Solucionario serway cap 17

17Sound Waves

CHAPTER OUTLINE

17.1 Speed of Sound Waves17.2 Periodic Sound Waves17.3 Intensity of Periodic Sound Waves17.4 The Doppler Effect17.5 Digital Sound Recording17.6 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 about343 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. Since d t= v , 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 44913794_17_ch17_p449-472.indd 449 1/3/07 8:08:12 PM1/3/07 8:08:12 PM

450 Chapter 17

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 r= 4 2π . 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 rh= 2π , 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:

4π r 1 2 P

1 and 4π r 2

2 P2 should agree. (300 m)2(2 μW�m2) and (950 m)2(0.2 μW�m2) 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 W�m2.

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 + vo)f�(v − v

s) we can consider the size of the fraction (v + v

o)�(v − v

s) in each case.

The positive direction is defi ned to run from the observer toward the source.

In (a), 340�340 = 1 In (b), 340�(340 − 25) = 1.08 In (c), 340�(340 + 25) = 0.932 In (d),(340 + 25)�340 = 1.07 In (e), (340 − 25)�340 = 0.926 In (f ), (340 + 25)�(340 + 25) = 1 In (g), (340 − 25)�(340 − 25) = 1. In order of decreasing size we have b > d > a = f = g > c > e.

450 Chapter 17


Sound Waves 451

*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 = v�f.

(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 v vlight sound>> we have d ≈ ( )( ) =343 16 2 5 56m s s 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

ρ2 80 10

13 6 101 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 t= v

td= = =v

3000 196

m

1 533 m ss.

P17.4 (a) At 9 000 m, ΔT = ⎛⎝

⎞⎠ −( ) = −9 000

1501 00 60 0. .° °C C

so T = −30 0. °C

Using the chain rule:

d

dt

d

dT

dT

dx

dx

dt

d

dT

dT

dx

v vv

vv= = = ( )⎛

⎝⎞0 607

1

150. ⎠⎠ = v

247, so dt

d= ( )247 sv

v

dt

d

t

t

f

i

i

f

0

247

247

∫ ∫= ( )

= ( ) ⎛⎝⎜

⎞⎠⎟

=

s

s

vv

v

v

v

v

ln 2247331 5 0 607 30 0

331 5 0 607 30 0s( ) + ( )

+ −ln

. . .

. . .(( )⎡⎣⎢

⎤⎦⎥

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

(b) th= =

+ ( )[ ] =v

9 000

331 5 0 607 30 025 7

. . .. s

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

Sound Waves 451


452 Chapter 17

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

3435 32 10 2. ..

m m

m ss

−( )= × −

so the warning should be shouted no later than 0 300 5 32 10 0 3532. . .s s s+ × =−

before the pot strikes.

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

22: 18 25

1

29 80 2. .m m s2= ( ) t

t = 1 93. s

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

into the fall, when the pot has fallen 1

29 80 1 58 12 22. . .m s s m2( )( ) =

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

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

(b) The distance the sound travels to the plane is d hh h

s = + ⎛⎝

⎞⎠ =2

2

2

5

2 The sound travels this distance in 2.00 s, so

dh

s = = ( )( ) =5

2343 2 00 686m s s m.

giving the altitude of the plane as h = ( )=2 686

5614

mm

(a) The distance the plane has traveled in 2.00 s is v 2 002

307. s m( ) = =h

Thus, the speed of the plane is: v = =307153

m

2.00 sm 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 2 1x . For the second, 2 2x . For the third,

2 21 2x x+ . For the fourth echo, 2 2 21 2 1x x x+ + .

Then

2 2

3401 922 1x x− =

m ss. and 2 2 2

3401 471 2 2x x x+ − =

m ss.

Thus

x1

1

2340 250= =m s 1.47 s m and

2

3401 92 1 472x

m ss s= +. . ; x2 576= m

(a) So x x1 2 826+ = m

(b) 2 2 2 2 2

3401 471 2 1 1 2x x x x x+ + − +( )

=m s

s.

452 Chapter 17


Sound Waves 453

Section 17.2 Periodic Sound Waves

*P17.8 (a) The speed gradually changes from v = (331 m �s)( 1 + 27°C�273°C)1�2 = 347 m �s to (331 m �s) (1 + 0�273°C)1�2 = 331 m �s, a 4.6% decrease. The cooler air at the samepressure 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 v�f = (347 m �s)�(4 000�s) = 86.7 mm to (331 m �s)�(4 000�s) = 82.8 mm. The crests are more crowded together when they move slower.

