Q17.3 If an object is

17

Sound Waves

CHAPTER OUTLINE

17.1 Speed of Sound Waves 17.2 Periodic Sound Waves

17.3 Intensity of Periodic Sound Waves 17.4 The Doppler Effect

17.5 Digital Sound Recording 17.6 Motion Picture Sound

ANSWERS TO QUESTIONS

*Q17.1 Answer (b). The typically higher density would by itself make the speed of sound lower in a solid compared to a gas.

Q17.2 We assume that a perfect vacuum surrounds the clock. The sound waves require a medium for them to travel to your ear. The hammer on the alarm will strike the bell, and the vibration will spread as sound waves through the body of the clock. If a bone of your skull were in contact with the clock, you would hear the bell. However, in the absence of a surrounding medium like air or water, no sound can be radiated away. A larger-scale example of the same effect: Colossal storms raging on the Sun are deathly still for us.

What happens to the sound energy within the clock? Here is the answer: As the sound wave travels through the steel and plastic, traversing joints and going around corners, its energy is con-verted into additional internal energy, raising the temperature of the materials. After the sound has died away, the clock will glow very slightly brighter in the infrared portion of the electromagnetic spectrum.

Q17.3 If an object is 1

2 meter from the sonic ranger, then the sensor would have to measure how long it would take for a sound pulse to travel one meter. Since sound of any frequency moves at about 343 m s, then the sonic ranger would have to be able to measure a time difference of under 0.003 seconds. This small time measurement is possible with modern electronics. But it would be more expensive to outfi t sonic rangers with the more sensitive equipment than it is to print “do not

use to measure distances less than 1

2 meter” in the users’ manual.

Q17.4 The speed of sound to two signifi cant fi gures is 340 m s. Let’s assume that you can measure time

to 1

10 second by using a stopwatch. To get a speed to two signifi cant fi gures, you need to measure a time of at least 1.0 seconds. Sinced=vt, the minimum distance is 340 meters.

*Q17.5 (i) Answer (b). The frequency increases by a factor of 2 because the wave speed, which is depen-dent only on the medium through which the wave travels, remains constant.

(ii) Answer (c).

*Q17.6 (i) Answer (c). Every crest in air produces one crest in water immediately as it reaches the interface, so there must be 500 in every second.

(ii) Answer (a). The speed increases greatly so the wavelength must increase.

449

13794_17_ch17_p449-472.indd 449

Q17.7 When listening, you are approximately the same distance from all of the members of the group. If different frequencies traveled at different speeds, then you might hear the higher pitched frequen-cies before you heard the lower ones produced at the same time. Although it might be interesting to think that each listener heard his or her own personal performance depending on where they were seated, a time lag like this could make a Beethoven sonata sound as if it were written by Charles Ives.

*Q17.8 Answer (a). We suppose that a point source has no structure, and radiates sound equally in all directions (isotropically). The sound wavefronts are expanding spheres, so the area over which the sound energy spreads increases according to A=4πr2. Thus, if the distance is tripled, the area increases by a factor of nine, and the new intensity will be one-ninth of the old intensity. This answer according to the inverse-square law applies if the medium is uniform and unbounded.

For contrast, suppose that the sound is confi ned to move in a horizontal layer. (Thermal strati-fi cation in an ocean can have this effect on sonar “pings.”) Then the area over which the sound energy is dispersed will only increase according to the circumference of an expanding circle:

A=2πrh, and so three times the distance will result in one third the intensity.

In the case of an entirely enclosed speaking tube (such as a ship’s telephone), the area perpen-dicular to the energy fl ow stays the same, and increasing the distance will not change the intensity appreciably.

*Q17.9 Answer (d). The drop in intensity is what we should expect according to the inverse-square law: 4πr₁2

P

1and 4πr22

P

2 should agree. (300 m)

2_{(2 μWⲐm}2_{) and (950 m)}2_{(0.2 μWⲐm}2_{) are 0.18 W and}

0.18 W, agreeing with each other.

*Q17.10 Answer (c). Normal conversation has an intensity level of about 60 dB.

*Q17.11 Answer (c). The intensity is about 10−13 WⲐm2.

Q17.12 Our brave Siberian saw the fi rst wave he encountered, light traveling at 3 00 108

. × m s. At the same moment, infrared as well as visible light began warming his skin, but some time was required to raise the temperature of the outer skin layers before he noticed it. The meteor produced compres-sional waves in the air and in the ground. The wave in the ground, which can be called either sound or a seismic wave, traveled much faster than the wave in air, since the ground is much stiffer against compression. Our witness received it next and noticed it as a little earthquake. He was no doubt unable to distinguish the P and S waves from each other. The fi rst air-compression wave he received was a shock wave with an amplitude on the order of meters. It transported him off his doorstep. Then he could hear some additional direct sound, refl ected sound, and perhaps the sound of the falling trees.

Q17.13 As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency; as you move away, the echo would be shifted down in frequency.

*Q17.14 In f ′=(v+v_o)fⲐ(v−v_s) we can consider the size of the fraction (v+v_o)Ⲑ(v−v_s) in each case. The positive direction is defi ned to run from the observer toward the source.

In (a), 340Ⲑ340 = 1 In (b), 340Ⲑ(340 − 25) = 1.08 In (c), 340Ⲑ(340 + 25) = 0.932 In (d),

(340 + 25)Ⲑ340 = 1.07 In (e), (340 − 25)Ⲑ340 = 0.926 In (f ), (340 + 25)Ⲑ(340 + 25) = 1 In (g), (340 − 25)Ⲑ(340 − 25) = 1. In order of decreasing size we have b > d > a = f = g > c > e.

13794_17_ch17_p449-472.indd 450

*Q17.15 (i) Answer (c). Both observer and source have equal speeds in opposite directions relative to the medium, so in f ′=(v+v_o)f(v−v_s) we would have something like (340 − 25)f(340 − 25) =f. (ii) Answer (a). The speed of the medium adds to the speed of sound as far as the observer is

concerned, to cause an increase in λ=vf.

(iii) Answer (a).

Q17.16 For the sound from a source not to shift in frequency, the radial velocity of the source relative to the observer must be zero; that is, the source must not be moving toward or away from the observer. The source can be moving in a plane perpendicular to the line between it and the observer. Other possibilities: The source and observer might both have zero velocity. They might have equal velocities relative to the medium. The source might be moving around the observer on a sphere of constant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequency heard will be equal to the frequency emitted by the source.

SOLUTIONS TO PROBLEMS

Section 17.1 Speed of Sound Waves

*P17.1 Since vlight>>vsound we have d≈

(

343 m s

)

(

16 2. s

)

= 5 56. km

We do not need to know the value of the speed of light. As long as it is very large, the travel time for the light is negligible compared to that for the sound.

P17.2 v= = ×

× =

B

ρ

2 80 10

13 6 10 1 43

10 3

.

