocity as the volume expansion (compression) wave. Alternative way to derive pressure wave equation We can derive the wave characteristics of the pressure . P in the following fashion as well. Differentiate (2.29) with respect to time twice. .

) ) )) ( ( ( ( ∂ ∂ 2ξy ∂ ∂ 2 ξz ∂2 P ∂ ∂ 2 ξx + + = −B ∂t 2 ∂ x ∂t 2 ∂ y ∂t 2 ∂z ∂t 2

(2.38)

The equation of motion yields the following equations. ρ

.

∂ 2ξy ∂ 2 ξz ∂ 2 ξx ∂P ∂P ∂P , ρ , ρ = − = − =− 2 2 2 ∂t ∂x ∂t ∂y ∂t ∂z

(2.39)

Using three relationships in (2.39) on the right-hand side of (2.38), we obtain the following equation. B ∂2 P . = ∂t 2 ρ

(

∂ ∂x

(

∂P ∂x

)

∂ + ∂y

(

∂P ∂y

)

∂ + ∂z

(

∂P ∂z

))

B = ρ

(

∂2 P ∂2 P ∂2 P + + ∂x2 ∂ y2 ∂z 2

) ,

or more compactly, in the following form. .

B ∂2 P = ∇2 P ∂t 2 ρ

(2.40)

Apparently, (2.40) is identical to the pressure wave equation (2.37). We derived (2.37) by considering the spatial differentiation of the equation of motion (2.32), and then using Hooke’s law (2.29). On the other hand, in deriving the pressure wave equation (2.40), we differentiated Hooke’s law (2.29) with respect to time first and then used the equation of motion. In either case, we derived the pressure wave equation based on Hooke’s law and Newton’s second law (the equation of motion).

2.2

Acoustic Wave Equations and Solutions

39

Density-change wave equation From the conservation of mass, we find the following equality. ρ0 V = ρV '

.

(2.41)

Here .ρ0 and .ρ are the density before and after the volume expansion. Define the relative volume change .s as follows. V' − V V' .s = (2.42) = −1 V V Substituting (2.41) into (2.42), s=

.

ρ0 ρ0 − ρ δρ −1= = ρ ρ ρ

(2.43)

From Hooke’s law (2.26), .

P = −Bs

(2.44)

Substituting (2.44) into pressure wave equation (2.40), we obtain the following wave equation. ∂ 2s B = ∇2s (2.45) . 2 ∂t ρ Using (2.43) we can write wave equation (2.45) ( ) ( ) B 2 δρ ∂ 2 δρ = ∇ . ∂t 2 ρ ρ ρ

(2.46)

We can view (2.46) as a wave equation of density change. Note that all the above four wave-equations have the following same phase velocity. √ B (2.47) .v p = ρ This phase velocity expression has the same form as the one for solids; the square root of the ratio elastic constant over density, as will be discussed in the next section.

2.2.2

Longitudinal and Transverse Vibrations in Solid

In elastic solids, longitudinal and transverse vibrations propagate as longitudinal and transverse (shear) waves. We can discuss the dynamics of these waves by considering equations of motion. First, consider longitudinal wave dynamics. Figure 2.7 illustrates a block of an elastic medium between .x and .x + d x (called the central block). This block is part of a long

40

2 Wave Dynamics

Fig. 2.7 Elastic force acting on central block

material extended along the .x axis. At the far-right end of this material, an external agent exerts a tensile force. The other external agent holds the opposite end (on the negative .x side). Consequently, the entire material experiences a stretch. Since the external force is dynamic, the block is not under static equilibrium. At the face at .x, the central block pulls the neighboring block on the negative .x side in the positive .x direction. In response, the neighboring block exerts a leftward reaction force . f el (x) on the central block. This force pair causes a local stretch at the boundary of the central and left neighboring blocks. Figure 2.8 illustrates the situation at the boundary in detail. Referring to the upper part of this figure, we can interpret this local stretch as differential displacement .dξ(x1 ) = ξ(x1 + δx) − ξ(x1 ). Here .δx represents the thin boundary area between the neighboring and central blocks, and .x1 is the coordinate of the neighboring block side of the boundary. Thus, using the spring constant .ke , we can express the leftward force exerted by the neighboring block on the central block as follows. ( ) ∂ξ(x) δx . f el (x 1 ) = − (k e dξ ) x = −k e = −ke ξ ' (x1 )δx (2.48) 1 ∂x x1 Here .ξ ' (x) is the spatial derivative of .ξ(x), i.e., .ξ ' (x) = ∂ξ/∂ x. On the other face at .x2 = x1 + Δx, the neighboring block located on the positive .x side exerts a rightward force . f el (x1 + Δx). Repeating the same argument as the face at .x1 , we can express the elastic force at the .x2 as follows. ( ) ∂ξ(x) δx . f el (x 1 + Δx) = (k e dξ ) x +Δx = k e = ke ξ ' (x1 + Δx)δx (2.49) 1 ∂x x1 +Δx Here we assume that .ke and .δx at .x = x1 + Δx are the same as at .x = x1 . Thus, the net external force acting on the central block is. f el (x1 + Δx) + f el (x1 ). According to Newton’s second law, the acceleration of the central block is equal to the net external force. From (2.48) and (2.49), we obtain the following equation of motion.

2.2

Acoustic Wave Equations and Solutions

41

Fig. 2.8 Local stretch at .x1 and .x2 = x1 + Δx

m

.

( ( ∂ξ ' ) ) ∂ 2ξ ∂ 2ξ ' ' ' ξ = k (x + Δx) − ξ (x ) δx = k dξ δx = k Δxδx Δx δx = k e 1 1 e e e ∂t 2 ∂x ∂x2 (2.50)

Expressing the mass of the central block with the density .ρ and the cross-sectional area . A as .m = ρ AΔx, we can rewrite the equation of motion (2.50) as follows. ρ AΔx

.

∂ 2ξ ∂ 2ξ = k Δxδx e ∂t 2 ∂x2

(2.51)

In the infinitesimal limit, we can put .δx = Δx = d x. Thus, canceling .Δx and rearranging terms, we can write (2.50) as follows. ρ

.

ke d x ∂ 2 ξ ∂ 2ξ = 2 ∂t A ∂x2

(2.52)

As discussed with (2.18), we can express the normal stress .σ with elastic force . f el as follows. f el ke dξ (2.53) .σ = = A A Normal stress is related to normal strain .∈ with Young’s modulus . E. σ = E∈ = E

.

dξ dx

(2.54)

Comparing the right-hand side of (2.53) and (2.54), we find the following relation. (This is the same argument as the derivation of (2.21).)

42

2 Wave Dynamics

ke d x = E A

.

(2.55)

Using (2.55), we can rewrite (2.52) as follows. .

E ∂ 2ξ ∂ 2ξ = ∂t 2 ρ ∂x2

(2.56)

Equation (2.56) is a wave equation that describes the dynamics of a longitudinal wave traveling through a solid that has Young’s modulus (the longitudinal elastic modulus) . E. A comparison of (1.14) and (2.56) indicates that the wave equation is, in fact, the equation of motion, and moreover, the phase velocity of the longitudinal elastic wave is given as follows. √ E long .v p = (2.57) ρ Similarly, considering the displacement behavior perpendicular to the .x axis and its variation with .x, we can derive a transverse wave equation. The lower part of Fig. 2.8 illustrates the .x dependence of the displacement in the . y direction, .η(x), at the .x1 and . x 2 = x 1 + Δx boundaries of the central block. In this case, the net force acting on the boundary area of thickness .δx is the shear force in the form of the spring constant times the differential displacement .dη(x). Thus, we can write the equation of motion for the shear dynamics as follows m

.

∂ 2η ∂ 2η = k d x d x, e ∂t 2 ∂x2

(2.58)

and stress-strain equation as follows. τ=

.

f el ke dη = A A

(2.59)

Here .τ is the shear stress and the other quantities are the same as (2.53). Equations (2.58) and (2.59) are the shear dynamics versions of (2.50) and (2.53), respectively. Similarly to (2.54), we can relate the shear stress with shear strain .γ . τ = Gγ = G

.

dη dx

(2.60)

Here.G is the shear modulus and is related to the spring constant as follows. This relationship is the shear deformation version of (2.55). ke d x = G A

.

(2.61)

With the same procedure as the longitudinal wave case, we can derive a wave equation and phase velocity for the shear dynamics as follows.

2.2

Acoustic Wave Equations and Solutions

G ∂ 2η ∂ 2η = 2 ∂t ρ ∂x2 √ G shear .v p = ρ .

43

(2.62) (2.63)

Note that as we discussed at the end of Sect. 2.1.3, the phase velocities of the longitudinal and transverse (shear) wave are material constants. They contain material constants. E,.G, and .ρ. When an acoustic wave of the frequency (.ν) determined by the source (e.g., the operation frequency of an acoustic transducer) propagates in an elastic material, the wavelength (.λ) is determined by the material’s unique phase velocity and the frequency (.λ = v p /ν).

2.2.3

Decaying Acoustic Wave Equations and Solutions

We derived the wave equations (2.36) and (2.37) from the equation of motion (2.35). While this equation of motion describes the mechanism in which the unit volume of air is accelerated by the differential pressure (the pressure gradient), it generates a solution that represents dynamics that lasts forever. In reality, waves decay due to various mechanisms such as velocity damping associated with collisions between air molecules. (Drag force in viscous media is proportional to the velocity only for lighter objects and lower speed [14, 15].) To the first-order approximation, we can represent this effect by including a velocitydamping force term in the equation of motion (2.33). ρ

.

∂ 2ξ ∂ξ +b − B∇ (∇ · ξ ) = 0 2 ∂t ∂t

(2.64)

Equation of motion (2.64) leads to the following decay wave equation of volume expansion. ∂ 2 (∇ · ξ ) ∂(∇ · ξ ) + 2β − v 2p ∇ 2 (∇ · ξ ) = 0 ∂t 2 ∂t Here .β is the decay constant and .v p is the phase velocity defined by (2.47). .

β=

.

b B , v 2p = 2ρ ρ

(2.65)

(2.66)

In the case of longitudinal and transverse waves in solids, we derived the longitudinal and transverse (shear) wave equations (2.56) and (2.62) from the corresponding equations of motion (2.50) and (2.58). Similarly to the wave equation in air, we can include the effect of velocity damping by adding the term proportional to the first-order time derivative of the displacement.

44

2 Wave Dynamics

∂ξ ( long )2 ∂ 2 ξ ∂ 2ξ + 2β =0 − vp ∂t 2 ∂t ∂x2 ∂η ( shear )2 ∂ 2 η ∂ 2η + 2β =0 . − vp ∂t 2 ∂t ∂x2 .

(2.67) (2.68)

As clear from the above arguments, the physical mechanism that generates acoustic waves in materials in air and solids, in common, is elasticity. The phase velocity has the form of the square root of the ratio of elastic constant over the density. When the medium is viscous, the velocity-damping term must be included. Thus, we can express the wave equation and phase velocity in the following general forms. 2 ∂ξ ∂ 2ξ 2∂ ξ + 2β =0 − v p ∂t 2 ∂t ∂x2 √ κ .v p = ρ

(2.69)

.

(2.70)

Here . f represents a quantity that exhibits the wave characteristics in general, .β is the damping coefficient, and .κ is the elastic constant. In the case of the waves in solids, .ξ is either longitudinal or shear displacement, and .κ is either Young’s modulus or shear modulus correspondingly. In the case of waves in air, . f can be the volume compression .∇ · ξ , pressure . p, or density change .δρ/ρ. The other types of elastic waves discussed in Chap. 4 can have a decay effect in the same fashion. Decaying acoustic wave solution Above we discussed that linear acoustic wave equations with a velocity damping mechanism can be expressed in the form of (2.69) commonly to in air and solids. In this section, we discuss general solutions to this type of wave equation. In Chap. 1, we used the cosine function (1.31) as a solution to the one-dimensional wave equation without a damping term (1.15). .

f (t, x) = A cos(ωt ± kx + φ)

(1.31)

The cosine function in (1.31) represents the oscillatory characteristics of the wave. In the wave equation, the combination of the secondary temporal and spatial derivative of the wave function represents the oscillatory characteristics. The square of the phase velocity that appears in front of the spatial secondary derivative term represents the elasticity, which is the physical origin of the oscillatory behavior. The damping term in (2.69) originates from the velocity damping effect, and does not cause the oscillatory behavior. The decaying wave equation (2.69) still has the secondary temporal and spatial derivative terms in the same form as the non-decaying wave equation.

2.2

Acoustic Wave Equations and Solutions

45

These observations indicate that a decaying wave solution can be expressed with a sine or cosine function, with the addition of a term that represents the decaying characteristics of the wave. So, it is natural to assume a solution that has a sinusoidal function as part of it. In this section, we use the exponential form of a sinusoidal function as it is more convenient to include the damping effect. Using Euler’s notation [16, 17], we can express (1.31) as follows. ξ(t, x) = Aei(ωt±kx) = A {cos(ωt ± kx) + i sin(ωt ± kx)}

.

(2.71)

We can view (1.31) as the real part of this exponential expression with initial phase .φ = 0. We set .φ = 0 because the initial phase does not affect the gist of the discussion. Substitution of (2.71) into (2.69) leads to the following equation. } { 2 2 2 ei(ωt±kx) = 0 . A (iω) + i2βω − v p (ik) (2.72) It follows that the content of the curly bracket is zero. ω2 − i2βω − v 2p k 2 = 0

.

(2.73)

In order for (2.73) to hold, it is necessary that either .ω or .k is a complex number. Let’s assume .k is complex, i.e., .k = kr + iki . ) ( ) ( 2 2 2 2 2 2 2 2 − i2 βω + v 2p kr ki = 0 (2.74) .ω − i2βω − v p (kr + iki ) = ω − v p kr + v p ki It follows that the following set of equations holds. ( ) ω2 = v 2p kr2 − ki2

.

βω =

.

−v 2p kr ki

(2.75) (2.76)

Elimination of .ki from (2.75) and (2.76) yields the following equation for .kr . v 2p (kr2 )2 − ω2 (kr2 ) −

.

(βω)2 =0 v 2p

(2.77)

Solve (2.77) as a quadratic equation of .kr2 taking only the real root to obtain the following expression. √ ( ⎡ ⎤ √ 2) √ ( )2 2 + ω4 + 4(βω)2 2 ω2 + ω4 1 + 4β 2 ω 2β ω ω 2 ⎦ = = 2 ⎣1 + 1 + .(kr ) = 2v 2p 2v 2p 2v p ω (2.78) From (2.75), we find .(ki )2 as follows.

46

2 Wave Dynamics

⎡ (ki )2 = (kr )2 −

.

ω2 v 2p

=

ω2

√

⎣−1 +

2v 2p

( 1+

2β ω

)2

⎤ ⎦

(2.79)

Thus, we find it as follows. ⎡ ω

kr = √ 2v p

.

ω

ki = √ 2v p

.

√

(

⎣1 +

1+

⎡

√

⎣−1 +

2β ω (

1+

)2

2β ω

⎤1 2

⎦

)2

(2.80) ⎤1 2

⎦

(2.81)

By separating the real and imaginary parts of .k, we can express the solution .ξ(t, x) (2.71) as follows. ξ(t, x) = Aei(iki x) ei(ωt±kr x) = Ae−ki x {cos(ωt ± kr x) + i sin(ωt ± kr x)}

.

(2.82)

The exponential term .ex p(−iki x) multiplied by amplitude . A represents the decaying effect as follows. As we discussed in Sect. 1.2.1, when the wave propagates in the positive direction, the phase term takes the form of .ωt − kx with a positive propagation constant .k. We set the complex propagation constant as .k = kr + iki . So, for a wave propagating in the positive .x direction,.ki > 0. This indicates as the wave propagates along the.x axis, i.e., with an increase in .x, the exponential term .ex p(−iki x) decreases. We can express a wave propagating in the negative .x direction by assuming .k is negative. In this case, the wave decays as it travels, too. As an increase of .x in the negative direction (i.e., as .x becomes more negative), the negative .ki < 0 makes the exponential term decay. So, regardless of the direction of propagation the wave decays as it travels. When the damping coefficient .b is negligible, we can put .β = 0 and (2.80) and (2.81) reduce to the following expressions. ]1 √ 1 ω [ ω ω 2 1+ 1+0 = √ kr = √ [2] 2 = =k vp 2v p 2v p ]1 √ 1 ω [ ω 2 −1 + 1 + 0 = √ .ki = √ [−1 + 1] 2 = 0 2v p 2v p

.

(2.83) (2.84)

Here (2.84) indicates that when .β = 0, .ki = 0, hence .kr = k. Using this fact, we expressed the phase velocity .v p as equal to .ω/k in (2.83). Thus, under the condition of .β = 0, the decay wave solution (2.82) reduces to the exponential form of the non-decaying solution (1.31).

2.2

Acoustic Wave Equations and Solutions

47

Three dimensional case Now quickly extend the above discussions regarding the decay wave solution into three dimensions by solving the three-dimensional wave equation. .

∂ξ ∂ 2ξ − v 2p ∇ 2 ξ = 0 + 2β ∂t 2 ∂t

(2.85)

As clear from the general discussion of extending a one-dimensional wave to a threedimensional wave made in Sect. 1.2.2, we can assume the solution in the following form. ξ (t, r) = Aei(ωt±k·r)

(2.86)

.

Use the Laplacian operator for the spatial differentiation of (2.86), as we did in Sect. 1.2.2. Expressing the magnitude of propagation vector .k with .k, we can express the second-order spatial differentiation as follows. ∇ 2 eik·r = ∇ 2 ei (k x x+k y y+kz z ) = (ik x )2 ei (k x x+k y y+kz z ) + (ik y )2 ei (k x x+k y y+kz z ) + (ik z )2 ei (k x x+k y y+kz z )

.