*P17.9 (a) If f = 2 4. MHz, λ = =×

=vf

1 500

2 4 100 6256

m s

smm

..

(b) If f = 1 MHz, λ = = =vf

1 500

101 506

m s

smm.

If f = 20 MHz, λ μ=×

=1 500

2 1075 07

m s

sm.

P17.10 ΔP smax max= ρ ωv

sP

maxmax

.

.= =

×( )( )

−Δρ ωv

4 00 10

1 20 343

3 N m

kg m

2

3 m s sm( )( ) ×( ) = ×−

−

2 10 0 101 55 10

3 110

π ..

P17.11 (a) A = 2 00. mμ

λ π= = =2

15 70 400 40 0

.. .m cm

v = = =ωk

858

15 754 6

.. m s

(b) s = ( )( ) − ( ) ×( )⎡⎣ ⎤−2 00 15 7 0 050 0 858 3 00 10 3. cos . . . ⎦⎦ = −0 433. mμ

(c) vmax . .= = ( )( ) =−Aω μ2 00 858 1 721m s mm s

P17.12 (a) ΔPx t

= ( ) −⎛⎝

⎞⎠1 27

340. sinPa

m s

π π (SI units)

The pressure amplitude is: ΔPmax .= 1 27 Pa

(b) ω π π= =2 340f s, so f = 170 Hz

(c) k = =2πλ

π m, giving λ = 2 00. m

(d) v = = ( )( ) =λ f 2 00 170. m Hz 340 m s

Sound Waves 453


454 Chapter 17

P17.13 k = =( ) = −2 2

0 10062 8 1π

λπ

..

mm

ω πλ

π= =

( )( ) = × −2 2 343

0 1002 16 104 1v m s

ms

..

Therefore,

ΔP x t= ( ) − ×⎡⎣ ⎤⎦0 200 62 8 2 16 104. sin . .Pa m s

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

0 500 13 0 10 6 50 1010 8. % . .( ) ×( ) = ×Pa Pa, the rod will break. Then, ΔP smax max= ρ ωv

sP

maxmax .

.= =

××( )

Δρ ωv

6 50 10

8 92 10 5 0

8

3

N m

kg m

2

3 110 2 5004 63

m s smm( )( ) =

π.

(b) From s s kx t= −( )max cos ω

v

v

= ∂∂

= − −( )

= = ( )

s

ts kx t

s

ω ω

ω π

max

max max

sin

2 500 4s .. .63 14 5mm m s( ) =

(c) I s= ( ) = = ×( )1

2

1

2

1

28 92 10 5

2 2 3ρ ω ρv vvmax max . kg m3 0010 14 52

m s m s( )( ).

= ×4 73 109. W m2

P17.15 ΔP s smax max max= = ⎛⎝

⎞⎠ρ ω ρ π

λv v

v2

λ πρ π= =

( )( ) ×( )−2 2 1 20 343 5 50 10

0

2 2 6v s

Pmax

max

. .

Δ ...

8405 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 5 0 10 5 00 1012 5 17. . .W m m W2 2

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

P = × = ×− −( . / )( . ) .1 0 5 0 10 5 00 102 5 2 5W m m W

P17.17 β =⎛⎝⎜

⎞⎠⎟

=××

⎛⎝

−

−10 104 00 10

1 00 100

6

12log log.

.

I

I ⎜⎜⎞⎠⎟

= 66 0. dB

454 Chapter 17


Sound Waves 455

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

P = ( ) =

= (

1

22

2 1 20

2 2 2 2

2

ρ ω π ρ

π

A s A f sv vmax max

. kg m3 )) ⎛⎝

⎞⎠ ( )( ) × −π 0 08

343 600 0 12 102

2 2..

m

2m s 1 s m(( ) =

221 2. W

P17.19 I s= 1

22 2ρω 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 6 252. .( ) = .

