Chapter 16: Sound and Hearing

Chapter 16: Sound and Hearing The most general definition of sound is that it is a longitudinal wave in a medium (air, water, etc.). When a sound wave...
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Chapter 16: Sound and Hearing The most general definition of sound is that it is a longitudinal wave in a medium (air, water, etc.). When a sound wave passes through air, it causes variations in pressure. Your ear perceives these changes in pressure as sound. The normal range of frequencies of sound waves that humans can hear is about 20 Hz to 20 kHz.

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Ch. 16: Sound and Hearing

Figure 1 shows a plunger driven horizontally in SHM, creating a sound wave in a cylinder of fluid (air, for example).

Figure 1

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Ch. 16: Sound and Hearing

The plunger creates a pattern of compressions (high-pressure regions) and rarefactions (low-pressure regions). This pattern propagates horizontally through the column of air. As this pattern of compressions and rarefactions passes, a particle of air (a given nitrogen molecule, e.g.) oscillates horizontally back and forth. The amplitude of this oscillation, shown as A in Fig. 1, is relatively small compared with the wavelength (the distance between successive compressions or rarefactions). Note also that the distance between any two particles that are one wavelength apart stays fixed. I will call the position of any particle in the fluid x ( t ) . I will call the

displacement of any particle away from its equilibrium position y ( x, t ) , as we did in the last chapter. (Note that this is not the same as x.) Remember that this displacement is in the horizontal direction. If the plunger is driven sinusoidally, it creates sinusoidal oscillations of any particle so that: y ( x, t ) = A cos ( kx − ωt ) (1)

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Ch. 16: Sound and Hearing

Sound Waves as Pressure Fluctuations We can describe the sound waves in Fig. 1 in terms of the displacements of the particles or in terms of variations in pressure. Figure 2 shows a cylindrical “slice” of the fluid in Fig. 1.

Figure 2

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Ch. 16: Sound and Hearing

When there is no wave present, the volume of this cylinder of fluid is S Δx . At a time t when a wave is passing through this slice, particles at x are displaced by y ( x, t ) while particles at x + Δx are displaced by y ( x + Δx, t ) . The change in volume of this slice is therefore:

ΔV = S ⎡⎣ y ( x + Δx, t ) − y ( x, t ) ⎤⎦ Since the original volume was V = S Δx , the fractional change in volume (i.e., the fraction that ΔV is of the original volume V ) is given by: ΔV S ⎡⎣ y ( x + Δx, t ) − y ( x, t ) ⎤⎦ y ( x + Δx, t ) − y ( x, t ) = = Δx V S Δx In the limit as Δx → 0 , ΔV → dV , and we get: ⎡ y ( x + Δx, t ) − y ( x, t ) ⎤ ∂y dV == lim ⎢ (2) ⎥= Δ x → 0 V Δx ⎣ ⎦ ∂x To see how this is related to the pressure, recall the bulk stress-strain relation from Ch. 11: ⎛ ΔV ⎞ Δp = − B ⎜ ⎟ ⎝ V ⎠

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Ch. 16: Sound and Hearing

(Note that I chose in the discussion above to call the original volume – with no wave present – simply V instead of V0 .) As Δx → 0 and ΔV → dV , Δp → dp . This small change in pressure is a fluctuation above and below the pressure when no wave is present (i.e., a fluctuation above and below atmospheric pressure, pa ). We now let p ( x, t ) represent this deviation away from atmospheric pressure. Therefore: ⎛ dV ⎞ p ( x, t ) = − B ⎜ ⎟, ⎝ V ⎠ or, using (2): ⎛ ∂y ⎞ (3) p ( x, t ) = − B ⎜ ⎟ ⎝ ∂x ⎠ Plugging in y ( x, t ) from (1), I get:

p ( x, t ) = BkA sin ( kx − ωt ) , (4) so the pressure varies sinusoidally between pmax = + BkA and − pmax = − BkA as the wave propagates down the x-axis. The amplitude of p ( x, t ) , which we call the pressure amplitude, is therefore: pmax = BkA (5) 6

Ch. 16: Sound and Hearing

Speed of Propagation We can derive an expression for the speed of propagation of the wave in the same way we did for waves on a string. Applying Newton’s second law to the “slice” of fluid in Fig. (2), I get: ∂2 y ∑ Fx = max = m ∂t 2 The force on each end of the slice is the pressure at that end times the cross-sectional area S. So: ∑ Fx = − ⎡⎣ F ( x + Δx, t ) − F ( x, t )⎤⎦ = − S ⎡⎣ p ( x + Δx, t ) − p ( x, t )⎤⎦ The mass of fluid in the slice is given by: m = ρV , (6) in which ρ is the density (mass per unit volume). The volume of the slice is V = S Δx . Plugging all this into Newton’s second law, I get: ∂2 y − S ⎡⎣ p ( x + Δx, t ) − p ( x, t ) ⎤⎦ = ( ρ S Δx ) 2 ∂t ⎡ p ( x + Δx, t ) − p ( x, t ) ⎤ ⎛ ∂2 y ⎞ −⎢ ⎥ = ρ⎜ 2 ⎟ Δx ⎝ ∂t ⎠ ⎣ ⎦ Taking the limit as Δx → 0 gives: 7

