OpenStax College Physics
Instructor Solutions Manual
Chapter 17
CHAPTER 17: PHYSICS OF HEARING
17.2 SPEED OF SOUND, FREQUENCY, AND WAVELENGTH
1.
When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its
wavelength if the speed of sound is 345 m/s?
Solution
v w  f , so that  
2.
What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?
Solution
vw  f  f 
3.
vw


vw
345 m/s

 0.288 m
f
1200 Hz
340 m/s
 3400 Hz
0.100 m
Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of
0.221 m.
Solution
v w  f  1500 Hz 0.221 m  331.5 m/s  332 m/s
4.
(a) What is the speed of sound in a medium where a 100-kHz frequency produces a
5.96-cm wavelength? (b) Which substance in Table 17.1 is this likely to be?
Solution (a) v w  f  (100  10 3 Hz)(5.96  10 2 m)  5.96  10 3 m/s
(b) steel (from value in Table 17.1)
5.
Solution
6.
Show that the speed of sound in 20.0C air is 343 m/s, as claimed in the text.
v w  (331 m/s)
T (K)
293 K
 (331 m/s)
 343 m/s
273 K
273 K
Air temperature in the Sahara Desert can reach 56.0C (about 134F ). What is the
speed of sound in air at that temperature?
OpenStax College Physics
Solution
7.
v w  (331 m/s)
Instructor Solutions Manual
Chapter 17
T (K)
329 K
 (331 m/s)
 363.4 m/s  363 m/s
273 K
273 K
Dolphins make sounds in air and water. What is the ratio of the wavelength of a
sound in air to its wavelength in seawater? Assume air temperature is 20.0C .
Solution We know vseawater  1540 m/s (from Table 17.1) and vair  343 m/s at 20.0C from
Problem 17.5, since vseawater  fseawater and vair  fair , we know
vair
vseawater

air
seawater

air
seawater

343 m/s
 0.223
1540 m/s
8.
A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance
to the object creating the echo? (Assume that the submarine is in the ocean, not in
fresh water.)
Solution
2d  v w t  d 
9.
(a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what
is the smallest difference in distances it can detect? (Assume that the submarine is in
the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on
the ability of the sonar system to detect the size and shape of the object creating the
echo.
Solution
(a) 2d  vw t  d 
v w t (1540 m/s)(1.20 s)

 924 m
2
2
vw t (1540 m/s)(0.010 0 s)

 7.70 m
2
2
(b) This means that sonar is good for spotting and locating large objects, but it isn’t
able to resolve smaller objects, or detect the detailed shapes of objects. Objects
like ships or large pieces of airplanes can be found by sonar, while smaller pieces
must be found by other means.
10.
A physicist at a fireworks display times the lag between seeing an explosion and
hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air
temperature is 24.0C and if you neglect the time taken for light to reach the
physicist? (b) Calculate the distance to the explosion taking the speed of light into
account. Note that this distance is negligibly greater.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 17
T (K)
297 K
 (331 m/s)
 345.24 m/s
273 K
273 K
d  v w t  (345.24 m/s)(0.400 s)  138.096 m  138 m
(a) v w  (331 m/s)
(b) The time for the light to travel is approximately
d
138.1 m

 4.60  10 7 s
c 3.00  10 8 m/s
d '  v w (t  t1 )  (345.24 m/s)(0.400 s  4.60  10 7 s)  138.096 m
t1 
The distance is the same to six significant figures.
11.
Solution
Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure
17.10.) (a) Calculate the echo times for temperatures of 5.00C and 35.0C . (b)
What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss
the significance of this uncertainty and whether it could cause difficulties for the bat.
(In practice, the bat continues to use sound as it closes in, eliminating most of any
difficulties imposed by this and other effects, such as motion of the prey.)
(a)
278 K
 334.0 m/s, so that
273 K
2d 2(3.00 m)


 0.0180 s
v w 334.0 m/s
v w  (331 m/s)
t 5.00C
308 K
 351.6 m/s, so that
273 K
2d 2(3.00 m)


 0.0171 s
v w 351.6 m/s
v w  (331 m/s)
t 35.0C
(b) % uncertainty =
Δt 0.0180 s  0.0171 s

 0.0500  5.00 %
t
0.0180s
(c) This uncertainty could definitely cause difficulties for the bat, if it didn’t continue
to use sound as it closed in on its prey. A 5% uncertainty could be the difference
between catching the prey around the neck or around the chest, which means
that it could miss grabbing its prey.
17.3 SOUND INTENSITY AND SOUND LEVEL
OpenStax College Physics
12.
Solution
13.
Solution
Instructor Solutions Manual
Chapter 17
What is the intensity in watts per meter squared of 85.0-dB sound?
 I 
  I  I 0 10  / 10  (10 12 N/m 2 )10 8.50  3.16  10 4 N/m 2
I
 0
 (dB)  10 log 
The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB.
What is this in watts per meter squared?
I 
 , where I 0  10 12 W/m 2 , so that I  I 0 10  / 10 , and
 I0 
  10 log 10 
I  (1.00  10 12 W/m 2 )1091.0 /10.0  1.26  10 3 W/m 2
14.
Solution
A sound wave traveling in 20C air has a pressure amplitude of 0.5 Pa. What is the
intensity of the wave?
I
(p) 2
(0.50 Pa) 2
=
= 3.04  10 -4 W/m 2
3
2 vm 2 1.20 kg  m 343 m/s 