Therefore, 6 25 0 600 3 75. . .( ) = W m2

(b) 0 600. W m2

P17.20 The original intensity is I s f s12 2 2 2 21

22= =ρω π ρmax maxv v

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

I f s22 2 22= ′( )π ρv max

so I

I

f s

f s

f

f2

1

2 2

2 2 2

22

2=

′( ) = ′⎛⎝⎜

⎞⎠⎟

π ρπ ρ

vv

max

max

or

If

fI2

2

1= ′⎛⎝⎜

⎞⎠⎟

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

2 while the displacement amplitude is doubled, the new

intensity is

If

s f s I22

22 2 2 2

122

2 2= ⎛⎝

⎞⎠ ( ) = =π ρ π ρv vmax max

or the intensity is unchanged .

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

343

146 82 34

m s

sm

..

For the high note λ = =343

8800 390

m s

sm.

We observe that the ratio of the frequencies of these two notes is 880

5 99Hz

146.8 Hz= . nearly

equal to a small integer. This fact is associated with the consonance of the notes D and A.

(b) β =⎛⎝⎜

⎞⎠⎟

=−1010

7512dBW m

dB2logI

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 m kg m2 3 443 0 161m s Pa( ) = .

for both low and high notes.

continued on next page

Sound Waves 455


456 Chapter 17 456 Chapter 17

(c) I s f s= ( ) =1

2

1

24

2 2 2 2ρ ω ρ πv vmax max

sI

fmax =2 2 2π ρv

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

3π

== × = ×−

−6 24 10

146 84 25 10

57.

..m m

for the high note, smax

..= × = ×

−−6 24 10

7 09 105

8

880m m

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

134 3

880

804 91 093

.

. ..= = ,

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

P17.22 We begin with β22

0

10=⎛⎝⎜

⎞⎠⎟

logI

I and β1

1

0

10=⎛⎝⎜

⎞⎠⎟

logI

I so β β2 1

2

1

10− =⎛⎝⎜

⎞⎠⎟

logI

I

Also, Ir2

224

=Pπ

and Ir1124

=Pπ

giving I

I

r

r2

1

1

2

2

=⎛⎝⎜

⎞⎠⎟

Then, β β2 11

2

2

1

2

10 20− =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

log logr

r

r

r

P17.23 (a) I112 10 121 00 10 10 1 00 101= ×( ) = ×(− ( ) −. .W m W m2 2β ))1080 0 10.

or I141 00 10= × −. W m2

I212 10 121 00 10 10 1 00 102= ×( ) = ×(− ( ) −. .W m W m2 2β ))1075 0 10.

or I24 5 51 00 10 3 16 10= × = ×− −. .. W m W m2 2

When both sounds are present, the total intensity is

I I I= + = × + × = ×− −1 2

4 51 00 10 3 16 10 1 32 10. . .W m W m2 2 −−4 W m2

(b) The decibel level for the combined sounds is

β = ××

⎛⎝⎜

⎞⎠⎟

=−

−101 32 10

1 00 1010

4

12log.

.

W m

W m

2

2 llog . .1 32 10 81 28×( ) = dB


Sound Waves 457 Sound Waves 457

P17.24 In Ir

=P

4 2π, intensity I is proportional to

12r

, so between locations 1 and 2: I

I

r

r2

1

12

22=

In I s= ( )1

22ρ ωv max , intensity is proportional to smax

2 , so I

I

s

s2

1

22

12=

Then, s

s

r

r2

1

2

1

2

2⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

or 1

2

21

2

2

⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

r

r giving r r2 12 2 50 0 100= = ( ) =. m m

But, r d22 250 0= ( ) +. m

yields d = 86 6. m

P17.25 (a) 120 1010 12 2dB dB

W m= ⎡

⎣⎢

⎤⎦⎥−log

I

Ir

rI

= =

= = ( ) =

1 004

4

6 000

2.

.

W m

W

4 1.00 W m

2

2

P

Pπ

π π..691 m

We have assumed the speaker is an isotropic point source.

(b) 0 1010 12dB dB

W m2=⎛⎝⎜

⎞⎠⎟−log

I

I

rI

= ×

= =×

−1 00 10

4

6 00

12.

.

W m

W

4 1.00 10 W

2

12

Pπ π − mm

km2( ) = 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 5 002 2. . .m m

Ir

= =×

( )= ×

−−P

4

1 00 10

4 5 003 18 102

3

26

π π.

..

W

mW m22

2

2dBW m

W mβ

β

=×⎛

⎝⎜⎞⎠⎟

=

−

−103 18 10

10

1

6

12log.

00 6 50 65 0dB dB. .=

(b) rBC = 4 47. m

I = ×( ) = ×

=

−−1 50 10

4 4 475 97 10

10

3

26.