. . km s

*P17.3 The sound pulse must travel 150 m before refl ection and 150 m after refl ection. We have d=vt

t= =d =

v

300

0 196 m

1 533 m s . s

P17.4 (a) At 9 000 m, ΔT = ⎛⎝ ⎞

⎠ −

(

)

= − 9 000

150 1 00. °C 60 0. °C so T = −30 0. °C Using the chain rule:

d dt

d dT

dT dx

dx dt

d dT

dT dx

v v

v v v

= = =

(

0 607

)

⎛_⎝ 1 ⎞ 150 .

⎠⎠ =247v , so dt

d

=

(

247 s

)

v

v

dt d

t

t

f

i

i f

0

247

247

∫

=

(

)

∫

=

(

)

⎛

⎝⎜ ⎞ ⎠⎟= s

s

v v

v v v v

ln 2247 331 5 0 607 30 0 331 5 0 607 30 0 s

(

)

+

(

)

+ −

ln . . .

. .

((

.

)

⎡

⎣

⎢ ⎤_⎦⎥

t= 27 2. s for sound to reach ground.

(b) t= =h

+

(

)

[

]

=

v

9 000

331 5. 0 607 30 0. . 25 7. s

It takes longer when the air cools off than if it were at a uniform temperature.

13794_17_ch17_p449-472.indd 451

P17.5 Sound takes this time to reach the man: 20 0 1 75

343 5 32 10

2

. .

.

m m

m s s

−

(

)

_{= ×} −

so the warning should be shouted no later than 0 300. s+5 32 10. × −2 s=0 353. s before the pot strikes.

Since the whole time of fall is given by y= 1gt

2

2

: 18 25 1 2 9 80

2

. m=

(

. m s2

)

t t=1 93. s

the warning needs to come 1 93. s−0 353. s=1 58. s

into the fall, when the pot has fallen 1

2 9 80 1 58 12 2

2

. m s2 . s . m

(

)

(

)

=

to be above the ground by 20 0. m−12 2. m= 7 82. m

P17.6 It is easiest to solve part (b) fi rst:

(b) The distance the sound travels to the plane is ds = h + ⎛⎝h⎞⎠ =h

2 2

2

5 2 The sound travels this distance in 2.00 s, so

ds = h =

(

)

(

)

=

5

2 343 m s 2 00. s 686m

giving the altitude of the plane as h=2 686

(

)

=

5 614

m

m

(a) The distance the plane has traveled in 2.00 s is v 2 00

2 307

. s m

(

)

= =h

Thus, the speed of the plane is: v=307m= 153 2.00 s m s

P17.7 Let x1 represent the cowboy’s distance from the nearer canyon wall and x2 his distance from the

farther cliff. The sound for the fi rst echo travels distance 2x1. For the second, 2x2. For the third,

2x1+2x2. For the fourth echo, 2x1+2x2+2x1.

Then

2 2

340 1 92

2 1

x − x

=

m s . s and

2 2 2

340 1 47

1 2 2

x + x − x

= m s . s

Thus

x1

1

2340 250

= m s 1.47 s= m and 2

340 1 92 1 47

2

x

m s= . s+ . s; x2=576m (a) So x1+x2= 826m

(b) 2 2 2 2 2

340 1 47

1 2 1 1 2

x + x + x −

(

x + x

)

=

m s . s

13794_17_ch17_p449-472.indd 452

Section 17.2 Periodic Sound Waves

*P17.8 (a) The speed gradually changes from v= (331 ms)( 1 + 27°C273°C)12_{= 347 ms to}

(331 ms) (1 + 0273°C)12_{= 331 ms, a 4.6% decrease. The cooler air at the same}

pressure is more dense.

(b) The frequency is unchanged, because every wave crest in the hot air becomes one crest without delay in the cold air.

(c) The wavelength decreases by 4.6%, from vf= (347 ms)(4 000s) = 86.7 mm to (331 ms)(4 000s) = 82.8 mm. The crests are more crowded together when they move slower.

*P17.9 (a) If f =2 4. MHz, λ = =

× =

v f

1 500

2 4 106 0 625

m s

s mm

. .

(b) If f =1 MHz, λ = =v =

f

1 500

106 1 50

m s

s . mm

If f =20 MHz, λ= μ

× =

1 500

2 107 75 0 m s

s . m

P17.10 ΔPmax=ρ ωv smax

smax P

max .

.

= =

₍

(

₎

×

)

− Δ

ρ ωv

4 00 10

1 20 343

3

N m

kg m

2

3

m s s m

(

)

( )

(

× −

)

= ×

− 2 10 0 103 1 1 55 10

10

π . .

P17.11 (a) A= 2 00. μm

λ= 2π = =

15 7. 0 400. m 40 0. cm

v=ω = =

k

858

15 7. 54 6. m s

(b) s=2 00 ⎡⎣

(

15 7 0 050 0

)

(

)

−

(

858 3 00 10

)

(

× −3

)

⎤

. cos . . . _{⎦⎦ = −}0 433. μm

(c) vmax = =

(

.

)

(

)

= .

−

Aω 2 00μ 858 1 1 72

m s mm s

P17.12 (a) ΔP=

(

)

⎛ x− t

⎝ ⎞⎠

1 27. Pa sin 340

m s

π π

(SI units)

The pressure amplitude is: ΔPmax= 1 27 Pa.

(b) ω=2πf =340π s, so f = 170 Hz

(c) k= 2π =

λ π m, giving λ = 2 00. m

(d) v=λf =

(

2 00. m

)(

170Hz

)

= 340 m s

13794_17_ch17_p449-472.indd 453

P17.13 k= =

(

)

= −

2 2

0 100 62 8

1

π λ

π

. m . m

ω π

λ π

= =

(

)

(

)

= × −

2 2 343

0 100 2 16 10

4 1

v m s

m s

. .

Therefore,

ΔP=

(

0 200. Pa

)

sin⎡⎣62 8. x m−2 16 10. × 4t s⎤⎦

P17.14 (a) The sound “pressure” is extra tensile stress for one-half of each cycle. When it becomes

0 500 13 0 1010 6 50 108

. % . .

(

)

(

× Pa

)

= × Pa, the rod will break. Then, ΔPmax=ρ ωv smax

smax P

max .

.

= = ×

×

(

)

Δ ρ ωv

6 50 10 8 92 10 5 0

8 3

N m kg m

2 3

1

10 m s 2 500s 4 63mm

(

)(

π

)

= .

(b) From s=smaxcos

(

kx−ωt

)

v

v

= ∂

∂ = −

(

−

)

= =

(

)

s

t s kx t

s

ω ω

ω π

max

max max

sin

2 500s

(

4..63mm

)

= 14 5. m s

(c) I= 1

(

s

)

= =

(

×

)

2

1 2

1

2 8 92 10 5

2 2 3

ρ ωv max ρvvmax . kg m

3

0

010 m s 14 5 m s 2

(

)(

.

)

= 4 73 10× 9

. W m2

P17.15 ΔPmax=ρ ωsmax=ρ ⎛⎝ ⎞⎠smax

π λ

v v 2 v

λ= πρ = π

(

)(

)

(

×

)

− 2 2 1 20 343 5 50 10

0

2 2 6

v s P

max max

. .

Δ ..840 = 5 81. m

Section 17.3 Intensity of Periodic Sound Waves

P17.16 The sound power incident on the eardrum is

P

= IA where I is the intensity of the sound and

A=5 0 10. × −5 m2 is the area of the eardrum.