= −(k x2 + k 2y + k z2 )ei (k x x+k y y+kz z ) = −k 2 ek·r

(2.87)

Substitution of (2.86) into (2.85) with the use of (2.87) yields the following equation. } { 2 2 2 ei(ωt±k·r) = 0 . A (iω) + i2βω + v p k (2.88) Clearly, (2.88) leads to the condition (2.73) for the one-dimensional case. ω2 − i2βω − v 2p k 2 = 0

.

(2.73)

Thus, we find the same expressions of the complex propagation constant as the onedimensional case. In other words, except for the direction of propagation, the threedimensional wave (2.86) behaves exactly the same as the one-dimensional wave (2.71). It is instructive to express the spatial part of the phase .k · r that appears in the exponent ˆ . j, ˆ and .kˆ (the unit vectors for .x, . y, of the three-dimensional wave function (2.86) using .i, and .z directions). Referring to Fig. 2.9, express .k · r in the following form. ˆ k · r = k lˆ · (x iˆ + y jˆ + z k)

(2.89)

.

Here .lˆ is the unit vector in the direction of the propagation vector .k. (I do not use .kˆ for this unit vector to avoid confusion with the unit vector for the .z direction.) ˆ .x, . y, and .z components are the direction As discussed in Appendix A, the unit vector .l’s cosines of the respective axes. lˆ = cos θx iˆ + cos θ y jˆ + cos θz kˆ

.

(A.8)

48

2 Wave Dynamics

Fig. 2.9 Propagation vector .k and its unit vector whose components are direction cosines

Using (A.8) we can rewrite (2.89) in the following form. { } k · r = k (cos θx )x + (cos θ y )y + (cos θz )z

.

= (k cos θx )x + (k cos θ y )y + (k cos θz )z = (k x )x + (k y )y + (k z )z

(2.90)

We can interpret the quantities .k x , .k y , and .k z that appear in the last term of (2.90) as the .x, y, and .z components of the propagation constant. Thus, this term indicates that the wave propagates in the direction .cos θx iˆ + cos θ y jˆ + cos θz kˆ gaining the phase (including the decay factor) along the .x, . y, and .z axes with the respective components of .k. By separating the propagation constant into the real (oscillatory) and imaginary (decaying) parts, we obtain the following explicit forms.

.

ξ (t, r) = Ae(i(ωt±k·r )) = Aeiωt eikr ·r ei

.

2 k ·r i

= Aei(ωt+kr ·r) e−ki ·r

(2.91)

where { } kr · r = kr (cos θx )x + (cos θ y )y + (cos θz )z ⎤1 ⎡ √ ( )2 2 } 2β ⎦ { ω ⎣ (cos θx )x + (cos θ y )y + (cos θz )z 1+ 1+ =√ ω 2v p { } .ki · r = ki (cos θ x )x + (cos θ y )y + (cos θz )z ⎤1 ⎡ √ ( )2 2 } 2β ⎦ { ω ⎣ (cos θx )x + (cos θ y )y + (cos θz )z −1 + 1 + =√ ω 2v p

.

(2.92)

(2.93)

When the damping coefficient .b is negligible, we can put .β = 0 and (2.92) and (2.93) reduce to the following expressions. ]1 √ ω [ ω 2 1+ 1+0 = kr = √ =k vp 2v p ]1 √ ω [ 2 −1 + 1 + 0 = 0 .ki = √ 2v p

.

(2.94) (2.95)

2.2

Acoustic Wave Equations and Solutions

49

As is the one-dimensional case, the condition .β = 0 reduces the solution (2.86) to the nondecaying solution where .k is a real number.

2.2.4

Nonlinear Wave Equations

So far we assumed that the elastic modulus does not depend on the space coordinate. If the elastic modulus is a function of the space coordinate, the wave equation becomes nonlinear. Consider this effect for the case of longitudinal waves in a solid. Reconsider the equation of motion (2.50) allowing the stiffness (spring constant) to depend on .x. In this case, we need to include differentiation of the stiffness in the evaluation of the net force on the central block. m

.

( ) ∂ 2ξ ∂(ke ξ ' ) ' ' = k (x + Δx)ξ (x + Δx) − k (x )ξ (x ) δx = Δxδx e 1 1 e 1 1 ∂t 2 ∂x

Using (2.55) allowing . E to depend on .x, we can put (2.97) in the following form. ( ) ( ) ( 2 ) ∂ 2ξ ∂(Eξ ' ) ∂ ∂ξ ∂ E ∂ξ ∂ ξ .ρ = = E = + E ∂t 2 ∂x ∂x ∂x ∂x ∂x ∂x2

(2.96)

(2.97)

This situation makes it impossible to put the wave equation in the form of (2.56) where the secondary differentiation of the wave function (.ξ(t, x)) with respect to time is proportional to the secondary differentiation of the function with respect to space. Thus, the phase velocity is not a constant. The situation is the same as the case when the displacement is a vector or the medium has viscosity. ∂ξ ∂ 2ξ +b (2.98) .ρ = (∇ E) (∇ · ξ ) + E∇ (∇ · ξ ) 2 ∂t ∂t As an example, we can discuss nonlinear wave equations in solids by allowing the spatial coordinate dependence in the spring constant in (2.50), i.e., Youg’s modulus or shear modulus depends on the coordinate variable. This type of nonlinearity is caused by the fact that the restoring (elastic force) has second or higher-order dependence on the space coordinates. A good example of such a case is the recovery force of residual stress discussed in Fig. 2.2. When residual stress causes the strain so large that the atom deviates from the bottom of the inter-atomic potential where the potential curve is approximately a quadratic function of the inter-atomic distance, its slope becomes nonlinear [18, 19]. Let’s take a moment to consider this case. Let .U (r ) be the inter-atomic potential and have a third-order dependence on the distance from the equilibrium (the bottom of the potential curve) .r . U (r ) = ar 3 + br 2 + cr + d

.

Here .a, .b, .c, and .d are constant, and .r corresponds to .ξ − ξ0 in Fig. 2.2.

(2.99)

50

2 Wave Dynamics

The spatial first-order differentiation of potential energy is the restoring force. So, in this case, the restoring force . fr has a second-order dependence on .r . .

fr (r ) =

U (r ) = 3ar 2 + 2br + c dr

(2.100)

Elastic force is expressed as the product of the stiffness (the spring constant) .ke and the displacement from the equilibrium. From (2.100), we find the following expression of the stiffness for the potential (2.99). ke =

.

fr = 3ar + 2b r

(2.101)

Apparently, .ke is not a constant. Although nonlinear acoustics and related technology [20] are interesting topics, we mostly discuss linear acoustic waves in this book.

2.3

Wave as a Flow of Energy

In this section, we discuss acoustic waves as a flow of acoustic energy.

2.3.1

Acoustic Energy

Consider in Fig. 2.10 that a cubic block of elastic medium experiences tensile force applied on the planes at .x and .x + Δx. This pair of forces stretch the block by the differential displacement in the .x direction .Δξ . We can express the work done by these forces as follows. ∫ Δξ ∫ Δξ ) ( 1 .W = f dξ = (2.102) ksp ξ dξ = ksp (Δξ )2 2 0 0 Here .ksp is the stiffness (spring constant) of the elastic medium. Express the force in two different ways so that we can rewrite (2.102) independent of the dimension. Fig. 2.10 Work done by force f is stored as potential (elastic) energy of the elastic medium

.

2.3 Wave as a Flow of Energy

51

.

f = ksp Δξ = ksp

.

f = σ A = E∈ A

∂ξ Δx = ksp ∈Δx ∂x

(2.103) (2.104)

Here .∈ is the normal strain, .σ is the normal stress, . E is Young’s modulus, and . A is the cross-sectional area of the volume element. By equating (2.103) and (2.104), we find the following relationship between the spring constant and Young’s modulus. EA .k sp = (2.105) Δx Substitute (2.105) into (2.102). 1 EA 1 EA 1 .W = ksp (Δξ )2 = (Δξ )2 = 2 2 Δx 2 Δx

(

∂ξ Δx ∂x

)2 =

1 2 E∈ (AΔx) 2

(2.106)

Noting that .(AΔx) is the volume of the block, we find the following expression for the energy density. W 1 .w p = (2.107) = E∈ 2 AΔx 2 This energy density represents the elastic potential energy per unit volume. The suffix . p stands for the potential energy. Instantaneous acoustic energy Above, we made the analysis for static equilibrium. Now consider that the force (stress) is dynamic in the context of an acoustic wave. As we discussed in Sect. 1.2.3, a one-dimensional wave solution can be put in the following form (see (1.31)). ξ(t, x) = ξ0 cos(ωt − kx)

.

(2.108)

Here .ξ represents the displacement from equilibrium. (In Fig. 2.10 we use .ξ to represent the particle displacement from the origin, not necessarily from the equilibrium. Here we define .ξ to be from the equilibrium. This difference does not affect the gist of the discussion here.) The particle velocity is the time differentiation of (2.108), and the normal strain .∈ is the spatial differentiation of (2.108). ∂ξ = −ωξ0 sin(ωt − kx) ∂t ∂ξ .∈(t, x) = = kξ0 sin(ωt − kx) ∂x v(t, x) =

.

(2.109) (2.110)

Figure 2.11 illustrates sample snapshots of .ξ(ωt − kx), .∈(ωt − kx), and .v(ωt − kx) at two instances corresponding to .t = 0 and one-eighth of the period later. Note that the strain

2 Wave Dynamics

(au)

(au)

(au)

52

(au) Fig. 2.11 Snapshot of displacement (top), strain (middle), and velocity (bottom) waves at two timesteps. The horizontal axis is common to all plots

wave is a quarter period shifted from the displacement wave, and the velocity wave is .π out of phase to the strain wave. These phase shifts are explained by (2.110) and (2.109), respectively. From (2.109) and (2.110), we find the following relationship between .v(t, x) and .∈(t, x). k ∈=− v ω

.

(2.111)

As a side note, the negative sign in (2.111) indicates that the velocity wave is .π out of phase to the strain wave. Also, .k/ω is the reciprocal of the phase velocity. Equation (2.111) indicates that the ratio of the amplitude of the velocity wave over the strain wave is equal to the phase velocity and that the velocity and strain waves are out of phase by .π . Substitution of (2.111) into (2.107) yields the following expression. wp =

.

1 2 E E∈ = 2 2

( )2 k v2 ω

(2.112)

From (1.11) we know .ω/k is the phase velocity and from (2.24) we know the phase velocity √ of a longitudinal elastic wave is related to the density .ρ as . (E/ρ). Thus, we find the right-hand side of (2.112) as follows.

2.3 Wave as a Flow of Energy

53

Fig. 2.12 Potential energy density in a solid (a) and air (b). In both cases, the external force increases the acoustic energy density

E . 2

( )2 k E 1 2 E ρ 2 1 v2 = v = v = ρv 2 ≡ wk 2 ω 2 vp 2 E 2

(2.113)

Expression (2.113) represents the kinetic energy of particles in the unit volume, .wk . How about the acoustic potential energy density in air? Fig. 2.12 illustrates the situation where the external force . f acting normally on the plane of area . A increases the acoustic energy density. In (a) the external force stretches the volume and in (b) it compresses the volume. In (a) the stress is .σ/A and in (b) the pressure is . p/A. Let .a and .b represent the shorter and longer sides of the cube in the direction parallel to the external force (.a < b). Using Hooke’s law expressions in solids (2.17) and air (2.29), we obtain the following equations. (For the in-air case, we use a one-dimensional expression of (2.29).) Here, in both cases, the external force does positive work, hence the change in the acoustic energy is positive. ∫

b

dw =

.

a

∫

a

dw =

.

b

∫

) E 2b E( 2 [∈ ]a = b − a2 > 0 2 2 ∫a a ) (−B) 2 a (−B) ( 2 pd∈ = (−B)∈d∈ = [∈ ]b = a − b2 > 0 2 2 b

σ d∈ =

b

E∈d∈ =

(2.114) (2.115)

In summary, we find the instantaneous elastic potential energy and the particle’s kinetic energy as follows. 1 2 1 E∈ ; wk = ρv 2 ; w p = wk ; wtotal = w p + wk = E∈ 2 = ρv 2 = B∈ 2 2 2 (2.116) The above discussions indicate that at a given moment, the acoustic wave has potential energy density .w p and kinetic energy density .wk , and they are equal to each other. In Sect. 2.3.2, we find that an acoustic wave carries these two types of energy at the phase velocity. This situation is analogous to that an electromagnetic wave carries electric and magnetic energy densities that are equal to each other. wp =

.

Average acoustic energy What is the actual magnitude of the energy that flows as a wave? It is determined by the initial condition, i.e., how much the external agent initially pulls the unit volume, i.e., the

54

2 Wave Dynamics

actual value of .ξ0 in (2.109) and (2.110). Putting the amplitude of the velocity and strain as .∈ p = kξ0 and .v p = ωξ0 , we can express the temporal variation of the velocity and strain as follows. These expressions correspond the temporal oscillation of .v and .∈ at a fixed .x in (2.109) and (2.110). v(t) = v0 sin ωt

(2.117)

∈(t) = ∈0 sin ωt

(2.118)

.

.

We can express the average values over one period (.τ = (2π )/ω) for these energies as follows. .

.

< v 2 (t) > = < ∈ 2 (t) > =

ω 2π ω 2π

∫

2π ω

(v0 sin ωt)2 dt =

0

∫

2π ω

(∈0 sin ωt)2 dt =

0

ω 2 v 2π 0 ω 2 ∈ 2π 0

∫ ∫

2π ω

sin2 ωtdt =

1 2 v 2 0

(2.119)

sin2 ωtdt =

1 2 ∈ 2 0

(2.120)

0 2π ω

0

Thus, the average total energy density becomes as follows. av wtotal = wkav + wav p =

.

ρ 2 E 2 v + ∈0 = ρv02 = E∈02 2 0 2

(2.121)

Some note regarding equality of kinetic and potential energy The equality between the kinetic and potential energy leads to some intriguing insight into the acoustic field. From (2.116) we obtain the following equality. .

ρ E 2 ∈ = v2 2 2

Rearranging (2.122) we find the ratio of .v over .∈ as follows. √ |v| E = = vp . |∈| ρ

(2.122)

(2.123)

√ Here, from (2.57) we know . E/ρ is the phase velocity of the longitudinal wave characterized by the material constants . E and .ρ. Knowing that .v = dξ/dt and .∈ = dξ/d x, we can view the ratio of the amplitude of the velocity wave over the amplitude of the strain wave as representing the phase velocity via the following argument. | | | | | | |v| || dξ || || dξ || || d x || = / = = vp . (2.124) |∈| | dt | | d x | | dt | Note that .v and .∈ in (2.123) represent the amplitude of the respective waves. That is why a negative sign does not appear in this equality unlike (2.111) which relates the instantaneous value of the velocity wave and strain wave.

2.3 Wave as a Flow of Energy

55

Acoustic impedance In the theory of electronic circuits, impedance represents the difficulty of current flowing through a medium. This concept corresponds to the resistance in a DC (direct current) circuit, which is defined as the ratio of the voltage drop across a resistor over the current. In an AC (alternating circuit) transmission line or electromagnetic wave transmission in air, impedance is significant because when an AC current or electromagnetic wave passes through a boundary where the impedance varies before and after, reflection occurs. Consequently, the power is not transmitted as expected. Therefore, various techniques to match the impedance are used. This operation is called impedance matching [21]. The acoustic impedance.z ac is defined as the ratio of the amplitude of the stress (pressure) wave to the amplitude of the velocity wave. In the case of a longitudinal wave in a solid, we obtain the following expression for the acoustic impedance. See (2.123) in going through the third equal sign in the below equation. .

z ac =

ρv 2p σ E∈ E = = ρv p = = v v vp vp

(2.125)

This definition is analogous to that of the electrical impedance. In electrodynamics, the scalar potential (voltage) is the spatial integration of the electric field, which exerts an electric force on the electric charges that flow as an electric current. Here in (2.125), the acoustic impedance is the ratio of stress (force) over the velocity (flow). In the case of air, the acoustic impedance takes the following form. √ B air (2.126) .v p = ρ Applying the equality between the potential and kinetic energy density (2.122) to the case of air, we obtain the following equation. √ B 2 B ρ 2 v (2.127) . ∈ = v , i.e., = = v air p 2 2 ∈ ρ So, .

air z ac =

−p B∈ B = = air = v v vp

)2 ( ρ v air p v air p

= ρv air p

(2.128)

When we excite an ultrasound in a solid with an ultrasound transmitter, it is necessary to use a buffer, typically distilled water or ultrasound gel, to increase the coupling. The role of the buffer is to match the impedance. (Tip: If an ultrasound gel is not available, honey may work well as a buffer. [22])

56

2 Wave Dynamics

2.3.2

Intensity of Acoustic Wave

In the preceding section, we discussed that acoustic energy density is the product of pressure (stress) and velocity. From this discussion, we can envision that a pair of pressure (stress) and velocity waves carry acoustic energy. In this section, we discuss this topic quantitatively. Let’s use the displacement wave in the form of (2.108) for this discussion. In this case, the velocity and strain waves are as shown in (2.109) and (2.110). ∂ξ = −ωξ0 sin(ωt − kx) ∂t ∂ξ ∈(t, x) = = kξ0 sin(ωt − kx) ∂x

v(t, x) =

.

(2.109) (2.110)

Remembering that .ω/k is the phase velocity .v p (see (1.11)), we find the following relationship from (2.109) and (2.110). .

v(t, x) −ωξ0 sin(ωt − kx) ω = = − = −v p ∈(t, x) kξ0 sin(ωt − kx) k

(2.129)

The relationship (2.129) is nothing new. It is an iteration of (2.111) that states that ratio of velocity wave over the strain wave is equal to the phase velocity with a negative sign. As I mentioned under (2.111), the negative sign indicates that the velocity and strain waves are out of phase with each other by .π . In a sinusoidal wave, out of phase by .π is equivalent to flipping the sign of the wave function. Viewing .∈ = ∂ξ/∂ x as the one-dimensional version of volume expansion .∇ · ξ and using (2.30) that relates pressure and volume expansion, we obtain the following expression for pressure . P. . P(x, t) = −B∈(x, t) (2.130) Thus, using (2.129) we can express the product of the pressure and velocity waves in two ways as follows. .

