Introduction to Acoustics

Warren Koontz

May 6, 2015

1 Introduction

If you pluck a string on an acoustic guitar, the string will vibrate rapidly for at least afew seconds. Since the string is anchored to the guitar at the bridge, the body of theguitar, especially the soundboard, will vibrate along with the string. The air around theguitar is displaced by these vibrations resulting in small variations in air pressure whichin turn affect more of the air around the guitar so that in a short time the variationsare occuring at some distance from the guitar. Upon reaching a healthy human ear, thevariations in air pressure create vibrations in the eardrum which are processed by theinner ear into electrical signals that are passed on to the brain and interpreted as sound.

What we have described here is a simple acoustic system consisting of three compo-nents:

SourceIn this example the source is the acoustic guitar. In general an acoustic source isa mechanism that sets the surrounding medium in motion. Mechanisms includemechanical vibration (guitar strings and soundboard, loudspeaker cone, etc.), ex-plosion (sudden release of gas by burst balloon, gun shot, etc.), and turbulence(forcing air through a small aperture). The human voice uses many of these mech-anisms to produce the vast variety of sounds of spoken language.

MediumA medium is necessary for the transmission of sound away from the source - soundcannot exist in a vacuum. The most common medium in human experience isthe air around us, but many media are possible, including liquid and solid mediaas well as gas. As stated above, the source sets the medium in motion and thismotion spreads away from the source as an acoustic wave. We will have more tosay about acoustic waves later in this chapter. For now, think of acoustic wavesas being analogous to the waves that form in still water when it disturbed by, forexample, a tossed stone.

ReceiverAlthough the combination of a source and a medium is sufficient to create sound,it is more interesting when the system includes a human listener (or other hearing

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Figure 1: Illustration of a Sound Wave

animal), a recording device, or other acoustic receiver. A receiver usually includesa membrane or other structure that is set in motion by the sound wave and amechanism to convert this motion into an electrical (usually) signal. In the case ofa human listener, the membrane is the eardrum and the inner ear creates electricalimpulses that are carried by the nervous system to the brain.

Figure 1 is a rough, two-dimensional illustration of an acoustic wave generated by ahypothetical omni-directional source (e.g., a vibrating sphere) in an air medium. Thealternating dark and light shading represents variation in the local air pressure caused bythe sound source. In the dark areas the air pressure is above the normal atmospheric leveland in the light areas the air pressure is below the normal atmospheric level. The highand low pressure areas are called compressions and rarefactions, respectively. The figure

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represents a moment in time. In an animated version of the figure1, the compressionand rarefaction waves propagate away from the source. Thus in the presense of sound,air pressure is varying (rapidly!) with both time and location.

2 Acoustic Pressure

An acoustic source disturbs the medium around it and causes properties of the mediumto vary with time and location. If the medium is air, one of the affected properties isthe air pressure, which we will measure in Pascals (Pa) where 1 Pa = 1 nt/m2. For nowwe will limit our discussion to the variation of pressure with time at some fixed point inspace

The total atmospheric pressure at a given point in the presence of sound can beexpressed mathematically as

P (t) = p(t) + P0 (1)

where P (t) is the total pressure, p(t) is the acoustic pressure caused by an acoustic waveand P0 is the normal atmospheric pressure. The notation indicates the time dependenceof the total pressure and the acoustic pressure. The normal atmospheric pressure istreated as a constant since it varies much more slowly than the acoustic pressure. Thenormal atmospheric pressure P0 is about 100,000 pascals (Pa) at sea level. During acompression the total pressure is greater than P0 and p(t) is positive. During a rarefac-tion the total pressure is less than P0 and p(t) is negative. Except for a relatively smallnumber of basic sounds, such as pure tones, we cannot completely specify p(t) in closedform. We can, however, define some useful overall measures of p(t), such as the RMS(root mean squared) amplitude.