15.
What intensity level does the sound in the preceding problem correspond to?
Solution
I 
Using the sound intensity level equation  dB  10 log 10   ,
 I0 
 2.90  10-4 
I
 = 85 dB .
(d ) = 10log 10 ( ) = 10log 10 
-12 
I0
 1.00  10 
16.
What sound intensity level in dB is produced by earphones that create an intensity of
4.00  102 W/m 2 ?
Solution
 I 
 4.00  10 2 W/m 2 
  106 dB
β  10log    10log 
12
2 
 1.00  10 W/m 
 I0 
17.
Show that an intensity of 1012 W/m 2 is the same as 1016 W/cm 2 .
OpenStax College Physics
Solution
Solution
Chapter 17
2
10
18.
Instructor Solutions Manual
12
 1m 
16
2
W/m 
  10 W/cm
 100 cm 
2
(a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b)
What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?
(a) From Table 17.3,
I2
 2  Δβ  3 dB. Thus β  Δβ  90 dB  3 dB  93 dB
I1
I
I
I
1
(b) 2   1  5 or  2
I1 5
I2
 I1
I
Δβ  10log 10  2
 I1



1



1
 5;
I
 10log 10  2
 I1

  7

Thus, β  Δβ  90 dB  7 dB  83 dB
19.
(a) What is the intensity of a sound that has a level 7.00 dB lower than a
4.00 109 W/m 2 sound? (b) What is the intensity of a sound that is 3.00 dB higher
than a 4.00 109 W/m 2 sound?
Solution (a) Using Table 17.3, 4.00 10 9 W/m 2 / 5  8.00 10 10 W/m 2
(b) 4.00 109 W/m 2  2  8.00 109 W/m 2
20.
Solution
(a) How much more intense is a sound that has a level 17.0 dB higher than another?
(b) If one sound has a level 23.0 dB less than another, what is the ratio of their
intensities?
(a) One factor of 10 (10.0 dB) and one factor of 5 (7.0 dB) make an overall factor of
1.7
50.1, i.e. 10
1
1
. The
(b) Two 10-decibel losses give two factors of
, or a total factor of
10
100
1
1
remaining 3 decibels give an additional factor of . So the ratio is a factor of
2
200
1
i.e. 10 2.3  0.005012 
200
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
21.
People with good hearing can perceive sounds as low in level as  8.00 dB at a
frequency of 3000 Hz. What is the intensity of this sound in watts per meter
squared?
Solution
 I 
β  10log  , so that
 I0 
I  I 0 10  / 10  (1.00  10 12 W/m 2 )10 8.00 / 10.0  1.58  10 13 W/m 2
22.
If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise
level of 1000 flies at that distance, assuming interference has a negligible effect?
Solution A factor of 1000 in intensity corresponds to an increase of 30.0 dB, i.e.,
10 log( 103 )  30.0 dB . Thus, 40.0 dB + 30.0 dB = 70.0 dB
23.
Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level
at the center of the circle. What is the average sound intensity level produced there by
each stereo, assuming interference effects can be neglected?
Solution A decrease of a factor of 10 in intensity corresponds to a reduction of 10 dB in sound
level. 120 dB  10 dB = 110 dB
24.
The amplitude of a sound wave is measured in terms of its maximum gauge pressure.
By what factor does the amplitude of a sound wave increase if the sound intensity
level goes up by 40.0 dB?
Solution The intensity I is proportional to X 2 where X is the amplitude. We know that 40 dB
4
is equivalent to an increase of a factor of 1.00 10 in intensity.
2
I2  X 2 
X
  1.00  10 4  2  100
 
I1  X 1 
X1
25.
If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge
pressure (sound amplitude) of 109 atm , what is the maximum gauge pressure in a 60dB sound? What is the maximum gauge pressure in a 120-dB sound?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 17
6
We know that 60 dB corresponds to a factor of 10 increase in intensity. Therefore,
2
1/2
X 
I 
I
I  X  2   2  , so that X 2  X 1  2   (10 9 atm)(10 6 )1/2  10 6 atm.
I1  X 1 
 I1 
120 dB corresponds to a factor of 1012 increase  109 atm(10 12 )1/2  103 atm .
2
26.
An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage.
What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?
Solution
E = Pt = IAt  E  IA
 3600 s 
3
E = (1.00  10 3 W/m 2 ) (0.400  10  2 m) 2  (8.00 h) 
  1.45  10 J
 1h 
27.
(a) Ear trumpets were never very common, but they did aid people with hearing losses
by gathering sound over a large area and concentrating it on the smaller area of the
eardrum. What decibel increase does an ear trumpet produce if its sound gathering
2
2
area is 900 cm and the area of the eardrum is 0.500 cm , but the trumpet only has
an efficiency of 5.00% in transmitting the sound to the eardrum? (b) Comment on the
usefulness of the decibel increase found in part (a).
Solution
(a) I 
P
I
A
so that 2  1 , so for a 5.00% efficiency:
A
I1 A2
 I 
I e At (0.0500)(9 00 cm 2 )
  :