P(x, t)v(t, x) = −B∈(t, x)v(t, x) = −B∈(t, x)∈(t, x)(−v p ) = B∈(t, x)2 v p = wtotal (t, x)v p

.

P(x, t)v(t, x) = Bkξ0 sin(ωt − kx)ωξ0 sin(ωt − kx) =

(2.131) (Bkωξ02 ) sin2 (ωt

− kx) (2.132)

Here going to the second line of (2.131), we use the total energy density expression (2.116). Dimensional analysis indicates that the . Pv wave is in (N/m.2 )(m/s). = (J/m.3 ) (m/s) = W/m.2 . This quantity is called intensity often expressed with a symbol . I . Expression (2.131) literally indicates that the . pv wave carries the total energy density at the phase velocity .v p . On the other hand, (2.132) indicates that the . Pv wave oscillates at the same frequency as the displacement wave and its amplitude is . Bkωξ0 . By expressing (2.132) with the pressure and velocity wave amplitude . B(kξ0 ) = B∈0 = p0 and .ωξ0 = v0 , we can find the average

2.3 Wave as a Flow of Energy

57

intensity as follows. See (2.119) and (2.120) for the equivalent expression for the velocity and strain waves. ∫ 2π ω ω 1 av p0 v0 .I = sin2 (ωt − kx)dt = p0 v0 (2.133) 2π 2 0 The above discussions indicate that the pressure and velocity wave together carry the acoustic energy. Figure 2.13 illustrates that the velocity and pressure waves are in phase, oscillating at the same frequency as the displacement wave. If we repeat the same procedure to express the product of the normal stress and velocity waves us .σ = E∈, we obtain the following pair of equations. σ (x, t)v(t, x) = E∈(t, x)v(t, x) = E∈(t, x)∈(t, x)(−v p ) = E∈(t, x)2 (−v p )

.

= wtotal (t, x)(−v p ) σ (x, t)v(t, x) = Ekξ0 sin(ωt − kx)(−ωξ0 sin(ωt − kx)) =

.

(2.134) −(Ekωξ02 ) sin2 (ωt

− kx) (2.135)

(au)

(au)

(au)

(au)

This time, unlike the product of pressure and velocity wave, we have a negative sign in front of the phase velocity .v p . Using Fig. 2.13, we can explain this discrepancy between the cases of the pressure wave and stress wave as follows. The stress wave is out of phase from the displacement wave by .π/2 and from the velocity and pressure waves by .π . The phase shift of .±π/2 between the displacement wave and the other three waves originates from the fact that these three waves are temporal or spatial derivatives of the displacement wave.

(au) Fig. 2.13 Displacement, stress, pressure, and velocity waves. Note that the pressure and velocity waves are in phase. They together carry acoustic energy according to (2.131)

58

2 Wave Dynamics

The.π phase difference between the stress and pressure waves, which have the same physical dimension of N/m.2 , is simply due to the difference in the definition; a stretch of an elastic medium corresponds to positive stress and negative pressure, i.e., .σ = E∈ vs . P = −B∇ · ξ . From this standpoint, we can say that.−σ0 sin(ωt − kx) carries acoustic energy together with the velocity wave .v0 sin(ωt − kx). In any event, the intensity is defined as the absolute value of the time average of the product of the component waves, i.e., the pressure and velocity waves or the stress and velocity waves. So, it is always a positive value.

2.4

Acoustic Wave Reflection and Transmission

2.4.1

Laws of Reflection and Refraction

Refer to Fig. 2.14. The acoustic wave is incident to the boundary AB from the left. The acoustic wave travels in the medium on the left of the boundary with a phase velocity .v1 , and in the medium on the right of the boundary with .v2 . The frequency of the wave is constant through the boundary. Therefore, the wavelength on the left of the boundary, .λ1 is different from that on the right, .λ2 . Lines N.1 A and N.2 B are normal to boundary AB. The angle made by these lines and AB is called the angle of incidence .θi . The dashed lines represent the wavefront of three waves, i.e., the incident wave, reflected wave, and transmitted wave (the crest of the respective waves). Lines S.1 A and S.2 B are perpendicular to the wavefronts, representing the propagation of the incident wave. Figure 2.14 depicts the moment when the left edge of the wavefront of the incident wave reaches the boundary at point A. Therefore, line DB is equal to wavelength .λ1 . Similarly, on the right of the boundary, line AE is equal to the wavelength .λ2 . S A

N S

E C D

N B

Fig. 2.14 Law of reflection and refraction

2.4

Acoustic Wave Reflection and Transmission

59

Consider triangle ABD. Angle ABD is complementary to angle DBN.2 . Angle BAD is complementary to angle ABD because angle ADB is a right angle. Therefore angle BAD is equal to angle DBN.2 , which is the angle of incidence .θi . Length BD is equal to the wavelength in medium 1, .λ1 . We obtain the following equation. BD = AB sin θi = λ1

.

(2.136)

Repeating the same argument for triangle ABC, we find the following equation. AC = AB sin θr = λ1

.

(2.137)

From (2.136) and (2.137) we find the following equality, which is known as the law of reflection; the angle of reflection is equal to the angle of incidence. θr = θi

.

(2.138)

Now pay attention to triangle ABE. We can easily find the following equation. AE = AB sin θt = λ2

.

(2.139)

From (2.136) and (2.139) we find the following equality, which is known as the law of refraction, or Snell’s law. .λ1 sin θt = λ2 sin θi (2.140) Since the wavelength is proportional to the phase velocity for the same frequency, we can express Snell’s law in the following form as well. v1 sin θt = v2 sin θi

.

(2.141)

Here .v1 and .v2 are the acoustic phase velocity in the medium on the left and right of the boundary. Now consider that a wave is incident to a boundary from medium 1 to medium 2 where the phase velocity in medium 1 is higher than medium 2, i.e., .v1 > v2 . Solving (2.141) for .sin θt , we obtain the following expression. .

sin θt =

v2 sin θi v1

(2.142)

Since the sine function .sin θi cannot take a value greater than unity and .v2 /v1 > 1, (2.142) does not hold for all .θi . The following condition is necessary. ( ) −1 v1 (2.143) .θc ≡ θi ≤ sin v2 This angle .θc is referred to as the critical angle.

60

2 Wave Dynamics

2.4.2

Behavior of Acoustic Waves Near Boundary

When an acoustic wave passes through a boundary that separates two elastic media of different elasticity and density, part of the wave energy is reflected and the rest is transmitted. If the angle of incidence is oblique, the direction of the transmitted wave is determined by Snell’s law. The amount of reflected and transmitted energy is determined by the boundary conditions. In this section, we discuss these topics. Formulation of incident, reflected, and transmitted waves We first consider acoustic waves in solids. In this section, we use the exponential form as it avoids complexity in the formulation. Set the incident, reflected, and transmitted displacement in the following form. ξi (t, x) = ξi0 ei(ωt−k1 x) ; ξr (t, x) = ξi0 ei(ωt+k1 x) ; ξt (t, x) = ξi0 ei(ωt−k2 x)

.

(2.144)

Here .ξ0i , .ξ0r , and .ξ0t , are the amplitude of the respective waves. Subscripts .1 and .2 for Young’s modulus . E and propagation constant .k denote medium 1 (the incident side) and medium 2 (the transmitted side). ∂ξi ∂ξi = E 1 (−ik1 )ξi (t, x); vi (t, x) = = (iω)ξi (t, x) ∂x ∂t ∂ξr ∂ξr .σr (t, x) = E 1 = E 1 (ik1 )ξr (t, x); vr (t, x) = = (iω)ξr (t, x) ∂x ∂t ∂ξt ∂ξt .σt (t, x) = E 2 = E 2 (−ik2 )ξt (t, x); vt (t, x) = = (iω)ξt (t, x) ∂x ∂t σi (t, x) = E 1

.

(2.145) (2.146) (2.147)

Remembering the definition of acoustic impedance (2.125), we obtain the following expressions. σi E 1 (−ik1 )ξi (t, x) −E 1 k1 E1 = = −z 1 = =− vi (iω)ξi (t, x) ω v p1 σr E 1 (ik1 )ξr (t, x) E 1 k1 E1 = = z1 . = = vr (iω)ξr (t, x) ω v p1 σt E 2 (−ik2 )ξt (t, x) −E 2 k2 E2 . = = −z 2 = =− vt (iω)ξt (t, x) ω v p2 .

(2.148) (2.149) (2.150)

Notice that the magnitude of the acoustic impedance is the same for the incident and reflected waves as they are in the same medium. Their signs are opposite because the propagations are opposite. The acoustic impedance of the transmitted wave is different from the impedance of the incident or reflected wave as the transmitted beam propagates in a different medium.

2.4

Acoustic Wave Reflection and Transmission

61

Now we make the same discussion as above for acoustic waves in air. According to (2.29) and (2.54), the only difference between the in-air and in-solids cases is the elastic modulus’s sign. (You may notice that while (2.29) is expressed in three dimensions for air, (2.54) is in one dimension for solids. We will discuss Hooke’s law in solids in a three-dimensional form in Chap. 4.) Replace . E with .−B and .σi with . pi in (2.145)–(2.150) to obtain the following equations. ∂ξi ∂ξi = −B1 (−ik1 )ξi (t, x); vi (t, x) = = (iω)ξi (t, x) ∂x ∂t ∂ξr ∂ξr . pr (t, x) = −B1 = −B1 (ik1 )ξr (t, x); vr (t, x) = = (iω)ξr (t, x) ∂x ∂t ∂ξt ∂ξt . pt (t, x) = −B2 = −B2 (−ik1 )ξt (t, x); vt (t, x) = = (iω)ξt (t, x) ∂x ∂t .

pi (t, x) = −B1

−B1 (−ik1 )ξi (t, x) pi B1 k1 B1 = = z1 = = vi (iω)ξi (t, x) ω v p1 pr −B1 (ik1 )ξr (t, x) −B1 k1 B1 = = −z 1 . = =− vr (iω)ξr (t, x) ω v p1 −B2 (−ik2 )ξt (t, x) pt B2 k2 B2 = = z2 . = = vt (iω)ξt (t, x) ω v p2 .

(2.151) (2.152) (2.153)

(2.154) (2.155) (2.156)

Similarly to the case in solids, the acoustic impedance for the incident and reflected waves have the same magnitude and opposite signs, whereas that for the transmitted wave is different. Boundary conditions and reflectivity At the boundary, the following two conditions must hold for the pressure and particle velocity. vi cos θi + vr cos θi = vt cos θt

.

σi + σr = σt

.

(2.157) (2.158)

Condition (2.158) represents the force balance at the boundary plane. If the balance is broken, the boundary plane will have acceleration, and consequently, it moves in the positive or negative .x direction. Condition (2.157) represents the law of momentum conservation along the .x axis, i.e., perpendicular to the boundary plane. Per condition (2.158), the net force acting on the particle along the .x is zero. Using (2.148)–(2.150) in (2.158) we obtain the following equation. .

− z 1 vi + z 1 vr = −z 2 vt

Eliminate .vt from (2.157) and (2.159).

(2.159)

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2 Wave Dynamics

vi (z 2 cos θi − z 1 cos θt ) + vr (z 2 cos θi + z 1 cos θt ) = 0

.

(2.160)

Similarly, eliminate .vr to obtain the following equation. 2vi z 1 cos θi = vt (z 1 cos θt + z 2 cos θi )

.

(2.161)

From (2.160) and (2.161), we find the reflection and transmission coefficient for the velocity wave as follows. z 1 cos θt − z 2 cos θi vr = vi z 1 cos θt + z 2 cos θi 2z 1 cos θi vt = .tv = vi z 1 cos θt + z 2 cos θi rv =

.

(2.162) (2.163)

Substituting (2.148)–(2.150) into (2.162) and (2.163), we can express the reflection and transmission coefficient in terms of the stress waves as follows. vr z 1 σr −z 1 cos θt + z 2 cos θi = = σi vi (−z 1 ) z 1 cos θt + z 2 cos θi σt vt (−z 2 ) 2z 2 cos θi .tσ = = = σi vi (−z 1 ) z 1 cos θt + z 2 cos θi rσ =

.

(2.164) (2.165)

Note that (2.162)–(2.165) represent the reflection and transmission of the stress and velocity waves. Therefore, each of these expressions does not represent the reflection or transmission of the acoustic energy. The product of the stress (N/m.2 ) and velocity (m/s) represents the intensity (N/m.2 · m/s.=W/m.2 ), i.e., the area density of acoustic power. As we discussed in Sect. 2.3.2, it is the product of the pressure (stress) and velocity waves that represents the flow of acoustic intensity. Energy conservation at boundary Now express the intensity of the incident wave striking the boundary plane by considering the product of the stress and velocity waves. From (2.144) and (2.145), we obtain the following expression. .

Ii = |σi vi | cos θi = |(−ik1 E 1 )ξi (iω)ξi | cos θi = (k1 E 1 ω)ξi2 cos θi

(2.166)

In (2.166), the factor.cos θi represents the component of the intensity parallel to the boundary plane (see Fig. 2.14). Since we are considering the event on a fixed .x where the boundary is located, the intensity is a function of time. In other words, the intensity . Ii oscillates with angular frequency .ω according to .ξi (t, x0 ) = ξi0 ei(ωt−k1 x0 ) . Here .x0 represents the .x coordinate where the boundary is located. Similarly, we can formulate the reflected and transmitted intensities.

References

.

63

Ir = |σr vr | cos θi = |(rσ σi )(rv vi )| cos θi |( | )( ) | −z 1 cos θt + z 2 cos θi | z 1 cos θt − z 2 cos θi (iω)ξi || cos θi = || (−ik1 E 1 )ξi z 1 cos θt + z 2 cos θi z 1 cos θt + z 2 cos θi | | | −(z 1 cos θt − z 2 cos θi )2 | (z 1 cos θt − z 2 cos θi )2 2| . = | | (z cos θ + z cos θ )2 (k1 E 1 ω)ξi | cos θi = (z cos θ + z cos θ )2 Ii (2.167) 1 t 2 i 1 t 2 i

.

It = |σt vt | cos θt = |(tσ σi )(tv vi )| cos θt |( | )( ) | | 2z 2 cos θi 2z 1 cos θi | (iω)ξi || cos θt =| (−ik1 E 1 )ξi z 1 cos θt + z 2 cos θi z 1 cos θt + z 2 cos θi | | | 4z 1 z 2 cos θi cos θt | 4z 1 z 2 cos θi cos θt 2| | =| (k E ω)ξ (2.168) 1 1 i | cos θi = (z cos θ + z cos θ )2 Ii (z 1 cos θt + z 2 cos θi )2 1 t 2 i

So, the total reflected and transmitted intensity is ) ( 4z 1 z 2 cos θi cos θt (z 1 cos θt − z 2 cos θi )2 Ii . Ir + It = + (z 1 cos θt + z 2 cos θi )2 (z 1 cos θt + z 2 cos θi )2 (z 1 cos θt )2 + (z 2 cos θi )2 − 2z 1 z 2 cos θi cos θt + 4z 1 z 2 cos θi cos θt Ii (z 1 cos θt + z 2 cos θi )2 (z 1 cos θt + z 2 cos θi )2 = Ii = Ii (2.169) (z 1 cos θt + z 2 cos θi )2 =

Equation (2.169) indicates that through the boundary the acoustic energy is conserved. By defining the reflectance . R = Ir /Ii and transmittance .T = It /Ii we obtain the following simple expression from (2.169). .

R+T =

Ir + It =1 Ii

(2.170)

In the case of acoustic waves in air, .vi , .vr , and .vt are the same as solids (compare (2.145)– (2.147) and (2.151)–(2.153)). Therefore .rv and .tv are the same as solids. For .r p and .t p , we need to replace . E 1 and . E 2 with .−B1 and .−B2 in the expression of the corresponding . p waves. This change results in the reflectivity and transmissivity as follows. −z 1 cos θt + z 2 cos θi pr vr (−z 1 ) = .r p = = (2.171) pi vi z 1 ) z 1 cos θt + z 2 cos θi pt vt z 2 2z 2 cos θi .t p = = = (2.172) pi vi z 1 z 1 cos θt + z 2 cos θi After all, both coefficients are the same as the solid’s case (2.164) and (2.165), if expressed with acoustic impedances.