The RMS amplitude of the acoustic pressure is given by

prms =

√1

T

∫ T

0p2(t)dt (2)

where T is a suitable averaging interval. If p(t) is measured in Pa, then prms is also inPa. The RMS amplitude of acoustic pressure, which is related to the perceived loudnessof the sound, is much smaller than P0 and is usually measured in micro-Pascals (µPa).

Sound pressure level (SPL) is a more common measure of acoustic pressure. SPL isa logarithmic measure given by

pdBSPL = 20 log10

(prms

pref

)(3)

where pref is the “reference level”, which is set at 20 µPa (the minimum level that ahuman can hear). The inverse formula to convert SPL to acoustic pressure is

prms = pref10pdBSPL/20 (4)

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Description RMS Level (Pa) dBSPL

Threshold of hearing 0.00002 0

Empty recording studio 0.0002 20

Average home 0.0063 50

Vacuum cleaner @ 1 m 0.063 70

Diesel truck @ 10 m 0.63 90

Chain saw @ 1 m 6.3 110

Threshold of pain 63.0 130

Table 1: Acoustic Pressure and dBSPL for Various Sounds

Table 1 lists the RMS and SPL measures of acoustic pressure for a variety of sounds.Table 1 shows that whereas the RMS level varies by orders of magnitude from the quietestto the loudest sounds, dbSPL increases more or less linearly. Also, dbSPL appears to bemore closely related to human perception of “loudness”.

3 Acoustic Waves

As we stated earlier, in the presence of sound, acoustic pressure varies with both timeand location as shown in Figure 1. Figure 2 is an illustration of another acoustic wave,called a plane wave, where the variation with location depends on only a single spacecoordinate. In this simple case we can express the acoustic pressure as p(t, x) where xis the position along the x-axis. We will eventually show that in general

p(t, x) = f(t− x/c) + g(t+ x/c) (5)

The function f(t − x/c) represents a wave propagating in the positive x direction andthe function g(t + x/c) represents a wave propagating in the negative x direction. Theterm c is called the wave velocity and its value is determined by the medium. The exactforms of f(·) and g(·) depend on the details of the source, but we can gain much insightby considering the basic case of a sinusoidal wave propagating only in the positive xdirection. In this case the acoustic pressure is given by

p(t, x) = A sin[ω(t− x/c)]= A sin(ωt− βx) (6)

where A is the amplitude in Pa, ω is the frequency in radians per second, and β = ω/cis called the wave number. Figure 2 is a plot of (6) with A = 100 µPa and t = 0 withthe distance x measured in wavelengths. The wavelength is the distance λ such that

βλ = 2π (7)

1Several examples can be found on the web

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Figure 2: Illustration of a Plane Acoustic Wave

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Figure 3: Illustration of Wave Velocity

Thus (6) can also be written as

p(t, x) = A sin (ωt− 2πx/λ) (8)

The wavelength is the distance between successive maxima, minima, or any other match-ing points on the acoustic wave. The wavelength and frequency are closely related since

λ =2π

β=

2πc

ω=c

f(9)

so thatλf = c (10)

Figure 3 is a plot of the acoustic wave of Figure 2 at two instants of time. Duringthis period of time dt the wave has moved in the positive direction of the x-axis by asmall amount dx. For any point on the wave, the argument of the sine function mustremain constant, i.e.,

ωt− βx = ω(t+ dt)− β(x+ dx) (11)

so thatωdt− βdx = 0 (12)

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Figure 4: Small Mass of Air

ordx

dt= ω/β = c (13)

Thus the wave is propagating in the positive x direction at velocity c. In air at sea levelc ≈ 340.27 m/s - the “speed of sound”.

4 Plane Acoustic Waves

In this section we will derive the equation describing a plane acoustic wave. This is one ofthe simplest examples of a wave equation, but it illustrates some of the physics behindacoustic waves and provides insights that are helpful in understanding more complexacoustic waves.