dB

10
log



90
.
Now,
using
10
I t Ae
0.500 cm 2
 I0 
 Ie 
I 
I 
  10 log  t   10log  e   10log(90)  19.54 dB  19.5 dB
 I0 
 I0 
 It 
 e   t  10 log 
(b) This increase of approximately 20 dB increases the sound of a normal
conversation to that of a loud radio or classroom lecture (see Table 17.2). For
someone who cannot hear at all, this will not be helpful, but for someone who is
starting to lose their ability to hear, it may help. Unfortunately, ear trumpets are
very cumbersome, so even though they could help, the inconvenience makes
them rather impractical.
OpenStax College Physics
28.
Instructor Solutions Manual
Chapter 17
Sound is more effectively transmitted into a stethoscope by direct contact than
through the air, and it is further intensified by being concentrated on the smaller area
of the eardrum. It is reasonable to assume that sound is transmitted into a
stethoscope 100 times as effectively compared with transmission though the air.
What, then, is the gain in decibels produced by a stethoscope that has a sound
2
gathering area of 15.0 cm , and concentrates the sound onto two eardrums with a
2
total area of 0.900 cm with an efficiency of 40.0%?
Solution
I 2 (100)(0.400)(15.0 cm 2 )

 666.67
I1
0.900 cm 2
 I2
 I0
 2  1  10 log 

I
  10 log  1

 I0

I
  10 log  2
 I1


  10log(666. 67)  28.2 dB

29.
Loudspeakers can produce intense sounds with surprisingly small energy input in spite
of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound
intensity level for a 12.0-cm-diameter speaker that has an efficiency of 1.00%. (This
value is the sound intensity level right at the speaker.)
Solution
P
IA
 100 IA  100(1.00  10 3 W/m 2 )π (0.600  10  2 m) 2  1.13  10 5 W
0.0100
17.4 DOPPLER EFFECT AND SONIC BOOMS
30.
(a) What frequency is received by a person watching an oncoming ambulance moving
at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound
on this day is 345 m/s. (b) What frequency does she receive after the ambulance has
passed?
Solution (a) Given vs  110 km/h  30.56 m/s,
f obs  f s
vw
345 m/s
 (800 Hz)
 877.75 Hz  878 Hz
v w  vs
345m/s  30.56 m/s
(b) f s  (800 Hz)
345 m/s
 734.9 Hz  735 Hz
345 m/s  30.56 m/s
OpenStax College Physics
31.
Instructor Solutions Manual
Chapter 17
(a) At an air show a jet flies directly toward the stands at a speed of 1200 km/h,
emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What
frequency is received by the observers? (b) What frequency do they receive as the
plane flies directly away from them?
Solution (a) Given vs  1200 km/h  333.33 m/s,
f obs  f s
vw
v w  vs
 (3500 Hz)
(b) f obs  f s
32.
Solution
33.
Solution
vw
342 m/s
 (3500 Hz)
 1772 Hz  1.77  10 3 Hz
v w  vs
342 m/s  333.33 m/s
What frequency is received by a mouse just before being dispatched by a hawk flying
at it at 25.0 m/s and emitting a screech of frequency 3500 Hz? Take the speed of
sound to be 331 m/s.
f obs  f s
vw
331 m/s
 (3500 Hz)
 3786 Hz  3.79  10 3 Hz
v w  vs
331 m/s  25.0 m/s
A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is
playing an 880-Hz note. At what speed is the musician approaching if the speed of
sound is 338 m/s?
f obs  f s
vs 
34.
342 m/s
 1.381  10 5 Hz  1.38  10 5 Hz
342 m/s  333.33 m/s
vw
, so that
v w  vs
v w ( f obs  f s ) (338 m/s)(888 Hz  880 Hz)

 3.05 m/s
f obs
888 Hz
A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of
sound is 335 m/s. (a) An observer waiting at the crossing receives a frequency of 208
Hz. What is the speed of the train? (b) What frequency does the observer receive as
the train moves away?
OpenStax College Physics
Solution
(a) f obs  f s
vs 
Solution
Chapter 17
vw

v w  vs
v w ( f obs  f s ) (335 m/s)(208 Hz  200 Hz)

 12.885 m/s  12.9 m/s
f obs
208 Hz
(b) f obs  f s
35.
Instructor Solutions Manual
vw
335 m/s
 (200 Hz)
 192.6 Hz  193 Hz
v w  vs
335 m/s  12.885 m/s
Can you perceive the shift in frequency produced when you pull a tuning fork toward
you at 10.0 m/s on a day when the speed of sound is 344 m/s? To answer this
question, calculate the factor by which the frequency shifts and see if it is greater than
0.300%.
f obs  f s
vw
f
vw
344 m/s
 obs 

 1.030
v w  vs
fs
vw  vs 344 m/s  10.0 m/s
Yes. The shift is just at 3% so it is audible by a factor of 10.
36.
Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0
m/s. Both screech, the first one emitting a frequency of 3200 Hz and the second one
emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound
is 330 m/s?
Solution The first eagle hears:
 v  v01  v w 


f obs  f s  w
v
v

v
w
02 

 w
330 m/s
 330 m/s  15.0 m/s 

3
 (3800 Hz) 

  4.23  10 Hz
330
m/s
330
m/s

20.0
m/s



The second eagle hears:
330 m/s
 330 m/s  20.0 m/s 

f obs  (3200 Hz) 