(a) At the threshold of hearing, I=1 0 10. × −12 W m2, and

P

=

(

× −

)

(

× −

)

= × −

1 0 10. 12 W m2 5 0 10. 5 m2 5 00 10. 17 W

(b) At the threshold of pain, I=1 0. W m2, and

P

=_{( .1 0 /} 2_{)( .5 0 10}× −5 2₎= _{5 00 10.} × −5

W m m W

P17.17 β = ⎛

⎝⎜ ⎞ ⎠⎟=

× × ⎛ ⎝

−

−

10 10 4 00 10

1 00 10

0

6 12

log log .

.

I

I ⎜⎜

⎞

⎠⎟= 66 0. dB

13794_17_ch17_p449-472.indd 454

P17.18 The power necessarily supplied to the speaker is the power carried away by the sound wave:

P

=

(

)

=

=

(

1

2 2

2 1 20

2 2 2 2

2

ρ ω π ρ

π

Av smax A f sv max

. kg m3

))

⎛

⎝ ⎞⎠

(

)(

)

× −

π 0 08 343 600 0 12 10

2

2 2

.

. m

2 m s 1 s

((

m

)

=

2

21 2. W

P17.19 I =1 s

2

2 2

ρω maxv

(a) At f =2 500 Hz, the frequency is increased by a factor of 2.50, so the intensity (at constant

smax) increases by 2 50 6 25 2

. .

(

)

= .

Therefore, 6 25 0 600.

(

.

)

= 3 75. W m2

(b) 0 600. W m2

P17.20 The original intensity is I1 s f s

2 2 2 2 2

1

2 2

= ρω maxv= π ρv max

(a) If the frequency is increased to f′ while a constant displacement amplitude is maintained, the new intensity is

I2 f s

2 2 2

2

= π ρv

(

′

)

max so

I I

f s f s

f f

2 1

2 2

2 2 2

2

2 2

=

(

′

)

=⎛ ′

⎝⎜ ⎞ ⎠⎟ π ρ

π ρ

v v

max max

or I f f I

2

2

1

=⎛ ′ ⎝⎜

⎞ ⎠⎟

(b) If the frequency is reduced to f′ = f

2 while the displacement amplitude is doubled, the new intensity is

I2 f s f s I

2

2

2 2 2 2

1

2

2 2 2

= π ρv⎛_⎝ ⎞_⎠

(

max

)

= π ρv max =

or the intensity is unchanged .

P17.21 (a) For the low note the wavelength is λ = =v =

f

343

146 8 2 34 m s

s m

. .

For the high note λ =343 = 880 0 390

m s

s . m

We observe that the ratio of the frequencies of these two notes is 880Hz 5 99 146.8 Hz= . nearly equal to a small integer. This fact is associated with the consonance of the notes D and A.

(b) β = ⎛

⎝⎜ − ⎞⎠⎟ = 10

1012 75

dB

W m2 dB

log I gives I=3 16 10× −5

. W m2

I P P

=

= × −

(

)

Δ

Δ

max

max . .

2

5

2

3 16 10 2 1 20 3

ρv

W m2 kg m3

4

43 m s 0 161Pa

(

)

= .

for both low and high notes.

continued on next page

13794_17_ch17_p449-472.indd 455

(c) I =1

(

s

)

= f s

2

1 2 4

2 2 2 2

ρ ωv max ρ πv max

s I

f

max=

2_{π ρ}2 2

v

for the low note, smax

.

. .

= 3 16 10× − 2 1 20 343

1 146 8

5 2

W m

kg m m s s

2 3

π

==6 24 10× − = × − 146 8 4 25 10

5

7

.

. m . m

for the high note, smax

.

. = 6 24 10× − = × −

7 09 10

5

8

880 m m

(d) With both frequencies lower (numerically smaller) by the factor 146 8 134 3

880

804 9 1 093 .

. = . = . , the wavelengths and displacement amplitudes are made 1.093 times larger, and the pressure amplitudes are unchanged.

P17.22 We begin with β₂ 2 0

10

= ⎛

⎝⎜ ⎞ ⎠⎟ log I

I and β1

1 0

10

= ⎛

⎝⎜ ⎞ ⎠⎟ log I

I so β2 β1

2 1

10

− = ⎛

⎝⎜ ⎞ ⎠⎟ log I

I

Also, I

r

2 2

2

4 =

P

π and I1 r1 2

4 =

P

π giving

I I

r r

2 1

1 2

2

=⎛ ⎝⎜

⎞ ⎠⎟

Then, β2 β1

1 2

2

1 2

10 20

− = ⎛

⎝⎜ ⎞ ⎠⎟ =

⎛ ⎝⎜

⎞ ⎠⎟ log r log

r

r r

P17.23 (a) I1

12 10 12

1 00 10 10 1 1 00 10

=

(

_. × − _{W m}2

)

(β )=

(

_. × − _{W m}2

))

1080 0 10.

or I1

4

1 00 10

= _. × − _{W m}2

I2

12 10 12

1 00 10 10 2 1 00 10

=

(

× −

)

( )=

(

× −

. W m2 β . W m2

))

1075 0 10.

or I2

4 5 5

1 00 10 3 16 10

= _. × −. = _. × −

W m2 W m2

When both sounds are present, the total intensity is

I= + =I1 I2 × − + × − = ×

4 5

1 00 10. W m2 3 16 10. W m2 1 32 10. −−4

W m2

(b) The decibel level for the combined sounds is

β = ×

× ⎛ ⎝⎜

⎞ ⎠⎟= −

− 10 1 32 10

1 00 10 10

4 12

log . .

W m W m

2

2 llog .1 32 10 81 2. 8

×

(

)

= dB

13794_17_ch17_p449-472.indd 456

P17.24 In I r

=

P

4_π 2, intensity I is proportional to

1

2

r , so between locations 1 and 2: I I r r 2 1 1 2 2 2 =

In I=1

(

s

)

2

2

ρ ωv max , intensity is proportional to smax 2

, so I

I s s 2 1 2 2 1 2 =

Then, s

s r r 2 1 2 1 2 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞

⎠⎟ or 1 2 2 1 2 2 ⎛

⎝ ⎞⎠ =⎛_⎝⎜rr⎞_⎠⎟ giving r2 =2r1=2 50 0

(

. m

)

=100m

But, r2 d

2 2

50 0

=

(

. m

)

+ yields d= 86 6. m

P17.25 (a) 120 10

10 12 2

dB dB W m = ⎡ ⎣ ⎢ − ⎤_⎦⎥ log I I r r I = = = =

₍

)

= 1 00 4 4 6 00 0 2 . . W m W 4 1.00 W m

2

2

P

P

π

π π ..691 m

We have assumed the speaker is an isotropic point source.

(b) 0 10

1012

dB dB

W m2

= ⎛ ⎝⎜ ⎞ ⎠⎟ − log I I r I = × = = × −

1 00 10

4 6 00 12 . . W m W 4 1.00 10 W

2

12

P

π π −

m

m2 km

(

)

= 691

We have assumed a uniform medium that absorbs no energy.