Figure 4 shows a small mass of air that is being displaced by a plane acoustic wave.The solid cylinder represents the equilibrium shape of this mass and the dotted cylinderrepresents the distortion induced by the acoustic wave. The length of the equilibriumcylinder is δx and the length of the displaced cylinder is

δx+∂ξ

∂xδx

where ξ(x, t) is the displacement. Since we are dealing with a fixed quantity of air, themass within the two cylinders must remain constant. The mass of the equilibrium (solid)cylinder is given by

δM = ρ0Sδx = ρ0δV0 (14)

where ρ0 is the equilibrium density of the air and S is the area of the face of the cylinder.

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The mass of the displaced (dotted) cylinder is given by

δM = ρS

[δx+

∂ξ

∂xδx

]= ρ

[1 +

∂ξ

∂x

]δV0 (15)

where ρ(x, t) is the density of the displaced cylinder. Since both equations must yieldthe same mass, we have

ρ

[1 +

∂ξ

∂x

]= ρ0 (16)

The total density ρ(x, t) can be expressed as the sum of the equilibrium density ρ0 anda variable incremental density ρ′(x, t), i.e., ρ = ρ0 + ρ′. Substituting this into (16) anddropping the second order term, we have

ρ′ = −ρ0∂ξ

∂x(17)

The total pressure and the total density are related by the adiabatic gas law as follows:

P = αργ (18)

The total pressure can also be expressed as the sum of the equilibrium pressure and theacoustic pressure as P = P0 + p. Applying a Taylor series approximation, we can write

P (ρ0 + ρ′) = P (ρ0) +∂P

∂ρ

∣∣∣∣ρ=ρ0

ρ′

= P0 + p (19)

so that the acoustic pressure can be expressed as

p =∂P

∂ρ

∣∣∣∣ρ=ρ0

ρ′ (20)

We can use (18) to determine the partial derivative resulting in the following expressionfor the acoustic pressure p.

p =γP0

ρ0ρ′ (21)

Substituting (17) into (21) yields a relationship between the incremental pressure andthe displacement.

p = −γP0∂ξ

∂x(22)

Figure 5 shows the forces acting on our small mass. These forces are due to pressuresimposed on the surface areas. Since the equilibrium pressure P0 is the same on both

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Figure 5: Forces on a Small Mass of Air

sides, the net force on the mass is entirely due to the incremental pressure p. The netforce can therefore be expressed as

∆F = p(x, t)S − p(x+ δx, t)S

= −∂p∂xSδx (23)

The net force accelerates the mass according to Newton’s second law, i.e.,

∆F = δM∂v

∂t

= ρ0∂v

∂tSδx (24)

where v is the velocity of the small body of mass. The velocity v(x, t), which we willcall the acoustic velocity, is related to the displacement as

v =∂ξ

∂t(25)

In (24) we use the equilibrium density since the incremental density is a second orderquantity. Combining (23) and (24), we have

∂p

∂x= −ρ0

∂v

∂t(26)

To complete the home stretch of this lengthy derivation, we first differentiate (26) withrespect to x to obtain

∂2p

∂x2= −ρ0

∂2v

∂x∂t(27)

Next we differentiate (22) twice with respect to t to obtain

∂2p

∂t2= −γP0

∂2

∂x∂t

(∂ξ

∂t

)= −γP0

∂2v

∂x∂t(28)

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and combine these two results to obtain

∂2p

∂x2=

1

c2∂2p

∂t2(29)

where

c =

√γP0

ρ0(30)

Equation (29) is a one-dimensional wave equation. In this case the quantity of interestis the acoustic pressure p. It turns out that the acoustic velocity v satisfies the samewave equation, i.e.,

∂2v

∂x2=

1

c2∂2v

∂t2(31)

Both the acoustic pressure p and the acoustic velocity v will be of interest to us.The solution of the wave equation depends upon the boundary conditions. In the

case of the plane acoustic wave, one of the primary boundary conditions is the acousticsource, which we will assume to be sinusoidal (or complex exponential) with radianfrequency ω = 2πf . The general solution to the wave equation is given by

p(x, t) = f(t− x/c) + g(t+ x/c) (32)

as can be easily verified by substitution into (29). As discussed earlier, functions f and gare waves propagating in the positive and negative x direction, respectively. The wavespropagate with wave velocity (not to be confused with acoustic velocity) c.