330 m/s

 330 m/s  15.0 m/s 
 3556 Hz  3.56  10 3 Hz
OpenStax College Physics
37.
Solution
Instructor Solutions Manual
Chapter 17
What is the minimum speed at which a source must travel toward you for you to be
able to hear that its frequency is Doppler shifted? That is, what speed produces a shift
of 0.300% on a day when the speed of sound is 331 m/s?
An audible shift occurs when
f obs  f s
vs 
f obs
 1.003 .
fs
vw
f
vw
 obs 

v w  vs
fs
v w  vs
v w ( f obs / f s )  1 (331 m/s)(1.003  1)

 0.990 m/s
f obs / f s
1.003
17.5 SOUND INTERFERENCE AND RESONANCE: STANDING WAVES IN AIR
COLUMNS
38.
A “showy” custom-built car has two brass horns that are supposed to produce the
same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is
produced?
Solution The beat frequency is f B  f1  f 2  263.8 Hz  264.5 Hz  0.7 Hz.
39.
What beat frequencies will be present: (a) If the musical notes A and C are played
together (frequencies of 220 and 264 Hz)? (b) If D and F are played together
(frequencies of 297 and 352 Hz)? (c) If all four are played together?
Solution (a) Using the equation f B  f1  f 2 :
f B,A&C  f1  f 2  264 Hz  220 Hz  44 Hz
(b) f B,D&F  f1  f 2  352 Hz  297 Hz  55 Hz
(c) We get beats from every combination of frequencies, so in addition to the two
beats above, we also have:
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
f B, F&A = 352 Hz  220 Hz = 132 Hz;
f B, F&C = 352 Hz  264 Hz = 88 Hz;
f B, D&C  297 Hz  264 Hz  33 Hz;
f B, D&A  297 Hz  220 Hz  77 Hz
40.
Solution
What beat frequencies result if a piano hammer hits three strings that emit
frequencies of 127.8, 128.1, and 128.3 Hz?
f B  f1  f 2
128.3 Hz  128.1 Hz  0.2 Hz ;
128.3 Hz  127.8 Hz  0.5 Hz;
128.1 Hz  127.8 Hz  0.3 Hz
41.
Solution
A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a
single piano string. What are the two possible frequencies of the string?
fB 
1
1

 0.500 Hz  f 1  f 2 so the other frequency is 263.5 Hz or 264.5 Hz.
TB 2.00
42.
(a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on
a day when the speed of sound is 344 m/s? (b) What is the frequency of its second
harmonic?
Solution
344 m/s
v 
(a) f n  n w  for n = 1, 2, 3…  f1 
 256 Hz
2(0.672 m)
 2L 
(b) 2 f1 = 2 (256 Hz) = 512 Hz
43.
Solution
If a wind instrument, such as a tuba, has a fundamental frequency of 32.0 Hz, what
are its first three overtones? It is closed at one end. (The overtones of a real tuba are
more complex than this example, because it is a tapered tube.)
v
v 
f n  n w  for n = 1, 3, 5… w  f1  32.0 Hz.
4L
 4L 
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
first overtone = f 3  3(32.0 Hz)  96.0 Hz;
second overtone = f 5  5(32.0 Hz)  160 Hz;
third overtone  f 7  7(32.0 Hz)  224 Hz
44.
Solution
What are the first three overtones of a bassoon that has a fundamental frequency of
90.0 Hz? It is open at both ends. (The overtones of a real bassoon are more complex
than this example, because its double reed makes it act more like a tube closed at one
end.)
v
 v 
f n  n
 f1  90.0 Hz
 for n = 1, 2, 3… 
2L
 2L 
first overtone = f 2  2(90.0 Hz)  180 Hz;
second overtone = f 3  3(90.0 Hz)  270 Hz;
third overtone  f 4  4(90.0 Hz)  360 Hz
45.
Solution
How long must a flute be in order to have a fundamental frequency of 262 Hz (this
frequency corresponds to middle C on the evenly tempered chromatic scale) on a day
when air temperature is 20.0C ? It is open at both ends.
vw
v
 v 
L w
f n  n
 for n  1,2,3...., so that f1 
2L
2 f1
 2L 
Since we know the air temperature:
vw  (331 m/s)
Therefore, L 
46.
Solution
T (K)
293 K
 (331 m/s)
 342.9 m/s.
273 K
273 K
342.9 m/s
 0.654 m  65.4 cm
2(262 Hz)
What length should an oboe have to produce a fundamental frequency of 110 Hz on a
day when the speed of sound is 343 m/s? It is open at both ends.
v
343 m/s
v 
f n  n w  for n  1,2,3...  L  w 
 1.56 m
2 f1 2(110 Hz)
 2L 
OpenStax College Physics
47.
Solution
48.
Solution
Instructor Solutions Manual
What is the length of a tube that has a fundamental frequency of 176 Hz and a first
overtone of 352 Hz if the speed of sound is 343 m/s?
v
v
v 
f n  n w  for n  1,2,3....  f 1  w and f 2  w
2L
L
 2L 
 1 1  vw
f 2  f1  v w  
, so that

 L 2L  2L
vw
343 m/s
L

 0.974 m
2( f 2  f 1 ) 2(352 Hz  176 Hz)
(a) Find the length of an organ pipe closed at one end that produces a fundamental
frequency of 256 Hz when air temperature is 18.0C . (b) What is its fundamental
frequency at 25.0C ?
T (K)
291 K
 (331 m/s)
 341.7 m/s.
273 K
273 K
v
341.7 m/s
v 
f n  n w  for n  1,3,5.....  L  w 
 0.3337 m  0.334 m
4 f1 4(256 Hz)
 4L 
(a) v w  (331 m/s)
f ' v ' T'
(b) 1  w   
f1 v w  T 
49.
Solution
Chapter 17
1/ 2
 298 K 
, so that f1 '  (256 Hz) 