P17.26 We presume the speakers broadcast equally in all directions.

(a) rAC = 3 00 +4 00 =5 00

2 2

. . m . m

I r = = ×

(

)

= × − −

P

4

1 00 10

4 5 00 3 18 10

2 3 2 6 π π . . . W

m W m

2 2

2 2

dB W m

W m β β = ⎛ × ⎝⎜ ⎞ ⎠⎟ = − − 10 3 18 10

10 1 6 12 log . 0

0dB6 50. = 65 0. dB

(b) rBC =4 47. m

I= ×

(

)

= ×

=

−

− 1 50 10

4 4 47 5 97 10

10 3 2 6 . . . W

m W m

2

π

β ddB

dB log .

.

5 97 10 10 67 8 6 12 × ⎛ ⎝⎜ ⎞ ⎠⎟ = − − β

(c) I=3 18. _μW m2+5 97. _μW m2

β = ⎛ × ⎝⎜ ⎞ ⎠⎟= − − 10 9 15 10

10 69 6

6 12

dBlog . . dB

13794_17_ch17_p449-472.indd 457

P17.27 Since intensity is inversely proportional to the square of the distance,

I4 I0 4

1 100

= . and I

P

0 4

2 2

2

10 0

2 1 20 343 0 121

.

max .

. .

= =

(

)

(

)(

)

=

Δ

ρv W m

2 2

The difference in sound intensity level is

Δβ = ⎛

⎝⎜ ⎞

⎠⎟=

(

−

)

= − 10 4 10 2 00 20 0

log I . .

I

km 0.4 km

d dB

At 0.400 km,

β0 4 10 12

0 121

10 110 8

. log

.

.

= ⎛

⎝⎜

⎞ ⎠⎟= −

W m

W m dB

2 2

At 4.00 km,

β4 =β0 4. +Δβ=

(

110 8. −20 0.

)

dB=90 8. dB

Allowing for absorption of the wave over the distance traveled,

′ = −

(

)

(

)

=

β4 β4 7 00. dB km 3 60. km 65 6. dB

This is equivalent to the sound intensity level of heavy traffi c.

P17.28 (a) E=

P

t=4_πr It2 =4_π

(

100

)

2

(

7 00 10× −2

)

0 200

m W m2 s

.

((

.

)

= 1 76. kJ

(b) β = ×

× ⎛ ⎝⎜

⎞ ⎠⎟= −

− 10 7 00 10

1 00 10 108

2 12

log .

. dB

P17.29 _{β =} ⎛

⎝ − ⎞⎠ 10

1012

log I I= ⎡⎣10(β10)⎤⎦

(

10−12

)

W m2

I120dB 1 00 2

W m

( )= . ; I100

2

1 00 10

dB

2

W m

( )= . × − ; I10

11

1 00 10

dB

2

W m

( )= . × −

(a)

P

=4πr I2 so that r I1 r I

2

1 2

2 2

=

r r I I

2 1

1 2

1 2

2

3 00 1 00

1 00 10 30 = ⎛

⎝⎜ ⎞

⎠⎟ =

(

.

)

× − =

.

. .

m 00 m

(b) r r I

I

2 1

1 2

1 2

11

3 00 1 00

1 00 10 9 = ⎛

⎝⎜ ⎞

⎠⎟ =

(

.

)

× − =

.

. .

m 449 10× 5 m

13794_17_ch17_p449-472.indd 458

P17.30 Assume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensity

of this sound is given by 100 10

1012

dB dB

W m2

= log −

I

; I=10−2

W m2

. If the lawnmower

radiates as a point source, its sound power is given by I r

=

P

4_π 2

P

=

(

)

− =

4π 1m210 2 W m2 0 126. W

Now let your neighbor have an identical lawnmower 20 m away. You receive from it sound with

intensity I=

(

)

= × −

0 126

2 5 10

2

5

.

. W

4 20 m W m

2

π . The total sound intensity impinging on you is

10−2 _{W m}2₊2 5 10_× −5 _{W m}2 ₌1 002 5 10_× −2 _{W m}2

. . . So its level is

10 1 002 5 10

10 100 01

2 12

dBlog . × = . dB −

−

If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as a change from 100 dB.

P17.31 (a) The sound intensity inside the church is given by

β = ⎛

⎝⎜ ⎞ ⎠⎟

=

(

)

⎛

⎝ − 10

101 10

10

0

12

ln

ln

I I

I

dB dB

W m2

⎜⎜ ⎞_⎠⎟

=

(

−

)

= − =

I 1010 1. 1012 W m2 10 1 90. W m2 0 012 6. W m2

We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is

P

=IA=

(

0 012 6. W m2

)

(

22 0. m2

)

=0 277. W

Are you surprised by how small this is? The energy radiated in 20.0 minutes is

E= t=

(

)

(

)

⎛

⎝ ⎞⎠

P

0 277. J s 20 0. min 60 0. s

1.00 min == 332 J

(b) If the ground refl ects all sound energy headed downward, the sound power,

P

=0 277. W, covers the area of a hemisphere. One kilometer away, this area is

A=2_πr2=2_π

(

1 000

)

2 =2_π×106

m m2

The intensity at this distance is

I A

= =

× = ×

−

P

0 277

4 41 108

.

. W

2 10 m6 2 W m

2

π and the sound intensity level is

β =

(

)

×

× ⎛ ⎝⎜

⎞ ⎠ −

− 10 4 41 10

1 00 10

8 12

dB W m

W m

2 2

ln .

. ⎟⎟ = 46 4. dB

13794_17_ch17_p449-472.indd 459

Section 17.4 The Doppler Effect

P17.32 (a) ω= π = π⎛ ⎝⎜

⎞ ⎠⎟=

2 2 115

60 0 12 0

f min

s min rad s

. .

vmax = =

(

.

)

(

. ×

)

= .

−

ωA 12 0 1 80 10 3 0 021 7

rad s m m s

(b) The heart wall is a moving observer.

′ = ⎛_⎝ + ⎞_{⎠ =}

(

)

⎛ +

⎝

f f v vO

v 2 000 000

1 500 0 021 7 1 500

Hz _⎜⎜ . ⎞

⎠⎟= 2 000 028 9. Hz (c) Now the heart wall is a moving source.

′′ = ′ − ⎛ ⎝⎜

⎞

⎠⎟=

(

)

−

f f

s v

v v 2 000 029

1 500 1 500 0 02 Hz

. 11 7 2 000 057 8 ⎛

⎝⎜

⎞

⎠⎟= . Hz

*P17.33 (a) ′ =

(

+

)

−

(

)

f f o

s v v v v

′ =

(

+

)

−

(

)

=

f 2 500 343 25 0

343 40 0 3 04 .

. . kHz

(b) ′ = + −

(

)

− − ⎛

⎝⎜ ⎞⎠⎟=

f 2 500 343 25 0 343 40 0 2

.

( . ) .08 kHzz

(c) ′ = + −

(

)

− ⎛

⎝⎜ ⎞⎠⎟=

f 2 500 343 25 0 343 40 0 2

.