For now we will assume that there is only a wave traveling in the positive x directionand the wave is driven by a complex exponential source located at x = 0. This leads tothe following boundary condition:

p(0, t) = f(t) = Pejωt (33)

This leads immediately to a general expression for p(x, t):

p(x, t) = Pejω(t−x/c)

= Pej(ωt−βx) (34)

where β = ω/c is called the wave number. It will be convenient to re-write (34) as

p(x, t) = P (x)e−jωt (35)

whereP (x) = Pe−jβx (36)

In many instances it suffices to consider only the variation of the wave with respect tothe space variable x, which is given by (36). If there is also a wave propagating in thenegative x direction, (36) becomes

P (x) = P+e−jβx + P−ejβx (37)

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4.1 Plane Acoustic Wave in Arbitrary Direction

Up to now we have limited our consideration to a plane wave propagating along the xaxis (in either direction). We will now consider a plane wave propagating in a moregeneral direction so that the acoustic pressure (or velocity) depends on more than onespace variable. In this case a more general version of the wave equation applies:

∇2p =1

c2∂2p

∂t2(38)

where

∇2p =∂2p

∂x2+∂2p

∂y2+∂2p

∂z2(39)

To simplify things a bit we will consider a plane wave propagating in the x-y plane sothat the last term of (39) is zero. If we again have a complex exponentail source thenthe pressure wave is given by

p(t, x, y) = P0ej(ωt−βxx−βyy) (40)

where

βx = β cos θ

βy = β sin θ (41)

and θ is the angle of propagation relative to the x axis. Figure 6 is an example of such aplane wave. We will consider this wave further when we discuss reflection and refractionof an acoustic wave.

5 Acoustic Impedance

In Section 4 we noted that the acoustic velocity v(x, t) also satisfies the wave equationand is thus another way to characterize an acoustic wave. In this section we will defineacoustic impedance in terms of the relationship between acoustic pressure and acousticvelocity. Acoustic pressure and velocity turn out to be analogous to voltage and currentin a transmission line or electric and magnetic fields in a radio wave (This analogy maybe more useful to some than to others!).

The most basic definition of acoustic impedance is given by

z =p

v(42)

which is called the intrinsic acoustic impedance. We can derive a relationship betweenp and v starting with (22).

p = −γP0∂ξ

∂x(43)

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Figure 6: Plane Wave Propagating at 30◦

We can differentiate this with respect to t and make some substitutions

∂p

∂t= −γP0

∂2ξ

∂t∂x

= −γP0∂v

∂x

= −ρ0c2∂v

∂x(44)

In the case of complex exponential waves, this equation can be used to derive a formulafor the intrinsic acoustic impedance. For a wave propagating in the positive x directionwe have

p = P+ej(ωt−βx)

v = V +ej(ωt−βx) (45)

Substituting this into (44) results in

jωP+ej(ωt−βx) = jβρ0c2V +ej(ωt−βx)

= jωρ0cV+ej(ωt−βx)

(46)

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Figure 7: Reflection and Transmission at an Impedance Boundary

Now we can calculate the intrinsic acoustic impedance as

z =P+ej(ωt−βx)

V +ej(ωt−βx)

= ρ0c (47)

For a wave propagating in the negative x direction it turns out that z = −ρ0c. Clearly,the intrinsic acoustic impedance is a property of the medium. For air at sea level, ρ0 =1.204 Kg/m3 and c = 343.2 m/s so that z = 413.2 Kg/m2-s. The MKS unit for acousticimpedance is the Rayl and is defined as 1 Kg/m2-s. (Unfortunately, the CGS unit foracoustic impedance, g/cm2-s, is also called the Rayl, but we will stick with MKS.) TheRayl can also be expressed as 1 Pa-s/m.