References 1. Addink CCJ (1951) Precise calibration of tuning forks. Philips Tech Rev 12(8):228–232 2. Harvard Natural Sciences Lecture Demonstrations, https://sciencedemonstrations.fas.harvard. edu/presentations/tuning-forks (accessed on July 17, 2023)

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3. https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/harmonic-motion.aspx 4. Blake RE (1996) Basic vibration theory. In: Harris CM (ed) Shcok and vibration handbook, 4th edn, Ch. 2. McGraw-Hill, New York 5. Margenau H, Murphy GM (1956) The mathematics of physics and chemistry, 2nd edn. D. Van Nostrand, Co. Inc., Toronto, New York, pp 50–53 6. King GC (2009) Vibrations and waves. Wiley, Chichester, UK, pp 33–48 7. øAström KJ, Murray RM (2021) Feedback systems, an introduction for scientists and engineers. Princeton University Press, Princeton, London, pp 27–29 8. Obando L (2018) Mass-spring-damper system, 73 exercises resolved and explained: systems dynamics and transfer function. Universidad Central de Venezuela, Caracas, Venezuela 9. Yoshida S (2017) Waves; fundamental and dynamics. Morgan & Claypool, San Rafael, CA, USA, IOP Publishing, Bristol, UK, Chap 1 10. Jones JE (1924) On the determination of molecular fields.—I. From the variation of the viscosity of a gas with temperature. Proc R Soc Lond Ser A 106(738):441–462 11. Toupin RA, Bernstein B (1961) Sound waves in deformed perfectly elastic materials. Acoustoelastic effect. J Acoust Soc Am 33:216 12. Mavko G, Mukerji T, Dvorkin J (2020) The rock physics handbook, 3rd edn. Cambridge University Press, New York, Chap. 2 13. Reddy JN (2013) An introduction to continuum mechanics, 2nd edn. Cambridge University Press, Cambridge, UK 14. Tonya Coffey, Drag, https://www.youtube.com/watch?v=xKfZ25yBxu8 (accessed on July 19, 2023) 15. Hooge C. 5.2 drag forces. In: BCIT physics 0312 textbook. https://pressbooks.bccampus.ca/ physics0312chooge/chapter/5-2-drag-forces/ 16. Stipp D (2017) A most elegant equation; Euler’s formula and the beauty of mathematics. Basic Books, New York, USA 17. Boas ML (2006) Mathematical methods in the physical sciences, 3rd edn. Wiley, London, p 61 18. Hughes DS, Kelly JL (1953) Second-order deformation of solids. Phys Rev 92(5):1145–1149 19. Yoshida S, Sasaki T, Usui M, Sakamoto S, Gurney D, Park IK (2016) Residual stress analysis based on acoustic and optical methods. Materials 9:112 20. Garrett SL (2020) Nonlinea acoustics. In: Understanding acoustics. Graduate texts in physics. Springer, Cham. https://doi.org/10.1007/978-3-030-44787-8_15 21. Rehman A (2018) Practical impedance matching, 1st edn. Signal Processing Group Inc., Tempe, AX, USA 22. Miyasaka C (2012) Private communication

3

Propagation of Acoustic Waves in Air

This chapter specifically discusses the propagation of acoustic waves in air. By considering the source of elasticity of air, we derive the equation of motion and corresponding wave equations. After these discussions of wave dynamics, we touch on the sounds we hear. We briefly discuss how human ears conceive and process sound waves. We consider how musical instruments generate their unique sounds. Toward the end of the chapter, we discuss amplitude and frequency modulations as methods to transmit sound waves over a distance.

3.1

Dynamics of Air Particles and Acoustic Wave Equation in Air

3.1.1

Particle Motions in Longitudinal Waves

In Sect. 2.2.1 we derived the longitudinal wave equations for volume compression and pressure from the equation of motion. There we expressed the equation of motion. In this process, we used the displacement of air particles to derive the equation of motion. These procedures indicate that the deviation of air particles from the equilibrium is the underlying mechanism to generate the wave dynamics of air particles. We start this chapter by visualizing the relationship between displacement and pressure gradient. Figure 3.1 is a snapshot of air particles where a speaker generates a sound wave in the air. The diaphragm of the speaker moves sinusoidally. The dots represent the position of layers of particles. In Fig. 3.1a, the horizontal arrows represent the deviation of layers from the equilibrium position. In those locations where the dots are dense (space), the pressure is high (low). Figure 3.1b illustrates the pressure variation as a function of .x, the coordinate variable along the axis of wave propagation.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_3

65

66

3 Propagation of Acoustic Waves in Air

Fig. 3.1 Snapshots of air particles a, pressure wave b, and displacement wave c, .λ and .τ represent the wavelength and period of the waves

Pay attention to the second positive peak of the pressure graph. In Fig. 3.1a, the horizontal arrows are inward to this .x location, indicating that the corresponding displacement vector changes the sign over this location. In fact, on the negative-x (positive-x) side of this location, the arrow is rightward (leftward), meaning that the displacement is positive (negative). Figure 3.1c explicitly illustrates this change in the sign. Algebraically, we can express the above relationship between pressure and displacement as . p = −∂ξ/∂ x; the negative slope of .ξ(x) corresponds to positive pressure (see (2.29)). The dashed lines in Fig. 3.1b, c are the spatial variation of the respective quantities .Δt later. Notice that the pressure wave travels to the right so does the displacement wave. Figure 3.1d illustrates this temporal change observed at the leftmost .x position. During the elapse of .Δt, the pressure at this point decreases while displacement increases.

3.1.2

Wave-Generating Mechanism in Air

Air may not sound like an elastic medium. So, in this section, we discuss what causes elasticity in air. Here we analyze the dynamics in one dimension to make the analysis easily visible. In short, we can say as follows: The pressure gradient is proportional to the density gradient (gas law). The density gradient is the displacement gradient. Therefore, the pressure

3.1

Dynamics of Air Particles and Acoustic Wave Equation in Air

67

Fig. 3.2 Change in air density due to pressure gradient

gradient is proportional to the displacement gradient. This is how Hooke’s law applies to air. Below, we will discuss these dynamics quantitatively. For simplicity, let’s consider the dynamics in one dimension. Figure 3.2 illustrates the situation where a volume of air expands along the .x-axis. Before the expansion, this volume has a cross-sectional area of . S and a short length along the .x axis. Due to the pressure dependence on .x, the cross-sectional plane at .x = x2 is displaced more than that at .x = x1 (.x2 > x1 ). Since we consider a one-dimensional case, the cross-sectional area is unchanged during the expansion. In the following sections, we discuss this phenomenon in detail. Density change due to volume expansion Let .ξ(x, t) be displacement as a function of coordinate variable .x and .l0 be the length of the volume along the .x-axis before the expansion. Thus, the volume before the expansion (.V0 ) and after (.V ) are given, respectively, as follows. .

V0 = Sl0

(3.1)

V = S{l0 + (ξ(x2 ) − ξ(x1 )} ≡ S(l0 + Δξ )

(3.2)

.

So, the volume change .ΔV becomes, ΔV = V − V0 = SΔξ

.

(3.3)

Since the process is an expansion, the mass of the volume is unchanged while the volume increases. Therefore, we can express the density before the expansion (.ρ0 ) and after (.ρ) as follows. m V0 m m = .ρ = V V0 + ΔV

ρ0 =

.

(3.4) (3.5)

Here .m is the mass. From (3.4) and (3.5), we find the following expression for the change in the density.

68

3 Propagation of Acoustic Waves in Air

Δρ =

.

m −ΔV m =m − V0 + ΔV V0 (V0 + ΔV )V0

(3.6)

Now consider the infinitesimal limit where we can express .l0 = d x and assume .ΔV sin θ2 . It follows that .θ2 < θ1 . From the law of reflection, the angle of reflection for the P-wave is the same as the angle of incidence. These arguments all together indicate that no matter how great the angle of incidence may be the incident P-wave is reflected back to the elastic medium. This situation

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

107

is contrastive to the case when an incident SV-wave is converted to a P-wave, as discussed below. (3) Incident SV wave We can repeat the same type of argument as the preceding section to discuss the case when the incident wave is an SV-wave. In this case, we can put . A1 = 0 in (4.85) as there is no forward-going component in the P-wave. Consequently, we obtain the following expressions. φ r e f l (x, y, t) = A2 e−ik1 cos θ1 y ei(ωt−k1 sin θ1 x)

(4.123)

. .

Hz (x, y, t) = B1 e

ik2 cos θ2 y i(ωt−k1 sin θ1 x)

e

+ B2 e

−ik2 cos θ2 y i(ωt−k1 sin θ1 x)

e

(4.124)

k12 sin 2θ1 A2 + k22 cos 2θ2 (B1 + B2 ) = 0

(4.125)

2 . − k 2 cos 2θ2 A 2

(4.126)

The boundary conditions (4.81) and (4.82) become as follows. .

− k22 sin 2θ2 (B1

− B2 ) = 0

By solving (4.125) and (4.126) we obtain the following reflection coefficients. .

k 2 sin 2θ1 sin 2θ2 − k22 cos2 2θ2 B2 = 12 B1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2

(4.127)

.

2k22 sin 2θ2 cos 2θ2 A2 =− 2 B1 k1 sin 2θ1 sin 2θ2 + k22 cos2 2θ2

(4.128)

Figure 4.8 plots the reflection coefficients (4.127) and (4.128). Here the right vertical axis indicates the reflection coefficient when the mode is unconverted on reflection whereas the left vertical axis indicates the reflection coefficient when the SV mode is converted to the P mode. The insert in Fig. 4.8 illustrates the rays of the incident and reflected waves. Note that in this case, unlike the case when a P-wave is converted to an SV-wave, the angle of reflection is greater than the angle of incidence on the SV- to P-wave conversion. This indicates that if we increase the angle of incidence, at a certain point the angle of reflection reaches .π/2. Beyond this point, according to (4.84) .sin θ2 exceeds unity and it becomes impossible to find the angle of reflection from this equation. The angle of incidence that makes the angle of reflection be .π/2 is called the critical angle .θc . (See (2.143).) When the angle of incidence is greater than .θc , the resulting P-wave becomes confined in the subsurface region of the elastic material and decaying exponentially away from the surface. We will discuss this phenomenon in the next section.

108

4 Propagation of Acoustic Waves in Solids

Fig. 4.8 Reflection coefficient as a function of incident angle for several Poison’s ratios when the incident wave is a SV-wave

4.2.3

Surface Waves

At the end of the last section, we touched upon the existence of a wave confined near the free surface in association with the SV- to P-wave mode conversion occurring at an angle of incidence greater than a critical angle. In this section, we discuss such surface waves in more general. Let .θ2c be the critical angle for an SV-wave. By definition, the corresponding angle of refraction for the transmitting P-wave is .π/2. According to (4.84) the situation can be expressed for the critical angle and angle .θ2' > θ2c as follows. k2 sin θ2c = k1 sin

.

k2 sin θ2' > k1

π = k1 2

.

(4.129) (4.130)

From (4.70) and (4.84) we know k12 =

.

ω2 c12

k1x = k1 sin θ1 k2 sin θ2 . sin θ1 = k1 .

(4.131) (4.132) (4.133)

Substituting (4.131)–(4.133) into (4.61) and using condition (4.130), we find as follows.

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

2 k1y

.

( )2 ) ( ω2 sin θ k 2 2 2 = 2 − k1x = k12 (1 − sin2 θ1 ) = k12 1 − c2 , it follows .k¯1y > k¯2y . Thus (4.155) indicates that |ξx | < |ξ y |

.

(4.158)

The type of surface wave expressed by (4.154) and (4.155) is known as the Rayleigh surface wave [5]. Figure 4.9 illustrates a sample Rayleigh surface wave traveling in the positive . x-direction. The spatial periodicity is seen to move to the right. Here, .c = ω/(k 1 sin θ1 ) can be interpreted as the wave velocity. (See the exponential term of (4.154) and (4.155).) Note that Rayleigh surface waves are non-dispersive. The insert to the right illustrates the elliptical pattern of the displacement vector as a function of time. Notice that when the Rayleigh wave travels in the positive .x direction, the trajectory of the displacement vector along the elliptical line is counterclockwise. The above discussions cover only the gist of the Rayleigh wave. For more information about Rayleigh waves, see Refs. [6, 7].

4.2.4

Waves Propagating in Plates

In this section, we extend the discussion on acoustic waves in an elastic medium with a free surface that we made in Sect. 4.2.2. The elastic medium has two free surfaces, instead of one. The governing equation (the wave equation) is the same as the one-free surface case

112

4 Propagation of Acoustic Waves in Solids

Fig. 4.9 Rayleigh wave traveling in positive .x-direction. The illustration on the right represents the elliptical pattern of the displacement vector as a function of time

but the other free surface gives an additional boundary condition, which makes the resultant displacement vector behave considerably different from the one free-surface case. The most important difference here is that the displacement wave exhibits dispersion. This feature has important engineering applications [8]. The aim of this section is to understand the mechanism that generates dispersion through some simple cases. For more general discussions about the dispersion of waves in plates, see Ref. [9, 11]. SH waves in a plate As is the case of the one free-surface case, we start with the SH-wave case since it is the simplest. The governing equation is the same as (4.106). ∇ 2 ξz =

.

1 ∂ 2 ξz c22 ∂t 2

(4.106)

As before, we consider a plane wave with the .∂/∂z = 0 condition. The solution can be put as follows. ( ) ˜ i k˜ y −i k˜2y y ei(ωt−k2x x) .ξz (x, y, t) = E 1 e 2y + E 2 e ) ( ˜ = E 1 (cos k˜2y y + i sin k˜2y y) + E 2 (cos k˜2y y − i sin k˜2y y) ei(ωt−k2x x) ) ( ˜ = (E 1 + E 2 ) cos k˜2y y + i(E 1 − E 2 ) sin k˜2y y ei(ωt−k2x x) ) ( ˜ ≡ E˜ 1 cos k˜2y y + E˜ 2 sin k˜2y y ei(ωt−k2x x) (4.159)

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

113

Fig. 4.10 Plate elastic medium with two free surfaces

where .

ω2 2 2 = k˜2x + k˜2y c22

(4.160)

Now continue the analysis with a plate of thickness.2b having infinitely large free surfaces at . y = ±b. The free surfaces are parallel to the .zx plane. Figure 4.10 illustrates the plate geometry. With the geometry shown in Fig. 4.10, the free surface boundary condition relevant to the SH-wave can be expressed as follows. ( ( ) ) ∂ξ y ∂ξz ∂ξz (4.161) .σ yz | y=±b = μ − | y=±b = μ | y=±b = 0 ∂y ∂z ∂y Here .∂/∂z = 0 is used for the plane wave solution. Substituting solution (4.159) into (4.161) we obtain the following equations. { } ˜2y − E˜ 1 sin k˜2y b + E˜ 2 cos k˜2y b ei(ωt−k˜2x x) = 0 .σ yz | y=b = k (4.162) { } ˜2y E˜ 1 sin k˜2y b + E˜ 2 cos k˜2y b ei(ωt−k˜2x x) = 0 .σ yz | y=−b = k (4.163) For the above conditions hold for any .x, the inside the curly brackets must be zero. This leads to the following condition. .

cos k˜2y b cos k˜2y b E˜ 1 =− = sin k˜2y b sin k˜2y b E˜ 2

(4.164)

It follows that .(cos k˜2y b) (sin k˜2y b) = −(cos k˜2y b) (sin k˜2y b) = 0, or (cos k˜2y b) (sin k˜2y b) = 0

.

(4.165)

leading to the following condition. nπ k˜2y b = ± , n = 0, 1, 2, 3, 4, . . . 2

.

(4.166)

114

4 Propagation of Acoustic Waves in Solids

Fig. 4.11 SH symmetric waves with even mode numbers and anti-symmetric waves with odd mode numbers in the plate

Condition (4.166) means as follows. When an SH-wave of a given frequency .ω travels through a plate in the .x-direction undergoing numerous reflections at the two free surfaces, the spatial frequency in the . y-direction (along the thickness) must satisfy condition (4.166). For a given thickness .b, the spatial frequency can be any of .k˜2y = ±(nπ )/(2b). Here, the integer .n is called the mode number and the corresponding wave is called the mode .n wave. When .n is an odd integer, .cos k˜2y b = 0 and .sin k˜2y b = 1 in (4.165). From (4.162) it follows that . E˜ 1 = 0. This makes the cosine term null in the .ξx (x, y, t) solution (4.159). Similarly, when .n is an even integer, the sine term of the .ξx (x, y, t) solution becomes null. Thus, we find the following expressions for .ξx (x, y, t) under the condition of free surfaces at . y = ±b. ) ( ˜ 2 sin nπ y ei(ωt−k˜2x x) , n = 1, 3, 5, 7, . . . .ξz (x, y, t) = E (4.167) ) 2b ( ˜ 1 cos nπ y ei(ωt−k˜2x x) , n = 0, 2, 4, 6, . . . (4.168) .ξz (x, y, t) = E 2b Apparently, solution (4.167) presents anti-symmetric waveforms with respect to . y-axis, and solution (4.168) presents symmetric waveforms. Figure 4.11 illustrates several antisymmetric and symmetric waves for low mode numbers. The displacement vector is polarized along the .z-axis and the waves travel along the .x-axis (normal to the page). Now consider the velocity of the SH plate wave. The wave solution (4.159) indicates that the SH plate-wave travels in the positive .x-direction at velocity .c = ω/k˜2x . From the frequency equations (4.160) and (4.166), the spatial frequency along the .x-axis takes the following form.

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

k˜2x =

/ (

.

ω c2

)2 −

) nπ (2 2b

115

(4.169)

Note that .c2 represents the phase velocity of the SH-wave incident to and reflected from the free surfaces parallel to the .x-axis, and is a constant. On the other hand, the phase velocity of the SH-wave traveling along the .x-axis .(c ≡ ω/k˜2x ) depends on .ω as shown below. ⎧/ ⎫−1 ( )2 ) ⎨ ( 2 1 ω nπ ⎬ .c = = − . ⎩ c2 2bω ⎭ k˜2x

(4.170)

In other words, this SH-wave exhibits dispersion. We can find the group velocity as .vg = ∂ω/∂ k˜2x [12]. Below, we consider this dispersion in some detail. First, normalize the spatial and temporal frequencies as follows so that they become dimensionless. 2bk2x π 2bω .Ω = c2 π ξ¯ =

.