 291 K 
1/2
 259 Hz
By what fraction will the frequencies produced by a wind instrument change when air
temperature goes from 10.0C to 30.0C ? That is, find the ratio of the frequencies at
those temperatures.
v 
 v' 
f n  n w , f ' n  n w  so that
 2L 
 2L 
T ' (K)
f ' n v' w
T'
273 K



fn
vw
T
T (K)
(331 m/s)
273 K
(331 m/s)
303 K
 1.0347  1.03 or 103%
283 K
OpenStax College Physics
50.
Solution
Instructor Solutions Manual
Chapter 17
The ear canal resonates like a tube closed at one end. (See Figure 17.39.) If ear canals
range in length from 1.80 to 2.60 cm in an average population, what is the range of
fundamental resonant frequencies? Take air temperature to be 37.0C , which is the
same as body temperature. How does this result correlate with the intensity versus
frequency graph (Figure 17.37) of the human ear?
T (K)
310 K
 (331 m/s)
 352.7 m/s.
273 K
273 K
v
v 
f n  n w  for n  1,3,5..... f1  w , therefore ,
4L
 4L 
352.7 m/s
f min 
 3392 Hz  3.39 kHz and
4(2.60  10  2 m)
352.7 m/s
f max 
 4899 Hz  4.90 kHz
4(1.80  10  2 m)
v w  (331 m/s)
For this frequency range, the graph shows a dip in the intensity of sound needed to
pass the hearing threshold, presumably because these frequencies are resonances.
51.
Solution
Calculate the first overtone in an ear canal, which resonates like a 2.40-cm-long tube
closed at one end, by taking air temperature to be 37.0C . Is the ear particularly
sensitive to such a frequency? (The resonances of the ear canal are complicated by its
nonuniform shape, which we shall ignore.)
v w  (331 m/s)
T (K)
310 K
 (331 m/s)
 352.7 m/s.
273 K
273 K
v 
Next, f n  n w  for n  1,3,5... so that:
 4L 
 352.7 m/s 
  1.10 10 4 Hz  11.0 kHz
f 3  3
 4(0.0240 m) 
The ear is not particularly sensitive to this frequency, so we don’t hear overtones due
to the ear canal.
OpenStax College Physics
52.
Solution
Instructor Solutions Manual
Chapter 17
A crude approximation of voice production is to consider the breathing passages and
mouth to be a resonating tube closed at one end. (See Figure 17.30.) (a) What is the
fundamental frequency if the tube is 0.240-m long, by taking air temperature to be
37.0C ? (b) What would this frequency become if the person replaced the air with
helium? Assume the same temperature dependence for helium as for air.
T (K)
310 K
 (331 m/s)
 352.7 m/s.
273 K
273 K
v
352.7 m/s
v 
f n  n w   f1  w 
 367 Hz
4 L 4(0.240 m)
 4L 
(a) v w  (331 m/s)
(b) f1 
 v'  (367 Hz)(965 m/s)
vw
v'
, f1 '  w so that f '  f1  w  
 1.07 kHz
4L
4L
(331 m/s)
 vw 
53.
(a) Students in a physics lab are asked to find the length of an air column in a tube
closed at one end that has a fundamental frequency of 256 Hz. They hold the tube
vertically and fill it with water to the top, then lower the water while a 256-Hz tuning
fork is rung and listen for the first resonance. What is the air temperature if the
resonance occurs for a length of 0.336 m? (b) At what length will they observe the
second resonance (first overtone)?
Solution
v
v 
(a) f n  n w  for n  1,3,5.....  f1  w  256 Hz
4L
 4L 
First resonance is at the fundamental, so vw  4 Lf1  (331 m/s)
T (K)
, and
273 K
2
16 L2 f1 (273 K) (16)(0.336 m) 2 (256 Hz) 2 (273 K)
T

 295 K  22.0C
(331 m/s) 2
(331 m/s) 2
(b) The second resonance occurs at three times the first resonance:
3L  30.336 m  1.01 m
54.
What frequencies will a 1.80-m-long tube produce in the audible range at 20.0C if:
(a) The tube is closed at one end? (b) It is open at both ends?
OpenStax College Physics
Solution
(a) v w  (331 m/s)
Instructor Solutions Manual
Chapter 17
T (K)
293 K
 (331 m/s)
 342.9 m/s.
273 K
273 K
 342.9 m/s
v 
f n  n w   n
 4L 
 4(1.80 m)
20,000 Hz
nmax 
 419.9
47.63 Hz

  n(47.63 Hz) for n  1,3, ...