..

W

mW m2

π

β ddB

dB

log.

.

5 97 10

10

67 8

6

12

×⎛⎝⎜

⎞⎠⎟

=

−

−

β

(c) I = +3 18 5 97. .W m W m2 2μ μ

β = ×⎛⎝⎜

⎞⎠⎟

=−

−109 15 10

1069 6

6

12dB dBlog.

.



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

I I4 0 4

1

100= .

and I

P0 4

2 2

2

10 0

2 1 20 3430 121.

max .

..= = ( )

( )( ) =Δ

ρvW m22

The difference in sound intensity level is

Δβ =⎛⎝⎜

⎞⎠⎟

= −( ) = −10 10 2 00 20 04log . .I

Ikm

0.4 km

ddB

At 0.400 km,

β0 4 12100 121

10110 8. log

..=

⎛⎝⎜

⎞⎠⎟

=−

W m

W mdB

2

2

At 4.00 km,

β β β4 0 4 110 8 20 0 90 8= + = −( ) =. . . .Δ dB dB

Allowing for absorption of the wave over the distance traveled,

′ = − ( )( ) =β β4 4 7 00 3 60 65 6. . .dB km km dB

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

P17.28 (a) E t r It= P = = ( ) ×( )−4 4 100 7 00 10 0 2002 2 2π π m W m s2. .(( ) = 1 76. kJ

(b) β = ××

⎛⎝⎜

⎞⎠⎟

=−

−107 00 10

1 00 10108

2

12log.

.dB

P17.29 β = ⎛⎝

⎞⎠−10

10 12logI I = ⎡⎣ ⎤⎦( )( ) −10 1010 12β W m2

I 120 1 00dB2W m( ) = . ; I 100

21 00 10dB2W m( )

−= ×. ; I 10111 00 10dB

2W m( )−= ×.

(a) P = 4 2πr I so that r I r I12

1 22

2=

r rI

I2 11

2

1 2

23 001 00

1 00 1030=

⎛⎝⎜

⎞⎠⎟

= ( )×

=−..

..m 00 m

(b) r rI

I2 11

2

1 2

113 001 00

1 00 109=

⎛⎝⎜

⎞⎠⎟

= ( )×

=−..

..m 449 105× m



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 1010 12dB dB

W m2= −logI

; I = −10 2 W m2. If the lawnmower

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

r=

P4 2π

P = ( ) =−4 1 10 0 1262 2π m W m W2 .

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

intensity I =( ) = × −0 126

2 5 1025.

.W

4 20 mW m2

π. The total sound intensity impinging on you is

10 2 5 10 1 002 5 102 5 2− − −+ × = ×W m W m W m2 2 2. . . So its level is

101 002 5 10

10100 01

2

12dB dBlog.

.× =

−

−

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 1010

0

12

ln

ln

I

I

IdB dB

W m2⎜⎜⎞⎠⎟

= ( ) = =− −I 10 10 10 0 012 610 1 12 1 90. . .W m W m2 2 W m2

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

P = = ( )( ) =IA 0 012 6 22 0 0 277. . .W m m W2 2

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

E t= = ( )( )⎛⎝

⎞⎠P 0 277 20 0

60 0. .

.J s min

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 r= = ( ) = ×2 2 1 000 2 102 2 6π π πm m2

The intensity at this distance is

IA

= =×

= × −P 0 2774 41 10 8.

.W

2 10 mW m6 2

2

π and the sound intensity level is

β = ( ) ××

⎛⎝⎜

⎞⎠

−

−104 41 10

1 00 10

8

12dBW m

W m

2

2ln.

. ⎟⎟ = 46 4. dB



Section 17.4 The Doppler Effect

P17.32 (a) ω π π= =⎛⎝⎜

⎞⎠⎟

=2 2115

60 012 0f

min

s minrad s

..

vmax . . .= = ( ) ×( ) =−ωA 12 0 1 80 10 0 021 73rad s m m s

(b) The heart wall is a moving observer.

′ = +⎛⎝

⎞⎠ = ( ) +⎛

⎝f f Ov v

v2 000 000

1 500 0 021 7

1 500Hz

.⎜⎜

⎞⎠⎟

= 2 000 028 9. Hz

(c) Now the heart wall is a moving source.