. .62 kHz while police car overtakes

′ = +

− −

(

)

⎛

⎝⎜

⎞ ⎠⎟=

f 2 500 343 25 0 343 40 0 2

.

. .40 kHz after police car passes

P17.34 (a) The maximum speed of the speaker is described by

1 2

1 2

20 0

5 00 0 500

2 2

m kA

k mA v

v

max

max

.

. .

=

= = N m

kg

(

mm

)

=1 00. m s

The frequencies heard by the stationary observer range from

′ = + ⎛ ⎝⎜

⎞ ⎠⎟

fmin f

max

v

v v to ′ = −

⎛ ⎝⎜

⎞ ⎠⎟

fmax f

max

v v v

where v is the speed of sound.

′ =

+ ⎛

⎝⎜

⎞ ⎠⎟=

fmin

. 440

343 1 00 439

Hz 343 m s

m s m s HHz

Hz 343 m s m s m s ′ =

− ⎛

⎝⎜

⎞ ⎠⎟=

fmax

. 440

343 1 00 4441 Hz

continued on next page

13794_17_ch17_p449-472.indd 460

(b) β= ⎛ π ⎝⎜ ⎞ ⎠⎟= ⎛ ⎝⎜ ⎞ ⎠⎟

10 10 4

0

2

0

dBlog I dBlog

I

r I

P

The maximum intensity level (of 60.0 dB) occurs at r=rmin =1 00 m. The minimum .

intensity level occurs when the speaker is farthest from the listener (i.e., when

r=rmax=rmin+2A=2 00 m)..

Thus, β β π max min min log log − = ⎛ ⎝⎜ ⎞ ⎠⎟− 10

4 ₀ 2 10

dB

P

dB

I r

P

P

4_{π 0} 2

I r_max

⎛ ⎝⎜ ⎞ ⎠⎟ or β β π π max min min max log − = ⎛ ⎝⎜ ⎞ ⎠ 10 4 4 0 2 0 2 dB

P

P

I r I r

⎟⎟ =10 ⎛_⎝⎜ ⎞_⎠⎟

2 2 dBlog max min r r

This gives:

60 0. dB−βmin =10dBlog

(

4 00.

)

=6 02. dB and βmin = 54 0 dB.

P17.35 Approaching ambulance: ′ = −

(

)

f f

S

1 v v

Departing ambulance: ′′ =

− −

(

)

(

)

f f

S

1 v v

Since f′ =560 Hz and f′′ =480 Hz 560 1⎛ − 480 1 ⎝ vv⎞⎠ = ⎛⎝ + ⎞⎠

v v

S S

1 040 80 0

80 0 343

1 040 26 4

v v v S S = =

(

)

= . . .

m s m s

P17.36 (a) v=

(

)

+

⋅

(

−

)

= 331 m s 0 6 m 10 325

s C C m s

.

° °

(b) Approaching the bell, the athlete hears a frequency of ′ = ⎛ +

⎝ ⎞⎠

f f v vO v

After passing the bell, she hears a lower frequency of ′′ = ⎛ + −

(

)

⎝⎜

⎞ ⎠⎟

f f v vO v

The ratio is ′′

′ = −+ = f f O O v v v v 5 6

which gives 6v−6v_o=5v+5v_o or vO= v = =

11 325

11 29 5 m s

m s .

P17.37 ′ =

− ⎛ ⎝⎜ ⎞ ⎠⎟ f f s v

v v 485 512

340 340 9 80 = − −

(

)

⎛ ⎝⎜ ⎞ ⎠⎟ . tfall

485 340 485 9 80 512 340

512 485

(

)

+

(

)

(

)

=

(

)(

)

= − . t t f f 4 485 340 9 80 1 93 ⎛

⎝ ⎞⎠ . = . s

d1 gtf

2

1

2 18 3

= = . m: treturn = = s

18 3

340 0 053 8 .

.

The fork continues to fall while the sound returns.

ttotal fall = +tf treturn=1 93. s+0 053 8. s=1 98. 55

1

2 19 3

s

m

total total fall 2

d = gt = .

13794_17_ch17_p449-472.indd 461

P17.38 (a) Sound moves upwind with speed 343 15

(

−

)

m s. Crests pass a stationary upwind point at frequency 900 Hz.

Then λ = =v =

f

328

900 0 364 m s

s . m

(b) By similar logic, λ = =v

(

+

)

=

f

343 15

900 0 398

m s

s . m

(c) The source is moving through the air at 15 ms toward the observer. The observer is station-ary relative to the air.

′ = +

− ⎛ ⎝⎜

⎞ ⎠⎟=

+ − ⎛

⎝ ⎞⎠ =

f f o

s v v

v v 900

343 0 343 15 94

Hz 11 Hz

(d) The source is moving through the air at 15 m s away from the downwind fi refi ghter. Her speed relative to the air is 30 m s toward the source.

′ = +

− ⎛ ⎝⎜

⎞ ⎠⎟=

+ − −

(

)

⎛

⎝⎜

f f o

s v v

v v 900

343 30 343 15

Hz ⎞⎞

⎠⎟=900 ⎛⎝ ⎞⎠ = 373

358 938

Hz Hz

P17.39 (b) sin

. θ = v =

vS

1

3 00; θ =19 5. °

tanθ =h

x; x h

= tanθ

x= 20 000 = × =

19 5 5 66 10 56 6

4

m

m km

tan . ° . .

(a) It takes the plane t x

S

= =

₍

×

₎

=

v

5 66 10

56 3

4

.

. m

3.00 335 m s s to travel this distance.

P17.40 θ =sin− =sin− =

. .

1 1 1

1 38 46 4

v vS

°

P17.41 The half angle of the shock wave cone is given by sinθ =v

v

light

S vS =v =

×

(

)

= ×

light m s

sin .

sin . .

θ

2 25 10

53 0 2 82 10

8

°

8 8

m s

t = 0

a.

h

Observer

b.

h

Observer hears the boom

x

FIG. P17.39(a)

13794_17_ch17_p449-472.indd 462

Section 17.5 Digital Sound Recording

Section 17.6 Motion Picture Sound

P17.42 For a 40-dB sound,

40 10

10

10

12

8

dB dB

W m

W m

2

2

= ⎡

⎣

⎢ ⎤_⎦⎥

= =

−

−

log I

I ΔPmmax

max .

2

8

2

2 2 1 20 343 10

ρ ρ

v

v

ΔP = I =

(

kg m2

)

(

m s

)

− W m2 =2 87 10. × −3 N m2

(a) code = × =

− 2 87 10

28 7 65 536 7

3

. .

N m N m

2 2

(b) For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fi delity.

(c) In a sound wave Δ P is negative half of the time but this coding scheme has no words available for negative pressure variations.