Since n must be an odd number, n max is 419 or 421 (depending on the
individual).
(b) f  n vw   n 342.9 m/s   n(95.25 Hz) for n  1,2, 3, ...
n
 2(1.80 m) 
 2L 


20,000 Hz
nmax 
 209.97  210
95.25 Hz
17.6 HEARING
55.
12
The factor of 10 in the range of intensities to which the ear can respond, from
threshold to that causing damage after brief exposure, is truly remarkable. If you
could measure distances over the same range with a single instrument and the
smallest distance you could measure was 1 mm, what would the largest be?
Solution
(1  10 3 m)(10 12 )  1  10 9 m  1  10 6 km
56.
The frequencies to which the ear responds vary by a factor of 103 . Suppose the
3
speedometer on your car measured speeds differing by the same factor of 10 , and
the greatest speed it reads is 90.0 mi/h. What would be the slowest nonzero speed it
could read?
Solution
90.0 mi/h
 0.0900 mi/h
103
57.
What are the closest frequencies to 500 Hz that an average person can clearly
distinguish as being different in frequency from 500 Hz? The sounds are not present
simultaneously.
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
Solution We know that we can discriminate between two sounds if their frequencies differ by
at least 0.3%, so the closest frequencies to 500 Hz that we can distinguish are
f = (500 Hz)(1 ± 0.003) = 498.5 Hz and 501.5 Hz .
58.
Solution
Can the average person tell that a 2002-Hz sound has a different frequency than a
1999-Hz sound without playing them simultaneously?
f 2  f1  2002 Hz  1999 Hz  3 Hz difference. However, the required difference is
0.3% of 2000.5 Hz or 6 Hz. Thus, an average person would not be able to distinguish
the sounds.
59.
If your radio is producing an average sound intensity level of 85 dB, what is the next
lowest sound intensity level that is clearly less intense?
Solution Although 1 dB sound level changes can be discerned, 3 dB changes are the threshold
for most people to notice. 85 dB  3 dB = 82 dB
60.
Can you tell that your roommate turned up the sound on the TV if its average sound
intensity level goes from 70 to 73 dB?
Solution
73 dB  70 dB  3 dB . Such a change in sound level is easily noticed.
61.
Based on the graph in Figure 17.36, what is the threshold of hearing in decibels for
frequencies of 60, 400, 1000, 4000, and 15,000 Hz? Note that many AC electrical
appliances produce 60 Hz, music is commonly 400 Hz, a reference frequency is 1000
Hz, your maximum sensitivity is near 4000 Hz, and many older TVs produce a 15,750
Hz whine.
Solution From Figure 17.36,
f (Hz)
Threshold (dB)
60
48
400
9
1000
0
4000
7
15,000
20
OpenStax College Physics
62.
Instructor Solutions Manual
Chapter 17
What sound intensity levels must sounds of frequencies 60, 3000, and 8000 Hz have in
order to have the same loudness as a 40-dB sound of frequency 1000 Hz (that is, to
have a loudness of 40 phons)?
Solution From Figure 17.36,
63.
f (Hz)
Sound Level (dB)
60
69
3000
37
8000
52
What is the approximate sound intensity level in decibels of a 600-Hz tone if it has a
loudness of 20 phons? If it has a loudness of 70 phons?
Solution From Figure 17.36: a 600 Hz tone at a loudness of 20 phons has a sound level of about
23 dB, while a 600 Hz tone at a loudness of 70 phons has a sound level of about 70
dB.
64.
(a) What are the loudnesses in phons of sounds having frequencies of 200, 1000,
5000, and 10,000 Hz, if they are all at the same 60.0-dB sound intensity level? (b) If
they are all at 110 dB? (c) If they are all at 20.0 dB?
Solution (a) From Figure 17.36,
f (Hz)
Loudness (phons)
200
51
1000
60
5000
58
10,000
47
(b) From Figure 17.36,
OpenStax College Physics
Instructor Solutions Manual
f (Hz)
Loudness (phons)
200
108
1000
110
5000
118
10,000
105
Chapter 17
(c) From Figure 17.36,
65.
f (Hz)
Loudness (phons)
200
4
1000
20
5000
22
10,000
10
Suppose a person has a 50-dB hearing loss at all frequencies. By how many factors of
10 will low-intensity sounds need to be amplified to seem normal to this person? Note
that smaller amplification is appropriate for more intense sounds to avoid further
hearing damage.
Solution 50 dB is five factors of 10.
66.
12
If a woman needs an amplification of 5.0  10 times the threshold intensity to
enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that
smaller amplification is appropriate for more intense sounds to avoid further damage
to her hearing from levels above 90 dB.
Solution
5.0  10 5 = five factors of 10 plus one factor of 5. In decibels, this is
50 dB  7 dB  57 dB (see Table 17.3).
67.
(a) What is the intensity in watts per meter squared of a just barely audible 200-Hz
sound? (b) What is the intensity in watts per meter squared of a barely audible 4000Hz sound?
Solution (a) At 200 Hz, 0 phons  23 dB
Using Table 17.3, 23 dB is twice (2.00) the intensity of 20 dB. From Table 17.2,
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
20 dB  1 10 10 W/m 2 . Thus, I  (1  10 10 W/m 2 )  2.00  2.00  10 10 W/m 2
(b) At 4000 Hz, 0 phons  7 dB
–7 dB is
I
1 1 