Ch. 16: Sound and Hearing

⎛ ∂2 y ⎞ ∂p − = ρ⎜ 2 ⎟ ∂x ⎝ ∂t ⎠ Using (3), this becomes:

∂2 y ρ ⎛ ∂2 y ⎞ = ⎜ 2 ⎟ 2 B ⎝ ∂t ⎠ ∂x This looks like the 1-D wave equation: ∂2 y 1 ⎛ ∂2 y ⎞ = 2 ⎜ 2 ⎟, 2 ∂x v ⎝ ∂t ⎠ from which we can read off the speed of propagation: B v=

ρ

(7)

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Ch. 16: Sound and Hearing

Standing Waves in a Fluid When sound waves propagate in a fluid in a pipe with finite length, the waves are reflected from the ends in a way similar to the way in which transverse waves on a string are reflected from the ends of the string. Just as for transverse waves on a string, standing waves in a pipe create sound waves in the surrounding air.

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Ch. 16: Sound and Hearing

Figure 3 shows the sinusoidal particle displacement y ( x, t ) and pressure fluctuation p ( x, t ) given by Eqs. (1) and (4), respectively. Notice that at

points where y ( x, t ) = 0 , the fluid particles are stationary. Neighboring particles are displaced either toward the stationary particle, leading to a compression, or away from the stationary particle, giving a rarefaction.

Figure 3 10

Ch. 16: Sound and Hearing

When a standing wave is established in a fluid, there are points where y ( x, t ) = 0 at all t. These are called displacement nodes. At these points, the pressure fluctuation is largest, giving a pressure antinode. A displacement node is always a pressure antinode, and a pressure node is always a displacement antinode.

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Ch. 16: Sound and Hearing

Organ Pipes and Wind Instruments Figure 4 shows a diagram of an organ pipe. A blower supplies air to the Figure 4

pipe at its bottom end. A stream of air emerges from the narrow opening at the edge of the horizontal surface and is directed against the top edge of the opening, called the mouth of the pipe. The column of air in the pipe is set into vibration and there is a series of possible standing waves. The mouth is open to the atmosphere and therefore always acts as a pressure node and therefore a displacement antinode. 12

Ch. 16: Sound and Hearing

Open Pipe If the pipe is open at both ends, then both ends are pressure nodes and displacement antinodes, as shown in Figure 5.

Figure 5

The condition to get standing waves is evidently: ⎛λ⎞ L = n ⎜ ⎟ , n = 1, 2,3,... ⎝2⎠ or, using v = λ f : nv , n = 1, 2,3,... fn = 2L

(8)

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Ch. 16: Sound and Hearing

Pipe Closed at One End If the pipe is closed at one end and open at the other, we get the situation shown in Figure 6.

Figure 6

The closed end is a displacement node because the end of the pipe does not allow the particles there to oscillate. The open end is, as before, a pressure node and therefore a displacement antinode. The condition for standing waves is now: ⎛λ⎞ L = n ⎜ ⎟ , n = 1,3,5,... ⎝4⎠ or: nv fn = , n = 1,3,5,... (9) 4L

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Ch. 16: Sound and Hearing

The Doppler Effect The Doppler effect is a change in the perceived pitch of a sound wave caused by the relative motion of the source and the “observer” (i.e, the listener). The effect gets its name from Christian Doppler, who explained it in the 1800s. Experimentally, we find that when the source and the observer are moving toward one another, the perceived pitch of the sound is higher; when source and observer are moving away from one another, the pitch is lower. What our ears perceive as the pitch of the sound is related to the frequency: high-pitched sounds have high frequencies; low-pitched sounds have low frequencies. This implies the following: When the source and observer are moving toward one another, the perceived frequency is higher; when source and observer are moving away from each other, the perceived frequency is lower. (Doppler effect)

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Ch. 16: Sound and Hearing

Case 1: Stationary Source, Moving Observer Figure 7 shows a stationary source (the “boom box”) producing sound waves while a moving observer (the boy on the skateboard) moves toward the source. In this case, we expect to find that the boy perceives a frequency higher than if he were at rest relative to the boom box.

Figure 7

The purple rings represent wave fronts (compressions). These propagate at 343 m/s relative to the air (the speed of propagation of sound through air at 20° C). The frequency of the source is f and the wavelength of the waves is λ . These are related by v = λ f . 16

Ch. 16: Sound and Hearing

Because he is moving toward the source with speed u , the observer perceives the waves to be coming at him with a speed of propagation: v′ = v + u I will use the “prime” to indicate perceived quantities throughout this discussion. The wavelength he perceives, however, is just λ ′ = λ , the wavelength he would perceive if he were standing next to the boom box. So we have: v′ = λ ′ f ′ v+u v+u f′= = λ (v f ) ⎛ u⎞ f ′ = f ⎜1 + ⎟ ⎝ v⎠ Repeating this with the observer moving away from the source effectively means replacing u with −u , so: ⎛ u⎞ f ′ = f ⎜1 ± ⎟ , (10) ⎝ v⎠ in which we should choose the + sign when the observer is moving toward the source and the – sign when the observer is moving away.