′′ = ′−

⎛⎝⎜

⎞⎠⎟

= ( )−

f fs

vv v

2 000 0291 500

1 500 0 02Hz

. 11 72 000 057 8

⎛⎝⎜

⎞⎠⎟

= . Hz

*P17.33 (a) ′ =+( )

−( )ff o

s

v vv v

′ = +( )−( ) =f 2 500

343 25 0

343 40 03 04

.

.. kHz

(b) ′ = + −( )− −

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

( . ).08 kHzz

(c) ′ = + −( )−

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

..62 kHz while police car overtakes

′ = +− −( )

⎛⎝⎜

⎞⎠⎟

=f 2 500343 25 0

343 40 02

.

..40 kHz after police car passes

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

1

2

1

2

20 0

5 000 500

2 2m kA

k

mA

v

v

max

max

.

..

=

= = N m

kgmm m s( ) = 1 00.

The frequencies heard by the stationary observer range from

′ =+

⎛⎝⎜

⎞⎠⎟

f fminmax

vv v

to ′ =−

⎛⎝⎜

⎞⎠⎟

f fmaxmax

vv v

where v is the speed of sound.

′ =+

⎛⎝⎜

⎞⎠⎟

=fmin .440

343 1 00439Hz

343 m s

m s m sHHz

Hz343 m s

m s m s′ =

−⎛⎝⎜

⎞⎠⎟

=fmax .440

343 1 004441 Hz




(b) β π=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

10 104

0

2

0

dB dBlog logI

I

r

I

P

The maximum intensity level (of 60.0 dB) occurs at r r= =min .1 00 m. The minimum intensity level occurs when the speaker is farthest from the listener (i.e., when r r r A= = + =max min .2 2 00 m).

Thus,

β βπmax min

min

log log− =⎛⎝⎜

⎞⎠⎟

−104

100

2dB dBPI r

PP4 0

2π I rmax

⎛⎝⎜

⎞⎠⎟

or

β βπ

πmax min

min

maxlog− =⎛⎝⎜

⎞⎠

104

4

02

02

dBP

PI r

I r⎟⎟ =

⎛⎝⎜

⎞⎠⎟

102

2dBlog max

min

r

r

This gives:

60 0 10 4 00 6 02. log . .mindB dB dB− = ( ) =β and βmin .= 54 0 dB

P17.35 Approaching ambulance: ′ =−( )f

f

S1 v v

Departing ambulance: ′′ =− −( )( )f

f

S1 v v

Since ′ =f 560 Hz and ′′ =f 480 Hz 560 1 480 1−⎛⎝

⎞⎠ = +⎛

⎝⎞⎠

vv

vv

S S

1 040 80 0

80 0 343

1 04026 4

vv

v

S

S

=

= ( )=

.

..m s m s

P17.36 (a) v = ( ) +⋅

−( ) =331 0 6 10 325m sm

s CC m s.

°°

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

⎞⎠f f Ov v

v

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

⎝⎜⎞⎠⎟

f f Ov vv

The ratio is ′′′

= −+

=f

fO

O

v vv v

5

6

which gives 6 6 5 5v v v v− = +o o

or vv

O = = =11

325

1129 5

m sm s.

P17.37 ′ =−

⎛⎝⎜

⎞⎠⎟

f fs

vv v

485 512340

340 9 80=

− −( )⎛

⎝⎜⎞

⎠⎟. tfall

485 340 485 9 80 512 340

512 485

( ) + ( )( ) = ( )( )

= −

. t

t

f

f 4485

340

9 801 93⎛

⎝⎞⎠ =

.. s

d gt f121

218 3= = . m: treturn s= =18 3

3400 053 8

..

The fork continues to fall while the sound returns.

t t tftotal fall return s s= + = + =1 93 0 053 8 1 98. . . 55

1

219 3

s

mtotal total fall2d gt= = .



P17.38 (a) Sound moves upwind with speed 343 15−( ) m s. Crests pass a stationary upwind point at frequency 900 Hz.

Then λ = = =vf

328

9000 364

m s

sm.

(b) By similar logic, λ = = +( )=v

f

343 15

9000 398

m s

sm.

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

′ = +−

⎛⎝⎜

⎞⎠⎟

= +−

⎛⎝

⎞⎠ =f f o

s

v vv v

900343 0

343 1594Hz 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 vv v

900343 30

343 15Hz

⎞⎞⎠⎟

= ⎛⎝

⎞⎠ =900

373

358938Hz Hz

P17.39 (b) sin.