Now consider an complex exponential acoustic wave propagating through one medium,such as air, and encountering a different medium, such as a wall. As shown in Figure7, the wave is both transmitted and reflected at the boundary between the two media.Both the acoustic pressure and the acoustic velocity are continuous at the boundary sothat

P+1 + P−1 = P+

2

V +1 + V −1 = V +

2 (48)

The second part of (48) can be re-written in terms of the acoustic pressure as

P+1

z1− P−1

z1=P+2

z2(49)

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Figure 8: Reflection and Transmission of a Non-Normal Plane Wave

We can use (48) and (49) to obtain expressions for the transmitted and reflected waves.

P+2 =

2z2z2 + z1

P+1

P−1 =z2 − z1z2 + z1

P+1 (50)

This can be re-written in terms of a parameter Γ known as the reflection coefficient

P+2 = (1 + Γ)P+

1

P−1 = ΓP+1 (51)

where

Γ =z2 − z1z2 + z1

(52)

It may appear from (50) that the sum of the reflected and transmitted waves exceedsthe incident wave. However, as we will see in Section 6, the sum of the transmitted andreflected power is equal to the incident power.

Figure 8 shows a plane wave with amplitude Pi incident on an impedance boundaryat an angle θi (the angle of incidence). There is both a reflected wave with amplitudePr and angle of reflection θr and a transmitted wave with amplitude Pt and angle of

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refraction θt. The three corresponding pressure wave amplitudes are given by

Pi = Pie−j(β1 cos θix+β1 sin θiy)

Pr = Pre−j(−β1 cos θrx+β1 sin θry)

Pt = Pte−j(β2 cos θtx+β2 sin θty) (53)

Since the pressure must be continuous at the boundary (x = 0), we have

Pie−jβ1 sin θiy + Pre

−jβ1 sin θry = Pte−jβ2 sin θty) (54)

Equation 54 must be true all along the boundary, i.e., for all values of y. This can onlybe true if

β1 sin θi = β1 sin θr = β2 sin θi (55)

This leads to the acoustic version of Snell’s Law:

θi = θr

β1 sin θi = β2 sin θt (56)

Since βx = ω/cx the second part of (56) can also be written as

sin θisin θt

=c1c2

(57)

From (54) we havePi + Pr = Pt (58)

just as in the case of normal incidence (θi = 0). We also require that the normalcomponent of the acoustic velocity be continuous across the boundary so that

Vi cos θi + Vr cos θr = Vt cos θt (59)

We can re-write (59) using pressures and impedances:

Piz1

cos θi −Prz1

cos θr =Ptz2

cos θt (60)

Using (56), (58) and (60) we can determine the following relationships among the inci-dent, reflected and refracted waves:

PrPi

=z2 cos θi − z1 cos θtz2 cos θi + z1 cos θt

PtPi

=2z2 cos θi

z2 cos θi + z1 cos θt(61)

Figure reffig:snell2 shows incident, reflected and refracted plane acoustic waves at theboundary between two media. The incident and reflected waves form an interferencepattern.

Note that if β1 > β2 the angle of refraction θt is 90◦ for θi > θc where

θc = sin−1 β2/β1 (62)

Thus if the angle of incidence is greater than the critical angle given by (62) then thewave is entirely reflected.

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Figure 9: Incident, Reflected and Refracted Acoustic Waves

6 Acoustic Power and Intensity

As an acoustic wave passes through a medium, it sets the medium into motion andthis clearly requires energy. Thus we must conclude that the acoustic wave carries someamount of energy that is stored in the medium and then retrieved as the medium returnsto its equilibrium state. Acoustic power and intensity are measures of this energy. Ashas been our custom, we will consider only plane waves.