(4.171) (4.172)

By substituting (4.171) and (4.172) into (4.169), we obtain the following dispersive relation. √ ¯ = ± Ω2 − n 2 (4.173) .ξ Here, the dimensionless frequency .Ω and the mode number .n are real. Therefore, if .Ω < n, the dimensionless spatial frequency .ξ˜ becomes imaginary, and if .Ω > n, .ξ˜ is real. Since .b is real, (4.171) indicates that an imaginary .ξ˜ and a real .ξ˜ correspond to an imaginary .k2x and a real .k2x , respectively. According to (4.167) and (4.168), the antisymmetric and symmetric SH plate wave travels when .ξ˜ is real. If .ξ˜ is imaginary, the oscillation from the source is not transferred as a wave, and instead, it decays exponentially as a function of .x. The condition .Ω > n or .Ω < n, which determines whether .ξ˜ is real or imaginary, tells us that for a given mode number the oscillation from the source travels as a wave if the oscillation frequency is higher than a certain value. We can intuitively understand this situation by imagining oscillating one end of an elastic string at various frequencies with the other end fixed to the wall. As we increase the oscillation frequency, the motion travels toward the fixed end with the shorter wavelength, i.e., the higher mode number. In other words, if we want to excite a wave of a certain mode, we need to oscillate the open end at a certain frequency. We can imagine that if we oscillate the open end very slowly the entire string oscillates up and down. Figure 4.12 shows the dispersion curve for various mode numbers .n. Here, the solid lines are for symmetric waves and the dashed lines are for anti-symmetric waves, and the right half is for the case when .ξ˜ is real and the left half is when it is imaginary. Notice that the

116

4 Propagation of Acoustic Waves in Solids

Fig. 4.12 Dispersion curves for SH-waves traveling through a plate

dispersion curve depends on the plate thickness via (4.171) and (4.172). This fact can be used to find a thickness of a layer from the dispersion curve [13]. P and SV waves in a plate and Rayleigh-Lamb frequency equation In this case, it is more convenient to use the potentials .φ and .H. Recall the discussions made under Sect. 4.2.2, Plane strain wave solution. The governing equations and the solutions we should deal with are as follows. ∇2φ =

.

1 ∂ 2 Hz 1 ∂ 2φ 2 , ∇ H = z c1 ∂t 2 c2 ∂t 2

(4.174)

and ( ) φ = A1 eik1y y + A2 e−ik1y y ei(ωt−k1x x ) ( ) = A˜ 1 cos k1y y + A˜ 2 sin k1y y ei(ωt−k1x x ) ( ) ik y −ik2y y ei(ωt−k1x x ) . Hz = B1 e 2y + B2 e ( ) = B˜ 1 cos k2y y + B˜ 2 sin k2y y ei(ωt−k1x x ) .

(4.175)

(4.176)

where . A˜ 1 = A1 + A2 , A˜ 2 = i(A1 − A2 ) and . B˜ 1 = B1 + B2 , B˜ 2 = i(B1 − B2 ). Similarly to the case of the free surface condition discussed in the Plane strain wave solution section, the boundary conditions relevant to the P and SV waves are as follows. σ yy = σx y = 0, at y = ±b

.

(4.177)

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

117

Here, the boundaries are assumed to be traction free and the relevant stress are expressed as follows. ) ( 2 ∂ 2 Hz ∂ φ ∂ 2 Hz − (4.178) .σ x y = μ 2 + ∂ x∂ y ∂ y2 ∂x2 ) ( ( 2 ) ∂ 2φ ∂ φ ∂ 2φ ∂ 2 Hz − 2μ .σ yy = (λ + 2μ) + + ∂x2 ∂ y2 ∂x2 ∂ x∂ y ) ( 2 ( ( 2 2 2 H )) ∂ ∂ φ φ φ ∂ ∂ z + 2 −2 + (4.179) = μ κ2 ∂x2 ∂y ∂x2 ∂ x∂ y At this point, express the displacement vector components .ξx and .ξ y using potentials (4.175) and (4.176) ∂φ ∂ Hz + ∂x ∂y ( ) = −ik1x A˜ 1 cos k1y y + k2y B˜ 2 cos k2y y ei(ωt−k1x x) ) ( + −ik1x A˜ 2 sin k1y y − k2y B˜ 1 sin k2y y ei(ωt−k1x x)

(4.180)

∂φ ∂ Hz − ∂y ∂x ( ) = −k1y A˜ 1 sin k1y y + ik1x B˜ 2 sin k2y y ei(ωt−k1x x) ) ( + k1y A˜ 2 cos k1y y + ik1x B˜ 1 cos k2y y ei(ωt−k1x x)

(4.181)

ξx (x, y, t) =

.

ξ y (x, y, t) =

.

Of the two terms enclosed by the parenthesis in expressions (4.180) and (4.181), the first one is symmetric about the .x-axis and the second one is anti-symmetric. The symmetric and anti-symmetric expressions of .ξx and .ξ y are shown below. Symmetric displacement components: ) ( (x, y, t) = −ik1x A˜ 1 cos k1y y + k2y B˜ 2 cos k2y y ei(ωt−k1x x) ) ( sym .ξ y (x, y, t) = −k1y A˜ 1 sin k1y y + ik1x B˜ 2 sin k2y y ei(ωt−k1x x) sym

ξx

.

(4.182) (4.183)

Anti-symmetric displacement components: ) ( (x, y, t) = −ik1x A˜ 2 sin k1y y − k2y B˜ 1 sin k2y y ei(ωt−k1x x) ) ( asym .ξ y (x, y, t) = k1y A˜ 2 cos k1y y + ik1x B˜ 1 cos k2y y ei(ωt−k1x x) asym

ξx

.

Figure 4.13 illustrates the symmetry.

(4.184) (4.185)

118

4 Propagation of Acoustic Waves in Solids

Fig. 4.13 Symmetric and anti-symmetric parts of .ξx and .ξ y components of plane P and SV waves traveling in a plate of thickness .2b

Substitute (4.182) and (4.183) into expressions (4.79) and (4.80) and find the stress due to the symmetric waves as follows. { sym .σ x y = μ − 2ik1x k1y (− A˜ 1 sin k1y y) } 2 2 + (k1x − k2y )(+ B˜ 2 sin k2y y) ei(ωt−k1x x) (4.186) { sym 2 2 2 .σ yy = μ (2k1x − κ 2 (k1x + k1y ))( A˜ 1 cos k1y y) } + 2ik1x k2y (+ B˜ 2 cos k2y y) ei(ωt−k1x x) { 2 2 = μ (k1x − k2y )( A˜ 1 cos k1y y) } (4.187) + 2ik1x k2y (+ B˜ 2 cos k2y y) ei(ωt−k1x x) Here the following substitution was made for .σ yy .

4.2

Propagation of Acoustic Waves in Solid Under Various Conditions

κ2 =

.

c12 c22

=

ω2 /c22 ω2 /c12

=

119

2 + k2 k2y 1x 2 + k2 k1y 1x

2 2 2 2 κ 2 (k1x + k1y ) = k2y + k1x 2 2 2 2 2 2k1x − κ 2 (k1x + k1y ) = k1x − k2y

(4.188)

Substitution of (4.186) and (4.187) into boundary conditions (4.177) yields the following equations. .

2 2 ± {2ik1x k1y ( A˜ 1 sin k1y b) + (k1x − k2y )( B˜ 2 sin k2y b)} = 0

2 .(k 1x

2 − k2y )( A˜ 1 cos k1y b) + 2ik1x k2y ( B˜ 2 cos k2y b)

=0

(4.189) (4.190)

Here the .± sign comes from the boundary condition .σx y = 0 at . y = ±b. We can view (4.189) and (4.190) as a system of equations for . A˜ 1 and . B˜ 2 , and express them in the form of a matrix as follows. ( ) ( ) 2 − k 2 ) sin k b ( ˜ ) 2ik1x k1y sin k1y b (k1x 0 A1 2y 2y = . (4.191) 2 − k 2 ) cos k b 2ik k cos k b ˜ 0 (k1x B 1y 1x 2y 2y 2 2y For the matrix equation (4.191) to hold for a nontrivial pair of . A˜ 1 and . B˜ 2 , it follows that the determinant of the matrix must be null. This observation yields the following frequency equation. 4k 2 k1y k2y tan k2y b . (4.192) = − 21x 2 2 tan k1y b (k1x − k2y ) Equation (4.192) is known as the Rayleigh-Lamb frequency equation for the propagation of symmetric waves in a plate. Remember that .k1x , .k1y and .k2y are related through angle .θ1 as follows. ω sin θ1 c1 ω cos θ1 = k1 cos θ1 = c1 ω = k2 cos θ2 = cos θ2 c2

k1x = k1 sin θ1 =

(4.70)

k1y

(4.71)

.

.

k2y

.

k1 sin θ1 = k2 sin θ2

.

(4.73) (4.84)

For a given elastic material (.c1 and .c2 ) with a plate thickness (.2b), once frequency .ω is determined (4.192) tells us the angle .θ1 and .θ2 for the symmetric waves (4.182) and (4.183). From the system of equations (4.189) and (4.190), we can derive the amplitude ratio as follows. 2 − k 2 ) sin k b (k1x 2y 2ik1x k2y cos k2y b A˜ 1 2y =− (4.193) . =− 2 2 ) cos k b 2ik1x k1y sin k1y b (k1x − k2y B˜ 2 1y

120

4 Propagation of Acoustic Waves in Solids

Repeating the same argument for the anti-symmetric case, we obtain the following equations. ( ) ( ) 2 − k 2 ) cos k b ( ˜ ) −2ik1x k1y cos k1y b (k1x 0 A2 2y 2y = (4.194) . 2 − k 2 ) sin k b −2ik k sin k b ˜ 0 (k1x B 1y 1x 2y 2y 1 2y .

2 − k 2 )2 (k1x tan k2y b 2y =− 2 tan k1y b 4k1x k1y k2y

(4.195)

Equation (4.195) is the Rayleigh-Lamb frequency equation of anti-symmetric waves. 2 − k 2 ) cos k b (k1x 2y 2ik1x k2y sin k2y b A˜ 1 2y =− 2 . =− 2 ˜ 2ik k cos k b (k B2 1x 1y 1y 1x − k2y ) sin k1y b

(4.196)

Equation (4.196) is the amplitude ratio for the anti-symmetric wave case. A note on vibrations and waves in a thin plate When a longitudinal wave is to be excited in a thin plate with the application of an acoustic transducer normally to the surface, we need to be careful about the relation between the wavelength and the thickness. If the thickness is significantly smaller than the wavelength, say less than a quarter of the wavelength, the phase difference between the leading and trailing edge of the longitudinal wave is less than .π/4. The plate is more or less oscillating as a whole. Recall that the wave velocity is intrinsic to the material. Therefore the wavelength is determined by the frequency. To make the wavelength shorter for a given wave velocity (material), it is necessary to increase the frequency. The same can be said about the spatial resolution. When we try to detect a void in a material, if the length of the void in the direction of the longitudinal wave’s propagation, the void falls in a small portion of the wavelength. It is necessary to increase the frequency. However, increase of frequency is not straightforward as the attenuation can be proportional to the square of the frequency [14–19].

References 1. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK 2. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK 3. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK, p 25 4. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 5 5. Rayleigh JWS (1887) On waves propagated along the plate surface of an elastic solid. Proc Lond Math Soc 17:4–11 6. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK, chap. 7 7. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 6

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8. Pilarski A, Rose JL (1988) A transverse wave ultrasonic oblique incidence technique for interfacial weakness detection in adhesive bonds. J Appl Phys 63(2):300–307 9. Graff KF (1975) Wave motion in elastic solids. Oxford University Press, Oxford, UK, Chap. 8 10. Maev RG (2008) Scanning acoustic microscopy. Physical principle and methods. Current development in R. G. Maev Acoustic microscopy: fundamentals and applications, Ch1. Wiley-VCH, Verlag. https://doi.org/10.1002/9783527623136.ch1 (accessed on July 28, 2023), pp 9–19 11. Rose JL (1999) Ultrasonic waves in solid media. Cambridge University Press, Cambridge, UK, chap. 8 12. Yoshida S (2017) Waves; Fundamentals and dynamics, Chap. 4. Morgan & Claypool, San Rafael, CA, USA 13. Parmon W, Bertoni H (1979) Ray interpretation of the material signature in the acoustic microscope. Electron Lett 15(21):684–686 14. Szabo TL, Wu J (2000) A model for longitudinal and shear wave propagation in viscoelastic media. J Acoust Soc Am 107(5):2437–2446 15. Szabo TL (1994) Time domain wave equations for lossy media obeying a frequency power law. J Acoust Soc Am 96(1):491–500 16. Chen W, Holm S (2003) Modified Szabo’s wave equation models for lossy media obeying frequency power law. J Acoust Soc Am 114(5):2570–2574 17. Chen W, Holm S (2004) Fractional Laplacian time-space models for linear and nonlinear lossy media exhibiting arbitrary frequency power-law dependency. J Acoust Soc Am 115(4):1424– 1430 18. Carcione JM, Cavallini F, Mainardi F, Hanyga A (2002) Time-domain modeling of constant-Q seismic waves using fractional derivatives. Pure Appl Geophys 159:1719–1736 19. D’astrous FT, Foster FS (1986) Frequency dependence of ultrasound attenuation and backscatter in breast tissue. Ultrasound Med Biol 12(10):795–808

5

Electrical-Mechanical Transduction

This chapter discusses acoustic transduction. We focus on the operation principle of typical acoustic transducers to understand how the devices work. To this end, we discuss the physics underlying the operation.

5.1

General Argument

Acoustic transduction generally means the conversion of an electric signal to an acoustic signal or an acoustic signal to an electric signal. The former is acoustic transmission, and the latter is acoustic sensing. A loudspeaker is an example of an acoustic transmitter, and a microphone is an example of an acoustic sensor. In the application to experimental mechanics dealing with solid objects, often acoustic sensors are vibration sensors, and acoustic transmitters are vibration actuators. Thus acoustic transduction is the interaction of an electric system and a mechanical system. The following pair of equations conveniently represent the process of the acoustic transaction [1]. v = z e0 i + Tem u

(5.1)

f = Tme i + z m0 u

(5.2)

. .

Here .v is the voltage on the electric system, .u is the velocity of the mechanical load, .z e0 is the electrical impedance, .Tem is the mechanical to electrical transduction coefficient, . f is the force on the mechanical system, . Tme is the electrical to mechanical transduction coefficient, and.z m0 is the mechanical impedance. (5.1) and (5.2) are the governing equations that describe the relationship between the electrical and mechanical systems. Call the former the electrical governing equation and the latter the mechanical governing equation. When the transducer acts as an acoustic sensor, the electrical governing equation expresses the © The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7_5

123

124

5 Electrical-Mechanical Transduction

acoustic (mechanical) input represented by the velocity of the sensing mechanism (e.g., a microphone’s diaphragm) represented by .u and the resultant current and voltage .i and .v detected by the electric system. When the transducer operates as a transmitter (actuator), voltage .v represents the input electric signal, and .i and .u represent the resultant current flowing through the electric system and the velocity of the actuation mechanism (e.g., a loudspeaker’s diaphragm). The mechanical governing equation describes the force acting on the vibrating part (e.g., the diaphragm) where the first term on the right-hand side represents the electric force and the second term represents the mechanical force. As these equations indicate, the conversion from acoustic to electric or electric to acoustic energy is not 100% efficient. For instance, when an acoustic signal oscillates the diaphragm in a microphone, part of the generated electric energy is lost in the circuit. The first term on the right-hand side of (5.1) represents this loss. Similarly, in the opposite process, part of the generated mechanical energy is lost by mechanical impedance expressed by the second term on the right-hand side of (5.2). Thus, the coupling factor . K is defined as follows. .

K2 =

Tem Tme z e0 z m0

(5.3)

Here the numerator is the product of the active terms on the right-hand side of (5.1) and (5.2), whereas the denominator is the product of the ineffective terms.

5.2

Reciprocal Transduction

This is the case when the following condition (reciprocal condition) holds. .

Tem = Tme = T

(5.4)

Define the transduction factor .φ as follows. φ=

.

T z e0

(5.5)

(5.1) and (5.2) become as follows. v = z e0 i + φz e0 u

(5.6)

f = φz e0 i + z m0 u

(5.7)

. .

In this case, the coupling factor takes the following form. .

K2 =

Tem Tme T2 1 T 2 z e0 z e0 = = 2 = φ2 z e0 z m0 z e0 z m0 z m0 z e0 z m0

(5.8)

5.2

Reciprocal Transduction

5.2.1

125

Electrostatic Transducer

An electrostatic transducer is an example of a reciprocal transducer. Figure 5.1 schematically illustrates a sample electrostatic transducer. This transducer consists of two stationary electrodes connected to an AC voltage supply and one movable plate (diaphragm) located in between the two stationary electrodes. Typically, the diaphragm consists of a thin plastic film coated with a conductive material. Initially, the diaphragm is positively charged with .q0 at a DC voltage .v0 . Throughout the operation, this DC voltage remains the same. (In other words, the diaphragm is AC-isolated.) The diaphragm forms two capacitors with each of the stationary electrodes, respectively. When this transducer operates as an acoustic transmitter, the AC voltage supply applies a sinusoidal voltage (an electric input) to the system. As explained below, the applied AC signal is proportional to the output acoustic signal. Figure 5.1 illustrates the moment when the applied AC voltage is negative on the left stationary electrode and positive on the right stationary electrode. The absolute value of these negative and positive voltage changes are the same. This application of the AC voltage causes two things; first, positive charges .q flow into the right electrode, and the same amount of positive charges, .q, flow out of the left electrode; second, these changes in the two electrodes generate a leftward electric field in the space between the electrodes. This electric field exerts Coulomb force on the positively charged diaphragm, causing it to move to the left by .x. Here, the charge .q caused by the applied AC voltage is much smaller than the total charge on the diaphragm, .q < q0 , and the corresponding displacement of the diaphragm is much smaller than the initial gap distance between the stationary electrode and diaphragm, .x v2 .) This means that the electric field vector is directed from the left stationary electrode towards the right stationary electrode through the diaphragm. Since the diaphragm is positively charged, this electric field exerts a rightward electric force on the diaphragm. This is a resistive force for the mechanical force. The external agent exerting the mechanical force feels a higher stiffness.