 the intensity of 0 dB. Thus,
5  5.00 
1  10 12 W/m 2
 2.00  10 13 W/m 2
5.00
68.
(a) Find the intensity in watts per meter squared of a 60.0-Hz sound having a loudness
of 60 phons. (b) Find the intensity in watts per meter squared of a 10,000-Hz sound
having a loudness of 60 phons.
Solution
(a) Using Figure 17.36, a 60.0 Hz sound at 60 phons has a sound level of
  77 dB  I  5.00 times the intensity of 70 dB.
I  5.00  (1  10 5 W/m 2 )  5.00  10 5 W/m 2
(b) Using Figure 17.36,   72 dB  I  2 times the intensity of 70 dB.
I  2  (1  10 5 W/m 2 )  2  10 5 W/m 2
69.
A person has a hearing threshold 10 dB above normal at 100 Hz and 50 dB above
normal at 4000 Hz. How much more intense must a 100-Hz tone be than a 4000-Hz
tone if they are both barely audible to this person?
Solution From Figure 17.36, the 0 phons line is normal hearing. So, this person can barely hear
a 100 Hz sound at 10 dB above normal, requiring a 47 dB sound level (  1 ). For a 4000
Hz sound, this person requires 50 dB above normal, or a 43 dB sound level (  2 ) to be
audible. So, the 100 Hz tone must be 4 dB higher than the 4000 Hz sound. To
calculate the difference in intensity, use the equation
I 
I 
1   2  10 log  1   10 log  2  and convert the difference in decibels to a ratio of
 I0 
 I0 
intensities.
 I1
 I0

I 
  10 log  1 
 I2 

I
 47 dB  43 dB  4 dB, or 1  10 4/10  2.5
I2
 1   2  10 log 

I
  10 log  2

 I0
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
So the 100 Hz tone must be 2.5 times more intense than the 4000 Hz sound to be
audible by this person.
70.
A child has a hearing loss of 60 dB near 5000 Hz, due to noise exposure, and normal
hearing elsewhere. How much more intense is a 5000-Hz tone than a 400-Hz tone if
they are both barely audible to the child?
Solution From Figure 17.36, hearing at 60 dB above normal (0 phons) at 5000 Hz requires a 55
dB sound level (  1 ). Normal hearing at 400 Hz requires a 10 dB sound level (  2 ).
 I1
 I0
1   2  55  10  45  10 log 

I
  10 log  2

 I0

I
  10 log  1
 I2


, so that

I1
 10 45/10  3.16  10 4
I2
71.
What is the ratio of intensities of two sounds of identical frequency if the first is just
barely discernible as louder to a person than the second?
Solution A 1 dB difference can barely be discerned.
 I1 
I 
I 
I
  10 log  2   10 log  1   1  101 / 10  1.26
I2
 I2 
 I0 
 I0 
1   2  1  10 log 
17.7 ULTRASOUND
72.
What is the sound intensity level in decibels of ultrasound of intensity 105 W/m 2 , used
to pulverize tissue during surgery?
Solution By Table 17.2, 160 dB corresponds to 110 4 W/m 2 , so 1105 W/m 2 has a sound level
of 170 dB.
73.
Is 155-dB ultrasound in the range of intensities used for deep heating? Calculate the
intensity of this ultrasound and compare this intensity with values quoted in the text.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 17
 I 
  I  I 0 10  / 10  (1  10 12 W/m 2 )(10155/10 )  3.16  10 3 W/m 2
I
 0
  10 log 
Yes, this is in the range 10 3 W/m 2 to 10 4 W/m 2 used for deep heating.
74.
Solution
Find the sound intensity level in decibels of 2.00 102 W/m 2 ultrasound used in
medical diagnostics.
 I 
 2.00  10 -2 W/m 2 
  10log 
  103 dB
-12
2
 10 W/m

 I0 
  10 log 
75.
The time delay between transmission and the arrival of the reflected wave of a signal
using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth
did this reflection occur?
Solution
2 x = vt = (1540 m/s) (0.13  10 -3 s) = 0.20 m; x = 0.10 m = 10 cm .
76.
In the clinical use of ultrasound, transducers are always coupled to the skin by a thin
layer of gel or oil, replacing the air that would otherwise exist between the transducer
and the skin. (a) Using the values of acoustic impedance given in Table 17.5 calculate
the intensity reflection coefficient between transducer material and air. (b) Calculate
the intensity reflection coefficient between transducer material and gel (assuming for
this problem that its acoustic impedance is identical to that of water). (c) Based on the
results of your calculations, explain why the gel is used.
Solution
Z 2  Z1 2
Use Table 17.5 for values and the equation  
Z1  Z 2 2
(a)  
30.8  10
30.8  10
30.8 10
(b)  
30.8 10
6
6

 429
 429
for  .
2
2
 1.00 , so 100% reflected.
6
 1.50 10 6
6
 1.50 10 6


2
2
 0.823
(c) Gel is used to facilitate the transmission of the ultrasound between the transducer
and the patient's body.
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
77.
(a) Calculate the minimum frequency of ultrasound that will allow you to see details
as small as 0.250 mm in human tissue. (b) What is the effective depth to which this
sound is effective as a diagnostic probe?
Solution
(a) v w  f so that f 
vw
1540 m/s

 6.16  10 6 Hz .
3
λ
0.250  10 m
(b) We know that the accepted rule of thumb is that you can effectively scan to a
depth of about 500 into tissue, so the effective scan depth is:
500  5000.250  10 3 m  0.125 m  12.5 cm
78.
(a) Find the size of the smallest detail observable in human tissue with 20.0-MHz
ultrasound. (b) Is its effective penetration depth great enough to examine the entire
eye (about 3.00 cm is needed)? (c) What is the wavelength of such ultrasound in
0C air?
Solution (a) Let v w = velocity of sound in tissue.
v w  f   
vw
1540 m/s