Let p and v be the acoustic pressure and velocity of a plane acoustic wave and considera small surface with area dS perpendicular to the wave. The force on the surface is pdSand during a small interval of time dt the surface moves a distance vdt. Thus the wavedoes a small amount of work dW given by

δW = pδS · vδt (63)

The acoustic intensity is the acoustic power per unit area and is given by

I =dW

dt/δS = pv =

p2

z(64)

The average intensity is proportional to the average value of the square of the pressure.If the pressure wave is sinusoidal, then the average intensity is given by

I =p2

z=P 2

2z(65)

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where P is the magnitude of the pressure wave. Now we can see what happens to acousticpower at a partially reflecting boundary. In the case of normal incidence the reflectedand transmitted pressure magnitudes are given by

P−1 =z2 − z1z2 + z1

P+1

P+2 =

2z2z2 + z1

P+1 (66)

The incident, reflected and transmitted intensities are given by

Ii =P+21

2z1

Ir =P−21

2z1=

(z2 − z1z2 + z1

)2

Ii

It =P+22

2z2=z1z2

(2z2

z2 + z1

)2

Ii (67)

It is straightforward to show with these equations that the sum of the reflected andtransmitted intensities equals the input intensity.

7 Plane Waves in Lossy Media

The plane wave described in Section 4 maintains all of its acoustic power as it propa-gates along. Real acoustic waves lose power due to absorption of acoustic power by themedium. More advanced texts model this power loss by adding terms to the basic waveequation, but we will take the simpler approach of adjusting the solution to include loss.

In a lossy medium (a medium that absorbs power from the acoustic wave) Equation37 becomes

P (x) = P+e−γx + P−eγx (68)

whereγ = α+ jβ (69)

Thus the amplitude of the acoustic pressure (and acoustic displacement) decays expo-nentially with propagation distance. Note that α is generally frequency dependent andhas units of nepers/meter.

8 Spherical Waves

An acoustic wave emanating from a point source can be modeled as a spherical wavethat varies with time and a single space variable - the distance from the source. If thedistance between the source and the receiver is large and the source is physically small,it can be approximated as a point source.

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When working with a point source, it is most convenient to use spherical coordinates(r, θ and φ) since in this coordinate system the only relevant coordinate is r - the distancefrom the source. The wave equation becomes

∂2p

∂r2+

2

r

∂p

∂r=

1

c2∂2p

∂t2(70)

A general solution to this equation is given by

p(r, t) =A

rf(t− r/c) (71)

as can be verified by substituting this solution into Equation 70. Equation 71 describesa spherical wave propagating away from the source. We generally assume that thereis no corresponding wave traveling toward the source, although this is mathematicallypossible. If the source is a complex exponential, then Equation 71 becomes

p(r, t) =A

rej(ωt−βr) (72)

where β = ω/c as in the case of a plane wave. Figure 10 is an example plot of (??)for a fixed point in time and for r greater than some threshold to avoid the obviousproblem at r = 0. Since the intensity of a wave is proportional to the square of theacoustic pressure, it is clear from (71) that the intensity of a spherical wave decreaseswith the square of the distance from the source. This loss of intensity is simply dueto the spreading of the wave and is in addition to any other loss mechanism, such asabsorption.

9 Summary

We have developed the notion of sound as a disturbance created by a source that prop-agates through a medium to one or more receivers. We have quantified sound in termsof acoustic pressure and acoustic velocity. Both of these quantities satisfy a wave equa-tion that provides a more rigorous measure of how these quantities vary with time andspace to form acoustic waves. We have also shown how acoustic pressure and velocityare related by acoustic impedance. This enabled us to see what happens when acousticwaves encounter a boundary between two different media. And finally we developed ameasure of acoustic intensity that quantifies the power carried by acoustic waves.

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Figure 10: Spherical Acoustic Wave

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