5.3

Anti-reciprocal Transducers

In this case, the mechanical-to-electrical and the electrical-to-mechanical transduction coefficients have mutually opposite signs. .

Tem = −Tme

(5.39)

Define the transformation factor .φ M as follows. φ M = Tem = −Tme

.

(5.40)

The subscript . M denotes “magnetic” because the magnetic field plays the main role in this transduction, as will be discussed later in this section. The electrical and mechanical governing equations take the following form. v = z e0 i + φ M u

(5.41)

f = −φ M i + z m0 u

(5.42)

. .

Similarly to the reciprocal case, consider short-circuiting the electric circuit to make v = 0. From (5.41), the current under this condition becomes .i = −φ M u/z e0 . Substituting this into (5.42) yields . f s and .z ms = f s /u as follows.

.

.

.

5.3.1

1 u + z m0 u z e0 φ2 = z m0 + M z e0

f s = φ 2M

z ms

(5.43) (5.44)

Moving-Coil Transducer

The most typical example of an anti-reciprocal transducer is the moving-coil transducer. In this case, .z e0 in (5.43) and (5.44) is the electrical impedance consisting of the DC resistance

5.3 .

Anti-reciprocal Transducers

133

R0 and inductance . L 0 . .

z e0 = R0 + jωL 0

(5.45)

Unlike the electrostatic case, the open-circuited condition makes the electric force inactive. This is because while in the electrostatic case, the electric force is the Coulomb force proportional to the voltage across the capacitor, in the moving-coil case the electric force is the magnetic force generated in the coil in proportion to the current. According to electrodynamics [2], the magnetic force due to the magnetic field . B acting on current .i flowing through a conductive wire of length .l is . f B = i Bl. On the other hand, when the electric current moves in the magnetic field with velocity .u, the magnetic induction induces an opposing voltage of . Blu. Consider the above-mentioned mechanism in more detail by referring to Fig. 5.3. In this figure, a permanent magnet applies a constant magnetic field into the page. Two conductive wires are connected to either an AC power supply (Fig. 5.3a) or a resistor (Fig. 5.3b). A conductor bar is placed on the two wires. Here (Fig. 5.3a) models an acoustic actuator and (b) models an acoustic sensor. The conductor bar represents a diaphragm in both cases. In the acoustic actuator configuration, the transducer induces a force on the diaphragm (represented with . f ) in response to an input electric current, whereas in the acoustic sensor configuration, the transducer induces an electric current in response to the motion of the diaphragm (represented with velocity .u) due to mechanical force. In Fig. 5.3, the dashed quantity depicts the induced quantity. When the transducer operates as an actuator, the AC power supply generates current. Figure 5.3a illustrates the moment when the current from the power supply flows upward in the conductor bar (length .l). Here, .xˆ etc. are the unit vector of the direction. The magnetic field due to the permanent magnet (.B = B yˆ ) exerts magnetic force leftward. The diaphragm moves to the left. In the next cycle when the current flips the direction, the magnetic force exerts rightward force on the diaphragm. In this fashion, the diaphragm oscillates. The magnetic force at the moment when the electric current is .i = i zˆ , the magnetic force .f B is .−Bli x ˆ .f B = i z ˆ × B yˆ l = Bli(ˆz × yˆ ) = −Bli xˆ (5.46) When the transducer operates as a sensor, the mechanical force causes velocity .u. Magnetic induction states that the induced voltage .v is the change in the magnetic flux .φ B = B A over time. Here . A is the area of the magnetic field. Since the area . A changes over time as the conductor bar moves at velocity .u while the magnetic field is constant, we can express the temporal change in the magnetic flux as follows. v=

.

dA dφ B =B = Blu dt dt

(5.47)

According to Faraday’s law, the induced current resulting from magnetic induction flows in such a way that the resultant magnetic field opposes the temporal change in the magnetic

134 Fig. 5.3 Conceptual illustration of a magnetic force generated by an acoustic actuator and b magnetic induction occurring in an acoustic sensor

5 Electrical-Mechanical Transduction

(a) Acoustic actuator

diaphragm

(b) Acoustic sensor

flux. When the magnetic force causes the rightward velocity as shown in Fig. 5.3b, the magnetic force increases the magnetic flux. Hence, the induced current flows in the direction that generates a magnetic field in the out-of-page direction inside the loop formed by the conductor bar and the two wires. Figure 5.4 illustrates a more realistic structure of a moving-coil transducer. Typically, a moving-coil transducer consists of a solenoid wound around a cylindrical core placed inside the inner space of a doughnut-shaped permanent magnet. The diaphragm (not shown in Fig. 5.4 to avoid confusion) is attached to the solenoid in such a way that the solenoid’s motion parallel to the axis of the core cylinder (called the vertical motion) oscillates the diaphragm. In this configuration, as the solenoid experiences vertical motion, the magnetic flux varies. In the situation illustrated in Fig. 5.3, the area that penetrates the permanent magnetic field changes as the conductor bar moves parallel to the plane. In the case of Fig. 5.4, the (active) Fig. 5.4 Schematic view of a moving-coil transducer

AC power supply Electric magnet (Solenoid) N

S

Permanent magnet Core

Magnetic field by permanent magnet

5.3

Anti-reciprocal Transducers

135

area changes as the conductor (the solenoid) moves perpendicular to the plane that the permanent magnet penetrates. In this situation, the circular area formed by each turn of the solenoid defines the active area involved in the magnetic flux. When the solenoid is moving out of the vertical field line of the permanent magnet, the induced voltage decreases as the circular area of each turn goes out of the effective zone of the permanent magnet. This is the phase when the induced voltage decreases. In the opposite phase, as each turn of the solenoid enters the effective zone of the permanent magnetic field, the induced voltage increases because the number of circular areas that contributes to the magnetic induction increases. Figure 5.5 illustrates another way of understanding the magnetic induction in the configuration illustrated in Fig. 5.4. The top illustration of this figure represents the situation where a solenoid carries an electric current provided by an AC voltage supply. A permanent magnet is partially located inside the cylindrical space of the solenoid with two poles as shown in the figure. In the phase when the current increases in the direction depicted by solid arrows (the current increases clockwise when the solenoid’s cross-sectional area is viewed from the left side of the solenoid), the solenoid increases the magnetic field vector rightward along the axis of the cylindrical space. Electromagnetic induction reduces this change. Since the magnetic field due to the permanent magnet is leftward, going deeper into the solenoid performs this task. Thus, the permanent magnet is pushed to the right. The solid big arrow in Fig. 5.5 indicates the direction in which the permanent magnet is pushed. We can understand the above effect in terms of the interaction between two magnets. When the current flows in the direction indicated by the solid line, the solenoid forms an electric magnet with the S-pole on the left and the N-pole on the right. (Consider this imaginary permanent magnet in the space of the solenoid’s core on the right of the permanent magnet.) On the other hand, the polarity of the permanent magnet is opposite, as the lower illustration of Fig. 5.5 indicates. The two S-poles facing each other repel the magnets from one another. Hence, the permanent magnet is pushed to the left. When the solenoid current increases, the electric magnet’s repelling force increases. Nature does not like this abrupt change, so it exerts a counterforce. This force corresponds to the above-mentioned opposing effect due to electromagnetic induction. Fig. 5.5 Explanation of temporal change in magnetic flux in moving-coil transducer

S

N

N

S

S

N

136

5 Electrical-Mechanical Transduction

In the other phase, when the current flows in the opposite direction (depicted by the thin dashed line in the top illustration), the direction of the magnetic field generated by the solenoid flips, whereas the direction of the permanent magnetic field remains the same. Consequently, the permanent magnet is pulled into the solenoid, as the thick arrow with the dashed line indicates. The above description illustrates the dynamics when the solenoid carries current. Hence it applies to the explanation of an acoustic actuator. Instead of supplying current to the solenoid, consider that a mechanical force pushes the permanent magnet into the solenoid. According to the law of magnetic induction, it will induce an electric current. This situation corresponds to the case when the transducer acts as an acoustic sensor. In the above explanation, we assumed that the solenoid is stationary in space, and the permanent magnet is movable. Since all the interactions between the permanent magnet and the solenoid are relative, we can argue the same effect by making the permanent magnetic field stationary and the solenoid movable. The dynamics illustrated in Fig. 5.4 corresponds to that case. The magnetic force and induced voltage yield the following two governing equations. The above argument indicates that for a common direction of the electric current, the direction of the induced force . f and the velocity .u caused by the magnetic force are opposite to each other. This explains the negative sign appearing in the first term on the right-hand side of the mechanical governing equation (5.49). v = z e0 i + Blu

(5.48)

f = −Bli + z m0 u

(5.49)

. .

Comparison of the set of equation (5.48) and (5.49) with (5.41) and (5.42) indicates the following expression for .φ M .φ M = Bl (5.50) Repeating the same type of argument as the electrostatic transducer case, we find the following force expression when the electric circuit is open. Note that, unlike the electrostatic case, the electromagnetic force is inactive when the electric circuit is open. This is because, in this case, the electromagnetic force is magnetic force, which is proportional to the current. Thus, calling this force . f o we obtain the following expressions for the force and the corresponding impedance .z m0 . ( ) k (5.51) . fo = γ u + j mω − u ω ) ( k (5.52) . z m0 = γ + j mω − ω When the electric circuit is closed and therefore .v = 0, from (5.48) and (5.50) we find i = −Blu/z e0 = −φ M u/z e0 . Substitution of this current into (5.49) with the use of (5.25) yields the expressions for the corresponding force . f s and impedance .z ms as follows.

.

5.3

Anti-reciprocal Transducers

.

.

) ( φ 2M k + φ 2 /L 0 u u + z m0 u = γ u + j mω − z e0 ω ) ( k + φ 2 /L 0 = γ + j mω − ω

fs =

z ms

137

(5.53) (5.54)

Here assuming . R0 0), it is negative. This situation corresponds to the fact that the gradient decreases with .xr , i.e., its positive value decreases as .xr increases positively; consequently, the ball falls faster as . xr increases. Similarly, we can explain the observation that the falling speed of the ball decreases as the negative reference coordinate increases toward .xr = 0, i.e., when .xr < 0, the gradient is positive. Now in Fig. B.2c, the gravitational potential has the same dependence on .x and . y. Consider the motion of a tennis ball when you release if at .(x, y) = (0, −1). You can easily imagine that the ball falls along a longitudinal line because in the tangential direction, the height does not change. As is the case of Fig. B.2b, let’s consider the gradient vector quantitatively. In this case, the function . f (x, y) has a dependence on .x and . y and can be expressed as follows. ( 2 )1 2 2 2 (B.5) . f (x, y) = 10 − x − y Substituting (B.5) into (B.2), we obtain the following expression for the gradient vector. −x −y ∇f = √ iˆ + √ jˆ 2 2 2 2 10 − x − y 10 − x 2 − y 2

.

(B.6)

This time, the gradient vector has both .x and . y components. This fact is consistent with the above qualitative observation that the ball should fall along a longitudinal line. In a top view, a longitudinal line contains both .x and . y components. However, (B.6) is not convenient to find the longitudinal direction that the ball falls from a reference point. It is better to use spherical coordinates. Using.rˆ ,.θˆ and.ψˆ to represent the unit vector for the radial, polar, and azimuthal component of . f (r , θ, ψ) we can express (B.2) and (B.5) as follows.

1 You may wonder what if we place the tennis ball at a point where the . x coordinate is 0. It is unclear

if the ball stays there forever or fall in either the positive or negative .x direction. In theory, the ball does not fall. In reality, it will fall either direction as another force, such as air pressure acts on it.

146

Appendix B: Gradient Operation

Fig. B.3 Coordinate point expressed with a Cartesian and b Spheric coordinate systems

.

f (r , θ, ψ) = r cos θ = 10 cos θ 1 ∂f ∂f 1∂f θˆ + ψˆ .∇ f = rˆ + ∂r r ∂θ r sin θ ∂ψ

(B.7) (B.8)

Here we use that the surface shown in Fig. B.2c is part of the surface of a sphere with a radius of 10. Substituting (B.7) into (B.8), we find the gradient vector as follows. ∇f =

.

1 (−r sin θ )θˆ = (− sin θ )θˆ r

(B.9)

The arrows in Fig. B.2c represent a tangential vector (the one parallel to the .x y-plane), and .∇ f expressed by (B.9) at a reference point. The tangential vector corresponds to the one on the ridge seen in Fig. B.2b. Referring to Fig. B.3 we find the following relationship between the spheric and Cartesian coordinate variables.

. .

x = r sin θ cos ψ, y = r sin θ sin ψ

cos θ θˆ = cos ψ iˆ + sin ψ jˆ

(B.10) (B.11)

Using .r = 10 and noting that from (B.10) .x 2 + y 2 = r 2 sin2 θ (cos2 ψ + sin2 ψ) = 102 2 .sin θ , we can write (B.6) as follows. −x

−10 sin θ cos ψ ˆ −10 sin θ sin ψ ˆ jˆ = i+ j 10 cos θ 10 cos θ 102 − (x 2 + y 2 ) 102 − (x 2 + y 2 ) (B.12) ( ) sin θ cos ψ iˆ + sin ψ jˆ (B.13) . = − cos θ

∇ f =√

.

ˆ √ i+

−y

Appendix B: Gradient Operation

147

Now use (B.11) in (B.9) to derive the following equation. ∇ f = (− sin θ )θˆ = (− sin θ )

.

) 1 ( cos ψ iˆ + sin ψ jˆ cos θ

Comparison of (B.12) and (B.14) indicates that these two equations are identical.

(B.14)

C

Appendix C

Orthogonality of Sine and Cosine Functions

According to Fourier’s theorem [1, 2] a periodic function can be expanded into a series of cosine and sine functions. .

a0 + a1 cos(ω0 t) + a2 cos(2ω0 t) + · · · + an cos(nω0 t) 2 + b1 sin(ω0 t) + b2 sin(2ω0 t) + · · · + bn sin(nω0 t)

f (t) =

(C.1)

Here .ω0 is the lowest angular frequency. If function . f (t) is defined on .[0 T ], the lowest frequency . f 0 and the corresponding angular frequency are as follows. .

f0 =

1 T

ω0 = 2π f 0 =

.

(C.2) 2π T

(C.3)

With this notation, we can express (C.1) as follows. .

f (t) =

) )) ( ( n ( a0 ∑ 2π k 2π k ak cos + t + bk sin t 2 T T k=1

=

a0 + 2

n ∑

{ak cos (ωk t) + bk sin (ωk t)}

(C.4)

k=1

The next step is to determine the coefficients .an and .bn . We can use the following property to determine the coefficients. This property is known as the orthogonality of the trigonometric functions.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7

149

150

Appendix C: Orthogonality of Sine and Cosine Functions

∫

t1 +T

.

cos(nω0 t) cos(mω0 t)dt =0 (n / = m)

t1 .

=

T π (n = m) = 2 ω0

(C.5) (C.6)

Here the following identities are used. .

1 cos(nω0 t) cos(mω0 t) = {cos((m − n)ω0 t) + cos((m + n)ω0 t)} 2 1 {cos((m − n)ω0 t) − cos((m + n)ω0 t)} . sin(nω0 t) sin(mω0 t) = 2 1 2 {1 + cos(2nω0 t)} . cos (nω0 t) = 2 1 2 {1 − sin(2nω0 t)} . sin (nω0 t) = 2

(C.7) (C.8) (C.9) (C.10)

Multiply .cos(2ω0 t) to both sides of Eq. (C.1) and integrate for one period (from .t = t1 to t = t1 + T ). From the orthogonality, all terms vanish except .a2 cos(2ω0 t) on the right-hand side. Thus, using Eq. (C.6), we find that coefficient .a2 can be expressed as follows.

.

a2 =

.

2 T

∫

t1 +T

f (t) cos(2ω0 t)dt

(C.11)

t1

It is clear that other coefficients can be evaluated with the same procedure. Generally, coefficients .an and .bn can be expressed as follows. 2 .an = T bn =

.

2 T

∫

t1 +T

t1 ∫ t1 +T t1

f (t) cos(nω0 t)dt

(C.12)

f (t) sin(nω0 t)dt

(C.13)

D

Appendix D

Notes on Standing Waves

When the amplitude of the forward-going wave (. A f ) is different from that of the backward-going wave (. Ab ), the superposed wave can be expressed as follows. A f sin(ωt − kz) + Ab sin(ωt + kz) .

= 2 Ab cos(kz) sin(ωt) + (A f − Ab ) sin(ωt − kz)

(D.1)

Equation (D.1) indicates that in this case the expression of the superposed wave consists of the standing-wave term (the first term on the right-hand side of Eq. (D.1)), and the traveling-wave term (the second term on the right-hand side of Eq. (D.1)). Because of this traveling-wave term, the superposed wave is behavior differs from the case when the forward-going and backward-going waves have the same amplitude. First, the superposed wave is not zeroed out even when the phase difference between the forward-going and backward-going waves is an odd integer multiple of .π . Figure D.1 illustrates the situation. Here, Fig. D.1a shows the forward-going (top), the back-ward-going (middle) and superposed (bottom) waves for three representative moments when the interference is destructive (left), constructive (right) and intermediate (center). Notice that under the destructive interference the superposed wave is completely zeroed out whereas under the constructive interference, the peak of the superposed wave is doubled as compared with the forward-going wave. The superposed wave oscillates as a standing wave. Figure D.1b exhibits the situation when the amplitude of the backward-going wave is 25% smaller than the forward-going beam (. Ab = 0.75A f ). It is seen that even when the two wave are completely out of phase (the phase difference is .π ), the superposed wave is not zeroed-out. Second, the superposed wave travels. According to the traveling-wave term of Eq. (D.1), the superposed wave travels forward if the amplitude of the forward-going wave is greater than the backward-going, and it travels backward if the amplitude of the forward-going wave is smaller than the backward-going wave. The superposed wave in Fig. D.1b travels but is © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7

151

152

Appendix D: Notes on Standing Waves

Fig. D.1 Superposition of two waves propagating in mutually opposite directions. a When forwardgoing and backward-going waves have the same amplitude. b When the forward-going wave has greater amplitude

not obvious. This is because the amplitude of the traveling-wave term in Eq. (D.1) increases in proportion to the difference in amplitude between the forward-going and backwardgoing waves. Figure D.2 shows that the superposed wave travels forward clearly when the amplitude of the backward-going wave is 25% of the forward-going wave (. Ab = 0.25A f ). With the increase in the difference between . A f and . Ab the amplitude of the travelingwave term increases. Thus, the traveling feature of the superposed wave is much clearer as compared with . Ab = 0.75A f case shown in Fig. D.1b.