Additional Problems

*P17.43 The gliders stick together and move with fi nal speed given by momentum conservation for the two-glider system:

0.15 kg 2.3 ms + 0 = (0.15 + 0.2) kg v v= 0.986 ms

The missing mechanical energy is

(1/ 2)(0.15 kg)(2.3 m /s)2 _{– (1/ 2)(0.35 kg)(0.986 m /s)}2 _{= 0.397 J – 0.170 J = 0.227 J}

We imagine one-half of 227 mJ going into internal energy and half into sound radiated isotropically in 7 ms. Its intensity 0.8 m away is

I = EAt= 0.5(0.227 J)

[4π(0.8 m)2 0.007 s] = 2.01 Wm2 Its intensity level is β= 10 log(2.0110−12) = 123 dB

The sound of air track gliders latching together is many orders of magnitude less intense. The idea is unreasonable. Nearly all of the missing mechanical energy becomes internal energy in the latch.

13794_17_ch17_p449-472.indd 463

*P17.44

The wave moves outward equally in all directions. (We can tell it is outward because of the negative sign in 1.36 r – 2030 t.) Its amplitude is inversely proportional to its distance from the center. Its intensity is proportional to the square of the amplitude, so the intensity follows the inverse-square law, with no absorption of energy by the medium. Its speed is constant at

v=f λ=ω k= (2030s)(1.36m) = 1.49 kms. By comparison to the table in the chapter, it can be moving through water at 25°C, and we assume that it is. Its frequency is constant at (2030s)2π= 323 Hz. Its wavelength is constant at 2π k= 2π (1.36m) = 4.62 m. Its pressure amplitude is 25.0 Pa at radius 1 m. Its intensity at this distance is

I= ΔPmax = 2

2ρv

(

)

( )(

2

25

2 1000 1490 N/m

kg/m m/

2

3

ss) W/m

2

=209μ

so the power of the source and the net power of the wave at all distances is

P

=I 4π r 2=

×

(

_{2 09 10}−4

)

_{4π 1} =_{2 63} . W/m2 ( m)2 . mW

.

*P17.45 Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose that, at the ambient temperature, sound moves at 340 ms; and suppose that the horizon-tal width of each row of seats is 60 cm. Then there is a time delay of

0 6

0 002 .

. m

340 m s s

(

)

=

between your sound impulse reaching each riser and the next. Whatever its material, each will refl ect much of the sound that reaches it. The refl ected wave sounds very different from the sharp pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty crests, each separated from the next in time by

2 0 6

340 0 004 .

. m

m s s

(

)

(

)

=

This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once again after its refl ection. Thus, you hear a sound of defi nite pitch, with period about 0.004 s, frequency

1

0 003 5. s=300Hz ~a few hundred Hz

wavelength

λ = =

(

)

(

)

=

v f

340

300 1 2 10

0

m s

s . m~ m

and duration

20 0 004 10 1

. s ~ s

(

)

−

(b) Yes. With the steps narrower, the frequency can be close to 1000 Hz. If the person clap-ping his hands is at the base of the pyramid, the echo can drop somewhat in frequency and in loudness as sound returns, with the later cycles coming from the smaller and more distant upper risers. The sound could imitate some particular bird, and could in fact con-stitute a recording of the call.

13794_17_ch17_p449-472.indd 464

*P17.46 _{(a) The distance is larger by 24060 = 4 times. The intensity is 16 times smaller at the larger}

distance, because the sound power is spread over a 42 _{times larger area.}

(b) The amplitude is 4 times smaller at the larger distance, because intensity is proportional to the square of amplitude.

(c) The extra distance is (240 – 60)45 = 4 wavelengths. The phase is the same at both points, because they are separated by an integer number of wavelengths.

P17.47 Since cos2 sin2

1

θ+ θ= , sinθ= ± −1 cos2θ

(each sign applying half the time)

ΔP=ΔPmaxsin

(

kx−ωt

)

= ±ρ ωv smax 1−cos

(

kx−ωt

)

2

Therefore ΔP= ±ρ ωv smax−smaxcos

(

kx−ωt

)

= ±ρ ωv smax−s

2 2 2 2 2

P17.48 (a) λ = =v ₋ =

f

343

1 480 1 0 232 m s

s . m

(b) β = = ⎡

⎣

⎢ − ⎤_⎦⎥

81 0 10

1012 . dB dBlog

W m2

I

I=

(

10−12 W m2

)

108 10. =10−3 90. W m2 =1 26 10. × −4 W mm W m

2

2

=

= =

(

×

)

−

1 2

2 2 1 26 10

2 2

2

4

ρ ω

ρ ω

v

v

s

s I

max

max

.

1

1 20 343 4 2 1 480 1 2 8 41 10

. kg m3 m s s .

(

)

(

)

(

−

)

= ×

− π

8 8

m

(c) ′ =

′= − =

λ v

f

343

1 397 1 0 246 m s

s . m Δλ λ λ= ′ − = 13 8. mm

P17.49 The trucks form a train analogous to a wave train of crests with speed v=19 7. m s and unshifted

frequency f = 2 = −

3 00 0 667

1

. min . min

(a) The cyclist as observer measures a lower Doppler-shifted frequency:

′ = ⎛_⎝ + ⎞_{⎠ =}

(

−

)

+ −

(

)

f f v vo

v 0 667

19 7 4 47 19 7

1

. . .

. min ⎛⎛

⎝⎜ ⎞⎠⎟= 0 515. min

(b) ′′ = ⎛ +

⎝⎜ ⎞⎠⎟ =

(

−

)

+ −

(

)

f f v vo v

′ _{0 667} 1 19 7 1 56

. min . .

1

19 7. 0 614. ⎛

⎝⎜ ⎞⎠⎟= min

The cyclist’s speed has decreased very signifi cantly, but there is only a modest increase in the frequency of trucks passing him.

P17.50 (a) The speed of a compression wave in a bar is

v= Y = × = ×

ρ

20 0 10

7 860 5 04 10

10

3

.

. N m

kg m m s

2 3

(b) The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time

t= L =

× = ×

−

v

0 800

1 59 104 .

. m

5.04 10 m s3 s

continued on next page

13794_17_ch17_p449-472.indd 465

(c) As described by Newton’s fi rst law, the rearmost layer of steel has continued to move forward with its original speed vi for this time, compressing the bar by

ΔL= it=

(

)

(

×

)

= × =

− −

v 12 0 1 59 104 1 90 103 1

. m s . s . m ..90 mm

(d) The strain in the rod is: ΔL

L =

× − _{= ×}

− 1 90 10

2 38 10

3

3

.

. m 0.800 m

(e) The stress in the rod is: σ = ⎛⎝ ⎞

⎠ =

(

×

)

(

× −

)

=

Y L L

Δ _{20 0 10}10 _{2 38 10} 3 ₄₇₆

. N m2 . M

P Pa

Since σ >400 MPa, the rod will be permanently distorted.

(f ) We go through the same steps as in parts (a) through (e), but use algebraic expressions rather than numbers:

The speed of sound in the rod is v= Y

ρ

The back end of the rod continues to move forward at speed vi for a time of t

L L

Y

= =

v

ρ_, traveling distance ΔL=vit after the front end hits the wall.