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Ch. 16: Sound and Hearing

Case 2: Moving Source, Stationary Observer Figure 8 shows a situation in which a pickup truck (the source) blows its horn as it moves toward a stationary observer (the woman).

Figure 8

The observer hears a higher frequency than she would if the source were not moving, but for a different reason than in Case 1. This time, the wave fronts get “bunched up” in front of the source and spread out behind the source. This happens because the source moves some distance between the time when one wave front is emitted and the time when the next wave front is emitted. 18

Ch. 16: Sound and Hearing

Consider Figure 9. This picture shows a wave front that was emitted at t = 0 just arriving at the observer’s ear at a time t = T when the next wave front is just about to be emitted. (Notice that this time is the period of the sound wave, the time between emission of successive wave fronts.)

Figure 9

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Ch. 16: Sound and Hearing

In the time T , the distance traveled by the wave front that was emitted at t = 0 is: λ = vT , in which v = 343 m/s . In this time, though, the source has moved a distance uT , so that when the source emits the next wave front, the spacing between the wave fronts will be: λ ′ = vT − uT This is what the observer will perceive the wavelength to be. Note that the observer perceives the waves to be coming at her at v = 343 m/s . Therefore: v′ v f′= = λ ′ vT − uT ⎛ 1 ⎞ f′= f ⎜ ⎟ 1 − u v ⎝ ⎠ If we repeated this analysis for the source moving away from the observer, it would be equivalent to replacing u with −u . Therefore:

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Ch. 16: Sound and Hearing

⎛ 1 ⎞ f′= f ⎜ (11) ⎟, ⎝1∓ u v ⎠ in which we should choose the – sign when the source is moving toward the observer and the + sign when the source is moving away from the observer. The General Case: Moving Source, Moving Observer If both the source and observer are moving, Equations (10) and (11) can be combined to give: ⎛ v ± uo ⎞ (12) f′= f ⎜ ⎟, ⎝ v ∓ us ⎠ in which uo is the speed of the observer and us is the speed of the source.

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Ch. 16: Sound and Hearing

Sound Intensity Propagating waves transfer energy from one region of the medium to another. We define the intensity, I , of the wave to be the average energy transferred per unit time per unit area perpendicular to the direction of propagation. ΔE I≡ , (13) AΔt in which ΔE is the energy transferred in the time Δt through the crosssectional area A . From Ch. 6, the quantity ΔE Δt is the average power, Pav , transferred by the wave, so: P (14) I = av A 2 The SI unit for the intensity is W m . Also from Ch. 6, we found that the instantaneous power could be written: P = F ⋅v In our case, the force that one “slice” of fluid exerts on the next slice and the velocity of the particles are in the x direction, so we can write this as: P = Fx vx 22

Ch. 16: Sound and Hearing

P = ⎡⎣ p ( x, t ) A⎤⎦

∂y ( x, t )

∂t For sinusoidal waves, y ( x, t ) and p ( x, t ) are given by (1) and (4), respectively. Plugging these in, I get: P = Bω kA3 sin 2 ( kx − ωt ) The amplitude Bω kA2 is a constant. Over one cycle of the wave, the average power transferred is therefore: Pav = Bω kA3 ⎡⎣sin 2 ( kx − ωt ) ⎤⎦ av

The average over one cycle of sin θ (where θ is any angle in radians) is found by doing the integral: 2π 1 2 2 ⎡⎣sin θ ⎤⎦ = sin θ dθ , ∫ av 2π 0 which works out to be 1/2. (I will let you do the integral. The trig identity sin 2 θ = (1 − cos 2θ ) 2 is useful.) Therefore: 1 Pav = Bω kA3 2 Plugging this into (14) gives: 2

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Ch. 16: Sound and Hearing

1 Bω kA2 (15) 2 Using v = ω k and v = B ρ , I can write (15) as: 1 ω2 ⎛ω ⎞ 2 1 I = Bω ⎜ ⎟ A = B A2 2 2 B ρ ⎝v⎠ 1 (16) I= ρ B ω 2 A2 2 Often it is more useful to express this in terms of the pressure amplitude pmax = BkA (Eq. 5): 2 1 ⎛ k 2v2 ⎞ 2 1 ⎡ k ( B ρ ) ⎤ 2 I = B⎜ ⎥A ⎟A = B⎢ 2 ⎝ v ⎠ 2 ⎣ v ⎦ 2 2 pmax pmax = (17) I= 2ρ v 2 ρ B I=

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Ch. 16: Sound and Hearing

The Decibel Scale Because the ear is sensitive to sound over a broad range of intensities, a logarithmic intensity scale is usually used. We define the sound intensity level, β , to be: ⎛ I ⎞ β ≡ (10 dB ) log ⎜ ⎟ , (18) ⎝ I0 ⎠ in which I 0 is the reference intensity, chosen to be 10−12 W/m 2 , about the threshold of human hearing. The unit for β is the decibel, dB, which is one-tenth of a bel, named for Alexander Graham Bell.

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