θ = =vvS

1

3 00; θ = 19 5. °

tanθ = h

x; x

h=tanθ

x = = × =20 000

19 55 66 10 56 64m

m kmtan .

. .°

(a) It takes the plane tx

S

= = ×( ) =

v5 66 10

56 34.

.m

3.00 335 m ss to travel this distance.

P17.40 θ = = =− −sin sin.

.1 1 1

1 3846 4

vvS

°

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

vlight

S

vv

S = =×( ) = ×light m s

sin

.

sin ..

θ2 25 10

53 02 82 10

8

°88 m s

t = 0

a.

h

Observer

b.

h

Observer hears the boom

x

FIG. P17.39(a)



Section 17.5 Digital Sound Recording

Section 17.6 Motion Picture Sound

P17.42 For a 40-dB sound,

40 1010

10

12

8

dB dBW m

W m

2

2

= ⎡⎣⎢

⎤⎦⎥

= =

−

−

logI

IPΔ mmax

max .

2

8

2

2 2 1 20 343 10

ρ

ρ

v

vΔP I= = ( )( ) −kg m m s2 W m N m2 2= 2 87 10 3. × −

(a) code =× =

−2 87 10

28 765 536 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 m �s + 0 = (0.15 + 0.2) kg v v = 0.986 m �s

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 = E�At = 0.5(0.227 J)�[4π(0.8 m)2 0.007 s] = 2.01 W�m2

Its intensity level is β = 10 log(2.01�10−12) = 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.



*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 = (2030�s)�(1.36�m) = 1.49 km �s. 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 (2030�s)�2π = 323 Hz. Its wavelength is constant at 2π �k = 2π �(1.36 �m) = 4.62 m. Itspressure amplitude is 25.0 Pa at radius 1 m. Its intensity at this distance is

IP

= =Δ max

2

2ρv

( )( )(

225

2 1000 1490

N/m

kg/m m/

2

3 ss)W/m2= 209 μ

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 1 2 634 π. ( .W/m m) mW2 2 .

*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 m �s; 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 ss( ) =

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

3400 004

..

m

m ss

( )( ) =

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 5300

.~

sHz a few hundred Hz=

wavelength

λ = =( )

( ) =vf

340

3001 2 100m s

sm 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.



*P17.46 (a) The distance is larger by 240�60 = 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 cos sin2 2 1θ θ+ = , sin cosθ θ= ± −1 2 (each sign applying half the time)

Δ ΔP P kx t s kx t= −( ) = ± − −( )max maxsin cosω ρ ω ωv 1 2

Therefore ΔP s s kx t s s= ± − −( ) = ± −ρ ω ω ρ ωv vmax max maxcos2 2 2 2 2

P17.48 (a) λ = = =−

vf

343

1 4800 2321

m s

sm.

(b) β = = ⎡⎣⎢

⎤⎦⎥−81 0 10

10 12. logdB dBW m2

I

I = ( ) = = ×− − −10 10 10 1 26 1012 8 10 3 90 4W m W m W2 2. . . mm

W m

2

2

=

= =×( )−

1

2

2 2 1 26 10

2 2

2

4

ρ ω

ρ ω

v

v

s

sI

max

max

.

11 20 343 4 1 4808 41 10

2 1 2.

.kg m m s s3( )( ) ( ) = ×

−−

π88 m

(c) ′ =′

= =−λ vf

343

1 3970 2461

m s

sm. 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

3 000 667 1

..

minmin

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

′ = +⎛⎝

⎞⎠ = ( ) + −( )−f f ov v

v0 667

19 7 4 47

19 71.

. .

.min

⎛⎛⎝⎜

⎞⎠⎟

= 0 515. min

(b) ′′ =+⎛

⎝⎜⎞⎠⎟ = ( ) + −( )−f f ov v

v′

0 66719 7 1 561.

. .min

119 70 614

..

⎛⎝⎜

⎞⎠⎟

= 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

ρ20 0 10

7 8605 04 10

103.

.N m

kg mm 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

tL= =

×= × −

v0 800

1 59 10 4..

m

5.04 10 m ss3




(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 ti= = ( ) ×( ) = × =− −v 12 0 1 59 10 1 90 10 14 3. . .m s s m ..90 mm

(d) The strain in the rod is: ΔL

L=

×= ×

−−1 90 10

2 38 103

3..

m

0.800 m

(e) The stress in the rod is: σ = ⎛⎝

⎞⎠ = ×( ) ×( ) =−Y

L

L

Δ20 0 10 2 38 10 47610 3. .N m M2 PPa

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

LL

Y= =

vρ

, traveling distance ΔL ti= v after the front end hits the wall.