The strain in the rod is: ΔL

L t

L Y

i i

=v =v ρ

The stress is then: σ= ⎛⎝ ⎞ ρ ρ

⎠ = =

Y L

L Y i Y i Y

Δ _{v v}

For this to be less than the yield stress, σy, it is necessary that

vi ρY <σy or vi y

Y

< σ ρ

With the given numbers, this speed is 10.1 ms. The fact that the length of the rod divides out means that the steel will start to bend right away at the front end of the rod. There it will yield enough so that eventually the remainder of the rod will experience only stress within the elastic range. You can see this effect when sledgehammer blows give a mushroom top to a rod used as a tent stake.

P17.51 (a) f′ = f

(

₋

)

v v vdiver

so 1− =

′

v v

diver f

f ⇒ = − ′

⎛ ⎝⎜ ⎞⎠⎟

vdiver v 1 f f

with v=343 m s, f =1 800 Hz and f′ =2 150 Hz

we fi nd vdiver= − m s

⎛

⎝⎜ ⎞⎠⎟= 343 1 1 800

2 150 55 8.

(b) If the waves are refl ected, and the skydiver is moving into them, we have

′′ = ′

(

+

)

⇒ ′′ = −

(

)

⎡ ⎣

⎢ ⎤

⎦

⎥ +

f f v v f f

v

v v v

v v

diver

diver

d diver

(

)

v

so ′′ =

(

+

)

−

(

)

=

f 1 800 343 55 8

343 55 8 2 500 .

. Hz

13794_17_ch17_p449-472.indd 466

P17.52 Let P(x) represent absolute pressure as a function of x. The net force to the right on the chunk of air is +P x A

( )

−P x

(

+Δx A

)

. Atmospheric pressure subtracts out, leaving

−

(

+

)

+

( )

⎡⎣ ΔP x Δx ΔP x ⎤⎦ = −∂ ∂

Δ _Δ

A P

x xA. The mass of the air is Δm=ρΔV=ρ ΔA x and its

acceleration is ∂ ∂

2 2

s

t . So Newton’s second law becomes

−∂

∂ =

∂ ∂ − ∂

∂ − ∂ ∂ ⎛

⎝⎜ ⎞⎠⎟= ∂∂

ΔP_{Δ Δ}

x xA A x s t x B

s x

s t

ρ

ρ

2 2 2

2 2

2 2

2 2

B s x

s t

ρ ∂

∂ =

∂ ∂

Into this wave equation as a trial solution we substitute the wave function

s x t

(

,

)

=smaxcos

(

kx−ωt

)

We fi nd

∂

∂ = −

(

−

)

∂

∂ = − −

s

x ks kx t

s

x k s kx t

max

max

sin

cos ω

ω

2 2

2

((

)

∂

∂ = +

(

−

)

∂ ∂ = −

s

t s kx t

s

t s kx

ω ω

ω

max

max

sin

cos

2 2

2

(

_−−ω

)

t B s

x s t

ρ∂∂ = ∂∂

2 2

2

2 becomes −

(

−

)

= −

(

−

)

B

k s kx t s kx t

ρ ω ω ω

2 2

maxcos maxcos

This is true provided B f

ρ π

λ π

4 4

2 2

2 2

=

The sound wave can propagate provided it has λ

ρ

2 2 2

f =v = B; that is, provided it propagates with speed v= B

ρ

P17.53 When observer is moving in front of and in the same direction as the source, ′ = − −

f f O

S v v v v where vO and vS are measured relative to the medium in which the sound is propagated. In this case the

ocean current is opposite the direction of travel of the ships and

vO =45 0. km h− −

(

10 0. km h

)

=55 0. km h=15 3. m s , annd

km h km h km h m

vS =64 0. − −

(

10 0.

)

=74 0. =20 55. ss

Therefore,

′ =

(

)

−

−

f 1 200 0 1 520 15 3 1 520 20 55

. .

.

Hz m s m s

m s mm s= 1 204 2. Hz

P17.54 Use the Doppler formula, and remember that the bat is a moving source. If the velocity of the insect is vx,

40 4 40 0 340 5 00 340 340 5 00 340

. . .

.

=

(

+

)

(

−

)

−

(

)

+

v v x x

((

)

Solving,

vx =3 31. m s

Therefore, the bat is gaining on its prey at 1.69 m s .

P(x)A P(x + Δx)A

FIG. P17.52

13794_17_ch17_p449-472.indd 467

P17.55 103 10

1012

dB dB

W m2

= ⎡

⎣

⎢ − ⎤_⎦⎥

log I

(a) I

r

= × = =

(

)

− 2 00 10

4 4 1 6

2

2 2

.

. W m

m

2

P

P

π π

P

= 0 642. W

(b) effi ciency =sound output power = total input power

0 642. WW

150 W = 0 004 28.

P17.56 (a)

(b) λ = =v ₋ =

f

343

1 000 1 0 343 m s

s . m

(c) ′ =

′= − ⎛

⎝ ⎞⎠ =

(

−

)

− =

λ v v v v

v f f

S 343 40 0

1 000 1 0

.

. m s

s 3303 m

(d) ′′ =

′′= + ⎛

⎝ ⎞⎠ =

(

+

)

− =

λ v v v v

v f f

S 343 40 0

1 000 1

. m s

s 00 383. m

(e) ′ = −

− ⎛ ⎝⎜

⎞

⎠⎟=

(

)

−

(

)

f f O

S v v

v v 1 000

343 30 0 34

Hz . m s

3

3−40 0 1 03

(

.

)

m s= . kHz

*P17.57 (a) The sound through the metal arrives first, bbecause it moves faster than sound in air.

(b) Each travel time is individually given by Δ t=L v. Then the delay between the pulses’ arrivals

is Δt=L⎛ − L

⎝⎜

⎞ ⎠⎟=

−

1 1

v v

v v

v v

air cu

cu air air cu

and the length of the bar is L= t

− =

(

)

(

×

)

v v

v v

air cu cu air

m s m s

Δ 331 3 56 10 3 5

3

. 6

60−331 365

(

)

m s Δt= ( m/s)Δt (c) L= (365 ms)(0.127 s) = 46.3 m

(d) The answer becomes L t

r

=

− Δ 1 331

1 m s v

where vr is the speed of sound in the rod. As vr

goes to infi nity, the travel time in the rod becomes negligible. The answer approaches (331 m s)Δ t, which is just the distance that the sound travels in air during the delay time.

P17.58

P

₂ 1

P

₁

20 0 =

. β β1 2

1 2

10 − = log

P

P

80 0 10 20 0 13 0

67 0

2

2

. log . .

.

− = = +

= β

β dB

FIG. P17.56(a)

13794_17_ch17_p449-472.indd 468

P17.59 (a) θ = ⎛ ⎝⎜

⎞

⎠⎟= ×

⎛

⎝ ⎞

− −

sin sin

.

1 1

3

331 20 0 10

v v

sound

obj ⎠⎠ =

0 948. °

(b) ′ =

× ⎛

⎝ ⎞⎠ =

− θ sin

. .

1

3

1 533

20 0 10 4 40°

P17.60 Let T represent the period of the source vibration, and E be the energy put into each wavefront. Then

P

_av=E

T . When the observer is at distance r in front of the source, he is receiving a spherical

wavefront of radius vt, where t is the time since this energy was radiated, given by vt−vSt=r.