Appendix D:Notes on StandingWaves

153

Fig. D.2 Superposition of two waves where backward-going wave’s amplitude is 25% of the forwardgoing wave

Appendix E

Generalized Hooke’s Law and Equation of Motion for Isotropic Media

E.1

Strain and Stress Matrices

The strain matrix consists of normal strain components and shear strain components. ⎛ ∂ξ ∂ξ ∂ξ ⎞ x x x ⎞ ⎛ ∈x x ∈x y ∈x z ⎜ ∂ x ∂ y ∂z ⎟ ⎜ ∂ξ y ∂ξ y ∂ξ y ⎟ ⎝ (E.1) .[∈] = ⎜ ⎟ = ∈ yx ∈ yy ∈ yz ⎠ ⎝ ∂ x ∂ y ∂z ⎠ ∈zx ∈zy ∈zz ∂ξz ∂ξz ∂ξz ∂x

∂y

∂z

Here the normal strain and shear strain components are defined as follows. ) ( 1 ∂ξi ∂ξi .∈ii = + 2 ∂ xi ∂ xi ) ( ∂ξi 1 ∂ξ j + .∈i j = 2 ∂ xi ∂x j

(E.2) (E.3)

where .i, j = 1, 2, 3 and .x1 = x, x2 = y, x3 = z and .ξ1 = ξx , ξ2 = ξ y , ξ3 = ξz . Similarly, we define the stress matrix as follows. ⎛ ⎞ σx x σx y σx z .[σ ] = ⎝σ yx σ yy σ yz ⎠ σzx σzy σzz

(E.4)

Here .σi j represents the stress on the plane .i in the direction of . j as indicated in Fig. E.1. The plane .i is the plane normal to the .i axis. Stress is defined as a force acting on a plane divided by the area of the plane. For instance, .σx x (x) is the force applied in the positive .x direction on the plane located at .x perpendicular to the .x axis. .σ yx is the force acting on the plane perpendicular to the . y axis at . y in the direction of positive .x. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7

155

E

156

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Fig. E.1 Normal and shear stress components. The first subscript denotes the plane and the second the direction of the stress (force) vector

E.2

Compliance and Stiffness Matrices

The compliance or stiffness matrix constitutes the basis of generalized Hooke’s law. These two matrices are mutually the inverse of the other and essentially represent the same stressstrain relationship. The compliance matrix . S and stiffness matrix .[C] are defined as follows. .

[∈] = [S][σ ]

(E.5)

[σ ] = [C][∈]

(E.6)

.

Here .[∈] and .[σ ] are the strain and stress matrices in Voigt’s notation [3]. In this notation the 3.×3 matrices (E.2) and (E.4) become a vector with 6 components, respectively. Voigt’s notation is applicable to a symmetric matrix in general. ( )† [∈] = ∈x x ∈ yy ∈zz ∈ yz ∈zx ∈x y ( )† .[σ ] = σ x x σ yy σzz σ yz σzx σ x y .

(E.7) (E.8)

Here .† represents transpose. For later uses, we express the compliance and stiffness matrices in the following component forms. ⎞ ⎛ S11 S12 S13 S14 S15 S16 ⎜S S S S S S ⎟ ⎜ 21 22 23 24 25 26 ⎟ ⎟ ⎜ ⎜ S31 S32 S33 S34 S35 S36 ⎟ (E.9) .[S] = ⎜ ⎟ ⎜ S41 S42 S43 S44 S45 S46 ⎟ ⎟ ⎜ ⎝ S51 S52 S53 S54 S55 S56 ⎠ S61 S62 S63 S64 S65 S66 ⎞ ⎛ C11 C12 C13 C14 C15 C16 ⎜C C C C C C ⎟ ⎜ 21 22 23 24 25 26 ⎟ ⎟ ⎜ ⎜C31 C32 C33 C34 C35 C36 ⎟ (E.10) .[C] = ⎜ ⎟ ⎜C41 C42 C43 C44 C45 C46 ⎟ ⎟ ⎜ ⎝C51 C52 C53 C54 C55 C56 ⎠ C61 C62 C63 C64 C65 C66

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

157

Fig. E.2 When expressed with the coordinate system of principal axes .x yz the shear stress is zero

E.3

Principal Axes

Before proceeding to further analysis, take a moment to consider the coordinate system to express stress and strain. Refer to Fig. E.2 and consider that an external force is acting on an elastic cubic object. In three dimensions, we can set up an .x yz coordinate system in any direction. In this figure, we consider two coordinate systems, .x yz and .x ' y ' z ' . Notice that the .x yz has the three coordinate axes normal to the corresponding plane of the cube. Consequently, there is no shear stress component. The coordinate axes that satisfy this condition are called principal axes [4]. The axes of the other coordinate system .x ' y ' z ' are not principal axes because when decomposed with this coordinate system, the force vector has shear components.

E.4

Isotropic Case

Now we are in a position to express the strain tensor components with the .x yz and .x ' y'z' coordinate systems. We restrict ourselves in an isotropic case. Here the isotropy means that the elasticity is independent of the orientation. Under this condition, the elasticity is plane-symmetric.

E.4.1

Axis Inversion

Using plane symmetry, we can reduce the number of independent compliance matrix elements. First, consider axis inversions. The elastic response is symmetric about the .x y plane. This means that if the .z axis is inverted the displacement components behave as follows. ξx' = ξx , ξ y' = ξ y , ξz' = −ξz

.

(E.11)

158

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

On the other hand, the spatial derivatives behave as follows. .

∂ ∂ ∂ ∂ ∂ ∂ = = =− , , ∂x' ∂ x ∂ y' ∂ y ∂z ' ∂z

(E.12)

Consequently, the strain components behave as follows through this inversion operation. ∈x' x =

.

' ∈ yy =

.

' ∈zz =

.

' ∈ yz =

.

' ∈zx =

.

∈x' y =

.

∂ξx ∂ξx' = = ∈x x ' ∂x ∂x ∂ξ y' ∂ξ y = = ∈ yy ∂ y' ∂y ∂ξz' ∂(−ξz ) ∂ξz =− = = ∈zz ∂z ' ∂z ∂z ( ) ( ) ∂ξ y' ∂ξ y ∂ξz 1 1 ∂ξz' − = −∈ yz = + − 2 ∂ y' ∂z ' 2 ∂y ∂z ( ( ) ) ∂ξz' ∂ξx 1 1 ∂ξx' ∂ξz − = −∈zx + ' = − 2 ∂z ' ∂x 2 ∂z ∂x ( ) ( ) ∂ξx' 1 ∂ξ y' ∂ξx 1 ∂ξ y = ∈x y + + = 2 ∂x' ∂ y' 2 ∂x ∂y

(E.13) (E.14) (E.15) (E.16) (E.17) (E.18)

Repeating the same procedure for the stress matrix components, we obtain the following equations. ' ' ' ' σx' x = σx x , σ yy = σ yy , σzz = σzz , σ yz = −σ yz , σzx = −σzx , σx' y = σx y

.

(E.19)

Now express the first row of (E.5) using the components of the compliance matrix in both .x yz and .x ' y ' z ' systems, respectively. ∈x x = S11 σx x + S12 σ yy + S13 σzz + S14 σ yz + S15 σzx + S16 σx y

.

∈x' x .

=

S11 σx' x

+

' S12 σ yy

+

' S13 σzz

+

' S14 σ yz

+

' S15 σzx

+

(E.20)

S16 σx' y

= S11 σx x + S12 σ yy + S13 σzz − S14 σ yz − S15 σzx + S16 σx y

(E.21)

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

⎞ ⎛ ∈x' x S11 ⎜∈ ' ⎟ ⎜ S ⎜ yy ⎟ ⎜ 21 ⎜ ∈' ⎟ ⎜ ⎜ zz ⎟ ⎜ S31 .⎜ ' ⎟ = ⎜ ⎜ ∈ yz ⎟ ⎜ S41 ⎜ ' ⎟ ⎜ ⎝ ∈zx ⎠ ⎝ S51 S61 ∈x' y ⎛ S11 ⎜S ⎜ 21 ⎜ ⎜S = ⎜ 31 ⎜ S41 ⎜ ⎝ S51 S61 ⎛

S12 S22 S32 S42 S52 S62

S13 S23 S33 S43 S53 S63

S14 S24 S34 S44 S54 S64

S12 S22 S32 S42 S52 S62

S13 S23 S33 S43 S53 S63

−S14 −S24 −S34 −S44 −S54 −S64

⎞⎛ ' ⎞ ⎛ σx x S11 S12 S13 S16 ⎜ ' ⎟ ⎜ S26 ⎟ ⎟ ⎜σ yy ⎟ ⎜ S21 S22 S23 ⎟⎜ ' ⎟ ⎜ S36 ⎟ ⎜ σzz ⎟ ⎜ S31 S32 S33 ⎟⎜ ' ⎟ = ⎜ ⎟ ⎜S S S S46 ⎟ ⎜ σ yz ⎟ ⎜ ' ⎟ ⎜ 41 42 43 ⎝ ⎠ S56 σzx ⎠ ⎝ S51 S52 S53 S66 S61 S62 S63 σx' y ⎞⎛ ⎞ ⎛ ⎞ σx x ∈x x −S15 S16 ⎜ ⎟ ⎜ ⎟ −S25 S26 ⎟ ⎟ ⎜σ yy ⎟ ⎜ ∈ yy ⎟ ⎟⎜ ⎟ ⎜ ⎟ −S35 S36 ⎟ ⎜ σzz ⎟ ⎜ ∈zz ⎟ ⎟⎜ ⎟ = ⎜ ⎟ −S45 S46 ⎟ ⎜ σ yz ⎟ ⎜−∈ yz ⎟ ⎟⎜ ⎟ ⎜ ⎟ −S55 S56 ⎠ ⎝ σzx ⎠ ⎝−∈zx ⎠ −S65 S66 σx y ∈x y

S15 S25 S35 S45 S55 S65

S14 S24 S34 S44 S54 S64

S15 S25 S35 S45 S55 S65

159

⎞⎛ ⎞ σx x S16 ⎜ ⎟ S26 ⎟ ⎟ ⎜ σ yy ⎟ ⎟⎜ ⎟ S36 ⎟ ⎜ σzz ⎟ ⎟⎜ ⎟ S46 ⎟ ⎜−σ yz ⎟ ⎟⎜ ⎟ S56 ⎠ ⎝−σzx ⎠ S66 σx y

(E.22)

By changing the sign of the fourth and fifth rows of (E.5) we obtain the following equation. ⎛ ⎞ ⎛ ⎞⎛ ⎞ σx x S11 S12 S13 S14 S15 S16 ∈x x ⎜ ⎜∈ ⎟ ⎜ S ⎟ ⎟ ⎜ yy ⎟ ⎜ 21 S22 S23 S24 S25 S26 ⎟ ⎜σ yy ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ∈zz ⎟ ⎜ S31 S32 S33 S34 S35 S36 ⎟ ⎜ σzz ⎟ .⎜ (E.23) ⎟=⎜ ⎟⎜ ⎟ ⎜−∈ yz ⎟ ⎜−S41 −S42 −S43 −S44 −S45 −S46 ⎟ ⎜ σ yz ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝−∈zx ⎠ ⎝−S51 −S52 −S53 −S54 −S55 −S56 ⎠ ⎝ σzx ⎠ ∈x y S61 S62 S63 S64 S65 S66 σx y Comparing (E.22) and (E.23), we obtain the following equation. ⎛ ⎞ ⎛ S11 S12 S13 S14 S11 S12 S13 −S14 −S15 S16 ⎜ S S S −S −S S ⎟ ⎜ S 24 25 26 ⎟ ⎜ 21 S22 S23 S24 ⎜ 21 22 23 ⎜ ⎟ ⎜ ⎜ S31 S32 S33 −S34 −S35 S36 ⎟ ⎜ S31 S32 S33 S34 .⎜ ⎟=⎜ ⎜ S41 S42 S43 −S44 −S45 S46 ⎟ ⎜−S41 −S42 −S43 −S44 ⎜ ⎟ ⎜ ⎝ S51 S52 S53 −S54 −S55 S56 ⎠ ⎝−S51 −S52 −S53 −S54 S61 S62 S63 −S64 −S65 S66 S61 S62 S63 S64

S15 S25 S35 −S45 −S55 S65

⎞ S16 S26 ⎟ ⎟ ⎟ S36 ⎟ ⎟ −S46 ⎟ ⎟ −S56 ⎠ S66

(E.24)

Note that on the left-hand side, the fourth and fifth columns (corresponding to the noninverted axes) the sign is flipped, and on the right-hand side, the fourth and fifth rows (corresponding to the non-inverted axes) the sign is flipped. From (E.24) we find as follows. .

S14 = S24 = S34 = S15 = S25 = S35 = S46 = S56

= S41 = S42 = S43 = S51 = S52 = S53 = S64 = S65 = 0

(E.25)

Note that the first and second lines of (E.25) list the matrix elements at mutually symmetric positions. Thus, by considering inverting the .z axis, we find the compliance matrix has the following form.

160

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

⎛

S11 ⎜S ⎜ 21 ⎜ ⎜ S31 .[S] = ⎜ ⎜ 0 ⎜ ⎝ 0 S61

S12 S22 S32 0 0 S62

S13 S23 S33 0 0 S63

0 0 0 S44 S54 0

0 0 0 S45 S55 0

⎞ S16 S26 ⎟ ⎟ ⎟ S36 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎠ S66

(E.26)

Similarly, by considering inverting the .x axis, we obtain the following equation. Those matrix elements found to be 0 from (E.25) are set to 0 in the below equation. ⎛ ⎞ ⎛ ⎞ S11 S12 S13 S11 S12 S13 0 0 −S16 0 0 S16 ⎜ ⎜S S S 0 −S26 ⎟ 0 0 S26 ⎟ ⎜ 21 22 23 0 ⎟ ⎜ S21 S22 S23 ⎟ ⎜ ⎟ ⎜ ⎟ 0 −S36 ⎟ ⎜ S31 S32 S33 0 0 S36 ⎟ ⎜ S31 S32 S33 0 .⎜ (E.27) ⎟=⎜ ⎟ ⎜ 0 0 0 S44 −S45 0 ⎟ ⎜ 0 0 0 S44 S45 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 0 0 S54 −S55 0 ⎠ ⎝ 0 0 0 −S54 −S55 0 ⎠ 0 −S66 0 −S66 S61 S62 S63 0 −S61 −S62 −S63 0 From (E.27), we find as follows. .

S45 = S16 = S26 = S36 = 0; S54 = S61 = S62 = S63 = 0

(E.28)

Substituting (E.27) into (E.26), we can further simplify the compliance matrix as follows. ⎞ ⎛ S11 S12 S13 0 0 0 ⎟ ⎜S S S ⎜ 21 22 23 0 0 0 ⎟ ⎟ ⎜ ⎜ S31 S32 S33 0 0 0 ⎟ .[S] = ⎜ (E.29) ⎟ ⎜ 0 0 0 S44 0 0 ⎟ ⎟ ⎜ ⎝ 0 0 0 0 S55 0 ⎠ 0 0 0 0 0 S66

E.4.2

Rotation Around Axis

Next, consider rotating the .x y plane around the .z axis by angle .θ . The plane symmetry condition indicates that the elastic property is intact by this operation. Call the coordinate system before this rotational operation .x yz and the one after .x ' y ' z ' . By expressing the same point on the plane by expressing it with the two coordinate systems, we obtain the following equation. ⎛ ⎞ ⎛ ⎞ ⎛ '⎞ x cos θ − sin θ 0 x . ⎝ y ⎠ = ⎝ sin θ (E.30) cos θ 0⎠ ⎝ y ' ⎠ z' z 0 0 1 Similarly, we can express the displacement vector components .ξx , .ξ y , and .ξz written in the old coordinate system with the new coordinate system as follows.

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

⎛ ⎞ ⎛ ξx cos θ . ⎝ξ y ⎠ = ⎝ sin θ ξz 0

⎞⎛ '⎞ ξx − sin θ 0 cos θ 0⎠ ⎝ξ y' ⎠ 0 1 ξz'

161

(E.31)

Next, express the spatial derivatives of the new coordinate system with the old one using (E.30) and (E.31). ∂ ∂ = ∂x' ∂x ∂ ∂ = . ' ∂y ∂x ∂ ∂ . = ∂z ' ∂z .

∂ ∂y ∂ ∂x ∂ + = cos θ + sin θ ∂x' ∂ y ∂x' ∂x ∂y ∂ ∂y ∂ ∂x ∂ + = − sin θ + cos θ ' ' ∂y ∂y ∂y ∂x ∂y

(E.32) (E.33) (E.34)

Using the above relationship between the quantities expressed with .x yz and .x ' y ' z ' systems, we can express .∈x' x in terms of the strain tensor components expressed with the .x yz system. ) ( ) ) ∂ ( ∂ ∂ξx' ∂ ( ' ξx cos θ + ξ y sin θ ξ = cos θ + ξ sin θ = cos θ .∈ x x = + sin θ x y ∂x' ∂x' ∂x ∂y ∂ξ y ∂ξ y ∂ξx ∂ξx = cos2 θ + cos θ sin θ + sin θ cos θ + sin2 θ ∂x ∂x ∂y ∂y = ∈x x cos2 θ + ∈ yy sin2 θ + 2∈x y cos θ sin θ

(E.35)

Similarly, we obtain the following equations. ' ∈ yy = ∈x x sin2 θ + ∈ yy cos2 θ − 2∈x y cos θ sin θ

(E.36)

' ∈zz = ∈zz

(E.37)

.