The strain in the rod is: ΔL

L

t

L Yi

i= =v

vρ

The stress is then: σ ρ ρ= ⎛⎝

⎞⎠ = =Y

L

LY

YYi i

Δv v

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

vi yYρ σ< or viy

Y<

σρ

With the given numbers, this speed is 10.1 m �s. 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

vv vdiver

so 1− =′

vvdiver f

f ⇒ = −

′⎛⎝⎜

⎞⎠⎟

v vdiver 1f

f

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

we fi nd vdiver m s= −⎛⎝⎜

⎞⎠⎟

=343 11 800

2 15055 8.

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

′′ = ′+( )

⇒ ′′ =−( )

⎡

⎣⎢

⎤

⎦⎥

+f f f f

v vv

vv v

v vdiver

diver

ddiver( )v

so ′′ = +( )−( ) =f 1 800

343 55 8

343 55 82 500

.

.Hz



P17.52 Let P(x) represent absolute pressure as a function of x. The net force to the right on the chunkof air is + ( ) − +( )P x A P x x AΔ . Atmospheric pressure subtracts out, leaving

− +( ) + ( )⎡⎣ ⎤⎦Δ Δ ΔP x x P x = −∂∂Δ

ΔAP

xxA. The mass of the air is Δ Δm V A x= =ρ ρ Δ and its

acceleration is ∂∂

2

2

s

t. So Newton’s second law becomes

−∂∂

=∂∂

−∂∂

−∂∂

⎛⎝⎜

⎞⎠⎟

=∂∂

ΔΔ Δ

P

xxA A x

s

t

xB

s

x

s

t

ρ

ρ

2

2

2

22

2

2

2

2

B s

x

s

tρ∂∂

=∂∂

Into this wave equation as a trial solution we substitute the wave function s x t s kx t, cosmax( ) = −( )ω We fi nd

∂∂

= − −( )

∂∂

= − −

s

xks kx t

s

xk s kx t

max

max

sin

cos

ω

ω2

22 (( )

∂∂

= + −( )

∂∂

= −

s

ts kx t

s

ts kx

ω ω

ω

max

max

sin

cos2

22 −−( )ωt

B s

x

s

tρ∂∂

= ∂∂

2

2

2

2 becomes − −( ) = − −( )Bk s kx t s kx t

ρω ω ω2 2

max maxcos cos

This is true provided B

fρ

πλ

π44

2

22 2=

The sound wave can propagate provided it has λρ

2 2 2fB= =v ; that is, provided it propagates with

speed v = B

ρP17.53 When observer is moving in front of and in the same direction as the source, ′ = −

−f f O

S

v vv 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 10 0 55 0 15 3. . . .km h km h km h m s , annd

km h km h km h mvS = − −( ) = =64 0 10 0 74 0 20 55. . . . ss

Therefore,

′ = ( ) −−

f 1 200 01 520 15 3

1 520 20 55.

.

.Hz

m s m s

m s mm sHz= 1 204 2.

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 0340 5 00 340

340 5 00 340. .

.

.=

+( ) −( )−( ) +

vv

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



P17.55 103 1010 12dB dB

W m2= ⎡⎣⎢

⎤⎦⎥−log

I

(a) Ir

= × = =( )

−2 00 104 4 1 6

22 2.

.W m

m2 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 = = =−

vf

343

1 0000 3431

m s

sm.

(c) ′ =′

= −⎛⎝

⎞⎠ = −( )

=−λ v v v vvf f

S 343 40 0

1 00001

..

m s

s3303 m

(d) ′′ =′′

= +⎛⎝

⎞⎠ = +( )

=−λ v v v vvf f

S 343 40 0

1 000 1

. m s

s00 383. m

(e) ′ = −−

⎛⎝⎜

⎞⎠⎟

= ( ) −( )f f O

S

v vv v

1 000343 30 0

34Hz

m s.

33 40 01 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 vv v

v vair cu

cu air

air cu

and the length of the bar is L t=−

=( ) ×( )v v

v vair cu

cu air

m s m sΔ

331 3 56 10

3 5

3.