 7.70  10 5 m  0.0770 mm
6
f
20.0  10 Hz
(b) Effective penetration depth =
500 λ  500(0.0770  10 3 m)  3.85  10 2 m  3.85 cm. Yes, it is enough.
(c)  
79.
vw
331 m/s
. In 0C air, v w  331 m/s. Thus λ 
 16.55  10 6 m  16.6 μm
f
20.0 MHz
(a) Echo times are measured by diagnostic ultrasound scanners to determine
distances to reflecting surfaces in a patient. What is the difference in echo times for
tissues that are 3.50 and 3.60 cm beneath the surface? (This difference is the
minimum resolving time for the scanner to see details as small as 0.100 cm, or 1.00
mm. Discrimination of smaller time differences is needed to see smaller details.) (b)
Discuss whether the period T of this ultrasound must be smaller than the minimum
time resolution. If so, what is the minimum frequency of the ultrasound and is that out
of the normal range for diagnostic ultrasound?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 17
2d 2(3.60  10 2 m  3.50  10 2 m)

 1.30  10 6 s
(a) 2d  v w t  t 
vw
1540 m/s
(b) The wavelength of the ultrasound must be larger than the size of the detail, so

since v  f  , the period of the ultrasound must be smaller than the
T
difference in time found in part (a). The minimum frequency of ultrasound is
1
therefore: f   (1.30  10 6 s) 1  7.70  10 5 Hz  0.770 MHz. Since 7 MHz
T
ultrasounds are possible, this frequency is attainable.
80.
Solution
(a) How far apart are two layers of tissue that produce echoes having round-trip times
(used to measure distances) that differ by 0.750 μs ? (b) What minimum frequency
must the ultrasound have to see detail this small?
(a) 2d  v w t  d 
(b) vw  f  f 
81.
Solution
vw


v w t (1540 m/s )(0.750  10 6 s)

 5.78  10 4 m
2
2
1540 m/s
 2.67  10 6 Hz
5.78  10 4 m
(a) A bat uses ultrasound to find its way among trees. If this bat can detect echoes
1.00 ms apart, what minimum distance between objects can it detect? (b) Could this
distance explain the difficulty that bats have finding an open door when they
accidentally get into a house?
(a) Assume T  20.0C, then v w  331 m/s
2d  v w t  d 
293 K
 342.9 m/s, and therefore since
273 K
v w t (342.9 m/s )(1.00  10 3 s)

 0.171 m.
2
2
(b) No, since doors are wider than 17 cm, this does not explain why bats cannot get
out of houses since their resolution is great enough to find the opening.
OpenStax College Physics
Instructor Solutions Manual
Chapter 17
82.
A dolphin is able to tell in the dark that the ultrasound echoes received from two
sharks come from two different objects only if the sharks are separated by 3.50 m,
one being that much farther away than the other. (a) If the ultrasound has a
frequency of 100 kHz, show this ability is not limited by its wavelength. (b) If this
ability is due to the dolphin’s ability to detect the arrival times of echoes, what is the
minimum time difference the dolphin can perceive?
Solution
(a) v w  1540 m/s  fλ  λ 
vw
1540 m/s

 0.0154 m. Since   3.50 m , the
f
100  10 3 Hz
ability to resolve the two sharks is not limited by its wavelength.
(b) 2d  v w t  t 
2d
2(3.50 m)

 4.55  10 3 s  4.55 ms
vw
(1540 m/s)
83.
A diagnostic ultrasound echo is reflected from moving blood and returns with a
frequency 500 Hz higher than its original 2.00 MHz. What is the velocity of the blood?
(Assume that the frequency of 2.00 MHz is accurate to seven significant figures and
500 Hz is accurate to three significant figures.)
Solution
 v  vs 
 gives
At first, the blood is like a moving observer, so the equation f obs  f s  w
v
 w 
the frequency it receives (with the plus sign used because the blood is approaching):
 v  vb 
 (where vb is the blood velocity)
f b  f s  w
v
w


Next, this frequency is reflected from the blood, which now acts as a moving source.
 vw 
 (with the minus sign used because the blood is still
The equation f obs  f s 
v

v
x 
 w
approaching) gives the frequency received by the scanner:
 vw 
 v  vb  vw 
 v  vb 
  f s  w

  f s  w

f 'obs  f b 
v

v
v
v

v
v

v
b 
w
b 
b 
 w

 w
 w
Solving for the speed of blood gives:
 f' f 
(1540 m/s)(500 Hz)
vb  v w  obs s  
 0.192 m/s
6
6
 f ' obs  f s  (2.00  10 Hz  500 Hz)  2.00  10 Hz
OpenStax College Physics
84.
Instructor Solutions Manual
Chapter 17
Ultrasound reflected from an oncoming bloodstream that is moving at 30.0 cm/s is
mixed with the original frequency of 2.50 MHz to produce beats. What is the beat
frequency? (Assume that the frequency of 2.50 MHz is accurate to seven significant
figures.)
Solution By Example 17.8,
 v  vb
f obs  f s  w
 vw

 vw
, f ' obs  f obs 

 vw  vb



 v  vb 
1540 m/s  0.300 m/s
  (2.500000  10 6 Hz)
 f s  w
 2.500974  10 6 Hz
1540 m/s  0.300 m/s
 vw  vb 
f B  f ' obs  f s  2,500,974 Hz  2,500,000 Hz  974 Hz
Note: extra digits were retained in order to show the difference.
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