.

' .∈ yz ' .∈zx ' .∈ x y

= ∈ yz cos θ − ∈zx sin θ

(E.38)

= ∈zx cos θ + ∈ yz sin θ ( ) = 2∈x y cos2 θ − sin2 θ + 2(∈ yy − ∈x x ) cos θ sin θ

(E.39) (E.40)

With the same procedure, we can obtain the following equation for .σx x . σx' x = σx x cos2 θ + 2σx y cos θ sin θ + σ yy sin2 θ

.

(E.41)

We can express .∈x x etc. on the right-hand side of (E.35) with stress matrix components by using the compliance matrix expression (E.29).

162

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

∈x' x = ∈x x cos2 θ + ∈ yy sin2 θ + 2∈x y cos θ sin θ ( ) ) ( = S11 σx x + S12 σ yy + S13 σzz cos2 θ + S21 σx x + S22 σ yy + S23 σzz sin2 θ

.

+ 2S66 σx y cos θ sin θ ( ( ) ) = σx x S11 cos2 θ + S21 sin2 θ + σ yy S12 cos2 θ + S22 sin2 θ ( ) + σzz S13 cos2 θ + S23 sin2 θ + σx y (2S66 ) cos θ sin θ

(E.42)

Now this time first express .∈x' x with .σi j using (E.22) and the compliance matrix (E.29). Note that because of the isotropic condition, the elastic behavior does not change by the rotational operation, hence we can use the same compliance matrix. ' ' ∈x' x = S11 σx' x + S12 σ yy + S13 σzz ( ) = S11 σx x cos2 θ + σ yy sin2 θ + 2σx y cos θ sin θ ( ) + S12 σx x sin2 θ + σ yy cos2 θ − 2σx y cos θ sin θ + S13 σzz ( ( ) ) = σx x S11 cos2 θ + S12 sin2 θ + σ yy S12 cos2 θ + S11 sin2 θ

.

+ σzz (S13 ) + σx y (2S11 − 2S12 ) cos θ sin θ

(E.43)

Comparing (E.42) and (E.43) for the same .σii terms, we find as follows. .

S12 = S21

(E.44)

.

S22 = S11

(E.45)

.

S66 = S11 − S12

(E.46)

Repeating the same procedure for rotation around other axis, we find the following conditions. S11 = S22 = S33 ; S12 = S13 = S23 = S21 = S31 = S32 ; S44 = S55 = S66 = S11 − S12 (E.47) Using (E.44)–(E.46) in (E.29), we obtain the following expression of compliance matrix. ⎞ ⎛ S11 S12 S12 0 0 0 ⎟ ⎜S S S 0 0 0 ⎟ ⎜ 12 11 12 ⎟ ⎜ 0 0 0 ⎟ ⎜ S12 S12 S11 .[S] = ⎜ (E.48) ⎟ ⎟ ⎜ 0 0 0 S11 − S12 0 0 ⎟ ⎜ ⎠ ⎝ 0 0 0 0 0 S11 − S12 0 0 0 0 0 S11 − S12 .

Expression (E.48) is the general form of compliance matrix for an isotropic elastic medium.

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

E.5

163

Expression of Compliance Matrix with Elastic Constants

We next express the compliance matrix with elastic modulus. Consider in Fig. E.3 that external force acts on a block of an elastic object along the .x axis. The elastic block experiences stretch in the .x direction and compression in the . y and .z directions according to Poisson’s effect. Clearly, there is no shear strain if we use the.x yz coordinate system. As shown in Fig. E.3, the external force along the.x axis causes three normal strains; tensile strain along the .x axis, and compressive strains along the . y and .z axes. The compressive strains are due to Poisson’s effect. The compressive strains can be related to the tensile strain via Poisson’s ratio .ν. The overall strain (the addition of all three strains) constitutes .∈x x . Thus, we obtain the following equation. ∈x x =

.

σ yy σx x σzz −ν −ν E E E

(E.49)

Here the first term on the right-hand side represents that the tensile strain in the .x direction is related to .σx x via Young’s modulus . E. The other terms represent that the normal strain is related to the normal stress and that the compressive strain is lower than the tensile strain by a factor of Poisson’s ratio .−ν where the negative sign indicates the strain is opposite to the one along the .x axis, which is the direct consequence of the applied force. Comparison of the compliance matrix (E.48) and (E.49) allows us to identify . S11 and . S12 as follows. 1 −ν . S11 = (E.50) , S12 = E E Using (E.50), we can express the compliance matrix for an isotropic elastic medium with Youg’s modulus and Poisson’s ratio, and thereby write the strain-stress relationship (E.5) as follows. ⎞⎛ ⎞ ⎛ ⎛ ⎞ σx x 1 −ν −ν 0 0 0 ∈x x ⎟ ⎜σ ⎟ ⎜−ν 1 −ν 0 ⎜∈ ⎟ 0 0 ⎟ ⎜ yy ⎟ ⎜ ⎜ yy ⎟ ⎟⎜ ⎟ ⎜ ⎟ 1 ⎜ 0 0 0 ⎟ ⎜ σzz ⎟ ⎜−ν −ν 1 ⎜ ∈zz ⎟ (E.51) .⎜ ⎟⎜ ⎟ ⎟= ⎜ ⎜ ∈ yz ⎟ 0 ⎟ ⎜ σ yz ⎟ E ⎜ 0 0 0 1+ν 0 ⎟⎜ ⎟ ⎜ ⎜ ⎟ ⎝ 0 0 0 ⎝ ∈zx ⎠ 0 1 + ν 0 ⎠ ⎝ σzx ⎠ 0 0 0 0 0 1+ν σx y ∈x y By finding the inverse matrix to the compliance matrix in (E.51), we can express the stressstrain relationship (E.6) in the following form. Fig. E.3 Elastic block experiencing normal strains

y

z

x

164

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Table E.1 Various elastic moduli and their relationships

.E

(.λ, μ)

(. E, G)

(. E, ν)

(.G, ν)

μ(3λ+2μ) . λ+μ

.E

.E

.2(1 .G

.G(μ)

.μ

.G

.ν

λ . 2(λ+μ)

E . 2(1+ν)

E−2G . 2G G(E−2G) . 3G−E

.ν

.ν

νE . (1+ν)(1−2ν)

.

.λ

.λ

+ ν)G

2νG 1−2ν

⎛ ⎞ ⎛ ⎞⎛ ⎞ σx x 1−ν ν ν 0 0 0 ∈x x ⎜σ ⎟ ⎜ ν 1−ν ν ⎜ ⎟ 0 0 0 ⎟ ⎜ yy ⎟ ⎜ ⎟ ⎜∈ yy ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ E 0 0 ⎟ ⎜ ∈zz ⎟ ⎜ σzz ⎟ ⎜ −ν −ν 1 − ν 0 .⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ σ yz ⎟ (1 + ν)(1 − 2ν) ⎜ 0 0 0 1 − 2ν 0 0 ⎟ ⎜ ∈ yz ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ σzx ⎠ ⎝ 0 0 0 0 1 − 2ν 0 ⎠ ⎝ ∈zx ⎠ 0 0 0 0 0 1 − 2ν σx y ∈x y (E.52) Conventionally, the stiffness matrix is expressed with Lamé’s first and second parameters .λ and .μ. ⎛ ⎞ ⎛ ⎞⎛ ⎞ σx x λ + 2μ λ λ 0 0 0 ∈x x ⎜σ ⎟ ⎜ λ ⎟ ⎜∈ ⎟ λ + 2μ λ 0 0 0 ⎜ yy ⎟ ⎜ ⎟ ⎜ yy ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ λ λ + 2μ 0 0 0 ⎟ ⎜ ∈zz ⎟ ⎜ σzz ⎟ ⎜ λ .⎜ (E.53) ⎟=⎜ ⎟⎜ ⎟ ⎜ σ yz ⎟ ⎜ 0 0 0 2μ 0 0 ⎟ ⎜ ∈ yz ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ σzx ⎠ ⎝ 0 0 0 0 2μ 0 ⎠ ⎝ ∈zx ⎠ σx y ∈x y 0 0 0 0 0 2μ Table E.1 summarizes the relationship among Lamé’s parameters, Young’s modulus, shear modulus, and Poisson’s ratio. Notice that we can always express an elastic constant with a pair of other elastic constants.

E.6

Equation of Motion

Refer to Fig. E.4 and consider the force acting on an elastic block of density .ρ having three sides of .d x, .dy, and .dz. The net force in the .x, . y, and .z directions is the addition of the normal and shear force in line with the respective direction. Thus, we obtain the following equation of motion for this elastic block. ∂ 2ξ .ρ = ∂t 2

(

) ( ) ∂σ yx ∂σx y ∂σ yy ∂σzy ∂σzx ˆ ∂σx x + + i+ + + jˆ ∂x ∂y ∂z ∂x ∂y ∂z ( ) ∂σ yz ∂σzz ˆ ∂σx z + + + k ∂x ∂y ∂z

(E.54)

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

165

y +

=

+

+ =

+

z

x

y

z + =

+

+

=

+

+

+

+

+

Fig. E.4 All external force acting on an elastic block. The three cubes in this figure represent the same elastic block. Each of the three cubes illustrates the net force exerted in the corresponding direction

ˆ . j, ˆ and .kˆ are the unit vector for the corresponding direction. Remember that the first Here, .i, subscript .i of the stress tensor component .σi j represents the plane on which the force is acting and the second subscript . j represents the direction of the force. The volume .d xd ydz is canceled on both-hand sides on deriving (E.54). Using (E.53) we can substitute the stress tensor components with the pertinent strain tensor components and obtain the following equation. ) ( ] ∂∈ yx ∂∈zx ˆ ∂ 2ξ ∂ [ (λ + 2μ)∈x x + λ∈ yy + λ∈zz + 2μ + 2μ i .ρ = ∂t 2 ∂x ∂y ∂z ( ) ] ∂σzy ∂∈x y ∂ [ + 2μ + 2λ∈x x + (λ + 2μ)∈ yy + 2λ∈zz + jˆ ∂x ∂y ∂z ( ) ] ∂∈ yz ∂∈x z ∂ [ + 2μ + 2μ + 2λ∈x x + 2λ∈ yy + (λ + 2μ)∈zz kˆ (E.55) ∂x ∂y ∂z The next step is to express the right-hand side of (E.55) in terms of displacement.ξ . Remember the following expressions of the strain tensor components. ∂ξi ∂ xi ) ( 1 ∂ξ j ∂ξi .∈i j = + 2 ∂ xi ∂x j ∈ii =

.

(E.56) (E.57)

166

Appendix E: Generalized Hooke’s Law and Equation of Motion for Isotropic Media

Using (E.56) and (E.57), consider expressing the .x component of the right-hand side of (E.55) expressing with the displacement vector components. .

] ∂∈ yx ∂ [ ∂∈zx (λ + 2μ)∈x x + λ∈ yy + λ∈zz + 2μ + 2μ ∂x ∂y ∂z [ ] ( ) ( ) ∂ξ y ∂ξx ∂ ∂ξ y ∂ ∂ξx ∂ξz ∂ξx ∂ξz ∂ (λ + 2μ) +μ +μ +λ +λ + + = ∂x ∂x ∂y ∂z ∂ y ∂x ∂y ∂z ∂z ∂x ( 2 ) ( ) 2 2 ∂ξ y ∂ ξx ∂ ∂ξx ∂ ξx ∂ ξx ∂ξz =μ + (λ + μ) + + (E.58) + + 2 2 2 ∂x ∂y ∂z ∂x ∂x ∂y ∂z

We can view the first term on the last line of (E.58) as the .x component of .∇ 2 ξ , and the second term as the .x component of .∇(∇ · ξ ). By repeating the same term rearrangement for the. y and.x components on the right-hand side of (E.55), we obtain the following expression. ρ

.

∂ 2ξ = μ∇ 2 ξ + (λ + μ)∇(∇ · ξ ) ∂t 2

(E.59)

Equation (E.59) is the equation of motion we wanted to derive.

References 1. Prof. Brad Osgood (2014) Lecture notes for EE 261 the Fourier transform and its applications. CreateSpace Independent Publishing Platform 2. Bracewell RN (1999) The Fourier transform and its applications, 3rd edn. McGraw-Hill, Boston, New York 3. Brannon RM (2018) Rotation, reflection, and frame changes, Chap. 26, Voigt and Mandel components. IOP, Bristol, UK 4. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. Dover Publications, New York

Index

A Acoustic energy conservation at boundary, 62 Acoustic impedance, 55, 60, 63 Acoustic radiation pressure, 69 Acoustic wave intensity, 56 Acoustoelasticity, 29 Amplitude demodulation, 81 Amplitude modulation, 79 Angular spatial frequency, 7 Angular temporal frequency, 7 Audible frequency, 73 Average acoustic energy, 53 Average intensity of acoustic wave, 57

B Boundary condition, 17–19 Bulk modulus, 35

C Complex propagation constant, 45 Constructive interference, 14 Critical angle, 59, 108

D Damping coefficient, 27 Deaying, longitudinal displacement wave equation, 44 Deaying, transverse displacement wave equation, 44

Decay constant, 27, 43 Decaying acoustic wave solution, 44 Decaying pressure wave in air, 71 Decay wave equation of volume expansion, 43 Destructive interference, 14 Direction cosines, 9 Dispersion, 112

E Eigen frequency, 17, 23 Elastic constant, 32 Elastic force, 24 Elasticity, 23 Elasticity of air, 66 Elastic potential energy, 26 Elastic potential energy density, 51 Electrical impedance, 55, 123, 127, 133 Electric conductivity, 33 Equation of motion due to pressure gradient, 70 Equation of motion for longitudinal vibration, 40 Equation of motion governing series of point masses, 30 Equation of motion of air compression, 36 Equation of motion of isotropic solids, 87 Equation of motion of spring-point mass system, 26 Equation of motion of unit volume, 37 Equation of motion of unit volume with velocity-damping force term, 43

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Yoshida, Fundamentals of Acoustic Waves and Applications, Synthesis Lectures on Wave Phenomena in the Physical Sciences, https://doi.org/10.1007/978-3-031-48200-7

167

168 Euler’s notation, 45

F Faraday’s law, 134 Fixed-end reflection, 18 Forced oscillation, 24, 28 Fourier series, 12 Fourier’s theorem, 12 Frequency demodulation, 83 Frequency modulation, 82 Frequency ranges of vowel and consonants, 74

G Gas law, 68 Gradient operation, 11

H Harmonic oscillation, 24, 27 Harmonics, 13 Hooke’s law, 32 Hooke’s law in air, 35 Hooke’s law of continuum, 33 Human ear, 72

I Impedance matching, 55 Initial condition, 16 Inverse piezoelectric effect, 137

K Kinetic energy density, 53

L Laplacian, 11 Law of reflection, 59 Law of refraction, 59 Linear differential equation, 28 Longitudinal and transverse vibrations in solid, 39 Longitudinal elastic wave equation, 42 Longitudinal elastic wave velocity, 42 Longitudinal wave, 4

Index M Magnetic force, 133 Mechanical impedance, 123 N Natural frequency, 27 Normal strain, 32 Normal stress, 33 O One-dimensional wave equation of series of point masses, 29 Open-end reflection, 19 Overtones, 77 P Particle velocity, 2 Period, 2 Phase velocity, 1, 8 Phase velocity as a delay of oscillation in a series of point masses, 30 Phase velocity as a material constant, 33, 43 Phase velocity of air expansion (compression) wave, 38 Phase velocity of air pressure wave, 38 Phase velocity of compression wave in isotropic media, 88 Phase velocity of longitudinal displacement wave, 34 Phase velocity of rotation wave in isotropic media, 88 Phase velocity of transverse displacement wave, 34 Piezoelectric effect, 137 Plane strain wave solution, 98 Plane wave, 15 Plane wave in elastic media, 91 Primary (P-) wave, 88, 89 Propagation constant, 9 Propagation vector, 9, 92 P-wave velocity, 96 R Radiation pressure, 69 Rayleigh-Lamb frequency equation, 116 Rayleigh surface wave, 111

Index Reflectance, 63 Reflection at free surface with various incident waves, 104 Reflection coefficient, 62 Resonance, 14 Resonance of wave, 20 Resonator, 14 Resonator mode number, 21 Rotation wave equation of solids, 88

S Scalar potential, 88 Scalar potential wave equation, 89 Secondary (S-) wave, 88, 89 Shear-Horizontal wave solution, 102 Shear modulus, 34 . S H waves in a plate, 112 Snell’s law, 59 Spatial frequency, 7 Spring constant, 27 Spring-mass system, 24 Standing wave, 13 Stiffness, 27 S-wave velocity, 96

T Temporal frequency, 7 Transmission coefficient, 62 Transmittance, 63

169 Transverse elastic wave equation, 42 Transverse elastic wave velocity, 42 Transverse waves, 5 Traveling waves, 8

U Unforced oscillation, 24, 26

V Variable separation, 15 Vector potential, 88 Vector potential wave equation, 89 Velocity damping, 26 Volume compression wave equation of solids, 88

W Wave equation, 9 Wave equation of air density change, 39 Wave equation of air pressure, 38 Wave equation of air volume expansion (compression), 37 Wave equation of density change, 39 Wavefront, 92 Wavelength, 2

Y Young’s modulus, 33