660 331365

−( ) =m s

m/s)Δ Δt t(

(c) L = (365 m �s)(0.127 s) = 46.3 m

(d) The answer becomes Lt

r

=−

Δ1

3311

m s� v

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

r

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 P2 1

1

20 0=

. β β1 2

1

2

10− = logPP

80 0 10 20 0 13 0

67 0

2

2

. log . .

.

− = = +

=

β

β dB

468 Chapter 17

FIG. P17.56(a)



P17.59 (a) θ =⎛

⎝⎜⎞

⎠⎟=

×⎛⎝

⎞− −sin sin.

1 13

331

20 0 10

vvsound

obj⎠⎠ = 0 948. °

(b) ′ =×

⎛⎝

⎞⎠ =−θ sin

..1

3

1 533

20 0 104 40°

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

Then Pav = 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 v vt t rS− = .

Then, t

r

S

=−v v

The area of the sphere is 442

2 2

2π πv

vv v

tr

S

( ) =−( )

. The energy per unit area over the spherical wavefront

is uniform with the value E

A

T

rS=

−( )Pav v v

v

2

2 24π. The observer receives parcels of energy with the

Doppler shifted frequency ′ =−

⎛⎝⎜

⎞⎠⎟

=−( )f f

TS S

vv v

vv v

, so the observer receives a wave with intensity

IE

Af

T

r TS=

⎛⎝⎜

⎞⎠⎟

′ =−( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ −

Pav

v v

vv

v v

2

2 24πSS

S

r( )⎛

⎝⎜

⎞

⎠⎟ =

−⎛

⎝⎜⎞

⎠⎟P

av

4 2πv 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 vv

L

T

= 8 00. , we have TY= μ

ρ64 0. where μ = m

L and

ρπ

= =mass

volume

m

r L2

This gives

Tr Y

= =×( ) ×( )−π π2 3 2 10

64 0

2 00 10 6 80 10

64.

. .m N m2

...

01 34 104= × N

Sound Waves 469



P17.62 (a) If the source and the observer are moving away from each other, we have: θ θS = =0 180°, and since cos180 1° = − , we get Equation (17.13) with negative values for both v

O and v

S.

(b) If vO

= 0 m /s then ′ =−

f fS S

vv 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 ′ =− ( ) ( )f343

343 0 800 25 0500

m s

m s m sHz

. .

or ′ =f 531 Hz

Note that as the train approaches, passes, and departs from the intersection, θS varies from

0° to 180° and the frequency heard by the observer varies between the limits

′ =−

=−

f fS

max cos .

vv v 0

343

343 25 0500

°

m s

m s m sHzz Hz

m s

m s

( ) =

′ =−

=

539

180

343

343f f

Smin cos

vv v ° ++

( ) =25 0

500 466. m s

Hz Hz

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

distance L at speed v is given by tL L

Y= =

v ρ Using this expression we fi nd

tL

Y

L1

1

1 1

1

107 00 10 2 7001 96= =

×( ) ( )=

ρ ..

N m kg m2 3××( )

= − = −

×

−10

1 50 1 50

1 60

41

21

2 2

1

L

tL

Y

L

s

m m. .

.ρ 110 11 3 1010 3N m kg m2 3( ) ×( ).

or t L23 4

11 26 10 8 40 10= × − ×( )− −. . s

t

t

3

3

1 50

8 800

4 24

=×( ) ( )

= ×

.

.

m

11.0 10 N m kg m10 3 3

110 4− s

We require t t t1 2 3+ = , or

1 96 10 1 26 10 8 40 10 4 24 1041

3 41

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

L1

2

6 45= .

(b) The ratio of lengths L

L1

2

is adjusted in part (a) so that t t t1 2 3+ = . Sound travels the two paths

in equal time and the phase difference Δφ = 0 .

470 Chapter 17

FIG. P17.63

L1 L2

L3


Sound Waves 471

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 m �s to 331 m �s. (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 10. × − m

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 73 109. × W m2

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

P17.18 21.2 W

P17.20 (a) If

fI2

2

1= ′⎛⎝⎜

⎞⎠⎟

(b) I I2 1=

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 km �s, 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 μW�m2, so the power of the source and the net power of the wave at all distances is 2.63 mW.


472 Chapter 17

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


P17.54 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


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


Solucionario serway cap 17

Engineering