Then,

t r

S

= −

v v

The area of the sphere is 4 2 4

2 2 2

π v πv

v v

t r

S

( )

=

−

(

)

. The energy per unit area over the spherical wavefront

is uniform with the value E

A T

r

S

=

P

av

(

v v−

)

v

2

2 2

4π . The observer receives parcels of energy with the

Doppler shifted frequency ′ = − ⎛ ⎝⎜

⎞

⎠⎟=

(

−

)

f f

T

S S

v v v

v

v v , so the observer receives a wave with

intensity

I E A f

T

r T

S

=⎛ ⎝⎜

⎞ ⎠⎟ ′ =

−

(

)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟

⎟ −

P

av v v

v

v v v

2

2 2

4π _SS

S

r

(

)

⎛ ⎝

⎜ ⎞

⎠

⎟ = ⎛ −

⎝⎜ ⎞ ⎠⎟

P

av 4π 2

v v v

P17.61 For the longitudinal wave vL

Y

=⎛ ⎝⎜

⎞ ⎠⎟ ρ

1 2

For the transverse wave vT

T

=⎛ ⎝⎜

⎞ ⎠⎟ μ

1 2

If we require v

v L T

=8 00. , we have T = μY

ρ

64 0. where μ =

m L and

ρ

π = mass =

volume

m r L2

This gives

T= r Y =

(

×

)

(

×

)

−

π 2 π 3 2 10

64 0

2 00 10 6 80 10 64

.

. m . N m2

..0 1 34 10.

4

= × N

13794_17_ch17_p449-472.indd 469

P17.62 (a) If the source and the observer are moving away from each other, we have: θS =θ0 =180°,

and since cos180°= −1, we get Equation (17.13) with negative values for both v_O and v_S.

(b) If v_O=0 m /s then ′ = −

f f

S S v v v cosθ

Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection,

cosθS =

4 5

so ′ =

−

(

)

(

)

f 343

343 0 800 25 0 500 m s

m s . . m s Hz

or f′ = 531 Hz

Note that as the train approaches, passes, and departs from the intersection, θ_S varies from 0° to 180° and the frequency heard by the observer varies between the limits

′ =

− = −

f f

S

max

cos .

v

v v 0

343

343 25 0 500 °

m s

m s m s Hzz Hz

m s m s

(

)

=

′ =

− =

539

180

343 343

f f

S

min

cos

v

v v ° ++25 0. m s

(

500Hz

)

=466Hz

P17.63 (a) The time required for a sound pulse to travel

distance L at speed v is given by t L L Y

= =

v ρ

Using this expression we fi nd

t L

Y

L

1 1 1 1

1 10

7 00 10 2 700

1 96

= =

×

(

)

(

)

=

ρ .

.

N m2 kg m3

(

××

)

= − = −

×

− 10

1 50 1 50

1 60

4 1

2

1

2 2

1

L

t L

Y

L

s

m m

. .

.

ρ 11010 11 3 103

N m2 kg m3

(

)

(

. ×

)

or t2 L

3 4

1

1 26 10 8 40 10 =

(

_. × − − _. × −

)

_s

t t

3

3

1 50

8 800

4 24 =

×

(

)

(

)

= ×

.

.

m

11.0 1010 N m3 kg m3

1 10−4

s

We require t1+ =t2 t3, or

1 96 104 1 1 26 10 8 40 10 4 24 10

3 4

1

4

. × − L + . × − − . × − L = . × − This gives L1=1 30. m and L2 =1 50 1 30. − . =0 201. m

The ratio of lengths is then L

L

1 2

6 45 = .

(b) The ratio of lengths L

L

1 2

is adjusted in part (a) so that t1+ =t2 t3. Sound travels the two paths

in equal time and the phase difference Δφ =0 .

FIG. P17.63

L1 L2

L3

13794_17_ch17_p449-472.indd 470

ANSWERS TO EVEN PROBLEMS

P17.2 1 43. km s

P17.4 (a) 27.2 s (b) longer than 25.7 s, because the air is cooler

P17.6 (a) 153 m s (b) 614 m

P14.8 (a) The speed decreases by 4.6%, from 347 ms to 331 ms. (b) The frequency is unchanged, because every wave crest in the hot air becomes one crest without delay in the cold air. (c) The wavelength decreases by 4.6%, from 86.7 mm to 82.8 mm. The crests are more crowded together when they move slower.

P17.10 1 55 10. × −10m

P17.12 (a) 1 27. Pa (b) 170 Hz (c) 2.00 m (d) 340 m s

P17.14 (a) 4.63 mm (b) 14 5. m s (c) 4 73 10. × 9 W m2

P17.16 (a) 5 00 10. × −17 W (b) 5 00 10. × −5 W

P17.18 21.2 W

P17.20 (a) I f f I

2

2

1

=⎛ ′

⎝⎜ ⎞⎠⎟ (b) I2 =I1

P17.22 see the solution

P17.24 86.6 m

P17.26 (a) 65.0 dB (b) 67.8 dB (c) 69.6 dB

P17.28 (a) 1 76. kJ (b) 108 dB

P17.30 no

P17.32 (a) 2 17. cm s (b) 2 000 028 9. Hz (c) 2 000 057 8. Hz

P17.34 (a) 441 Hz; 439 Hz (b) 54.0 dB

P17.36 (a) 325 m s (b) 29 5. m s

P17.38 (a) 0.364 m (b) 0.398 m (c) 941 Hz (d) 938 Hz

P17.40 46 4. °

P17.42 (a) 7 (b) For sounds of 40 dB or softer, too few digital words are available to represent the wave form with good fi delity. (c) In a sound wave ΔP is negative half of the time but this coding scheme has no words available for negative pressure variations.

P17.44 The wave moves outward equally in all directions. Its amplitude is inversely proportional to its distance from the center so that its intensity follows the inverse-square law, with no absorption of energy by the medium. Its speed is constant at 1.49 kms, so it can be moving through water at 25°C, and we assume that it is. Its frequency is constant at 323 Hz. Its wavelength is constant at 4.62 m. Its pressure amplitude is 25.0 Pa at radius 1 m. Its intensity at this distance is 209 μWm2_,

so the power of the source and the net power of the wave at all distances is 2.63 mW.

13794_17_ch17_p449-472.indd 471

P17.46 (a) The intensity is 16 times smaller at the larger distance, because the sound power is spread over a 42 _{times larger area. (b) The amplitude is 4 times smaller at the larger distance, because}

intensity is proportional to the square of amplitude. (c) The phase is the same at both points, because they are separated by an integer number of wavelengths.

P17.48 (a) 0 232. m (b) 84 1. nm (c) 13.8 mm

P17.50 (a) 5 04. km s (b) 159 sμ (c) 1.90 mm (d) 0.002 38 (e) 476 MPa (f ) see the solution

P17.52 see the solution

P17.54 The gap between bat and insect is closing at 1.69 m s.

P17.56 (a) see the solution (b) 0.343 m (c) 0.303 m (d) 0.383 m (e) 1 03. kHz

P17.58 67 0. dB

P17.60 see the solution

P17.62 (a) see the solution (b) 531 Hz

13794_17_ch17_p449-472.indd 472