Physics 107 HOMEWORK ASSIGNMENT #13
Cutnell & Johnson, 7th edition
Chapter 16: Problem 76
Chapter 17: Problems 6, 8, 12, 54
*76 Multiple-Concept Example 11 presents a model for solving this type of problem. A bungee
jumper jumps from rest and screams with a frequency of 589 Hz. The air temperature is 20 °C.
What is the frequency heard by the people on the ground below when she has fallen a distance of
11.0 m? Assume that the bungee cord has not yet taken effect, so she is in free-fall.
6 Suppose that the two speakers in Figure 17-7 are separated by 2.50 m and are vibrating
exactly out of phase at a frequency of 429 Hz. The speed of sound is 343 m/s. Does the observer
at C observe constructive or destructive interference when his distance from speaker B is (a) 1.15
m and (b) 2.00 m?
*8 Refer to Interactive Solution 17.8 to review a method by which this problem can be solved.
Two loudspeakers on a concert stage are vibrating in phase. A listener is 50.5 m from the left
speaker and 26.0 m from the right one. The listener can respond to all frequencies from 20 to 20
000 Hz, and the speed of sound is 343 m/s. What are the two lowest frequencies that can be
heard loudly due to constructive interference?
12 Sound exits a diffraction horn loudspeaker through a rectangular opening like a small
doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is
273 K, the diffraction angle θ has a value of 15.0°. What is the diffraction angle for the same
sound on a summer day when the temperature is 311 K?
**54 Two loudspeakers are mounted on a merry-go-round whose radius is 9.01 m. When
stationary, the speakers both play a tone whose frequency is 100.0 Hz. As the drawing illustrates,
they are situated at opposite ends of a diameter. The speed of sound is 343.00 m/s, and the merrygo-round revolves once every 20.0 s. What is the beat frequency that is detected by the listener
when the merry-go-round is near the position shown?
76. REASONING AND SOLUTION The speed of the Bungee jumper, vs, after she has fallen a
distance y is given by
vs2  v02  2ay
Since she falls from rest, v0 = 0 m/s, and


vs  2ay  2 9.80 m/s 2  11.0 m   14.7 m/s
Then, from Equation 16.11
 1 
1


f o  fs 
 (589 Hz) 
  615 Hz .
 1  v / v 
1  14.7 m/s  /  343 m/s  
s


6.
REASONING The two speakers are vibrating exactly out of phase. This means that the
conditions for constructive and destructive interference are opposite of those that apply
when the speakers vibrate in phase, as they do in Example 1 in the text. Thus, for two wave
sources vibrating exactly out of phase, a difference in path lengths that is zero or an integer
number (1, 2, 3, …) of wavelengths leads to destructive interference; a difference in path
lengths that is a half-integer number

1
1
1
, 1 , 2 , ...
2
2
2
 of wavelengths leads to constructive
interference. First, we will determine the wavelength being produced by the speakers.
Then, we will determine the difference in path lengths between the speakers and the
observer and compare the differences to the wavelength in order to decide which type of
interference occurs.
SOLUTION According to Equation 16.1, the wavelength  is related to the speed v and
frequency f of the sound as follows:

v 343 m/s

 0.800 m
f
429 Hz
Since ABC in Figure 17.7 is a right triangle, the Pythagorean theorem applies and the
difference d in the path lengths is given by
2
2
d  d AC  d BC  d AB
 d BC
 d BC
We will now apply this expression for parts (a) and (b).
a. When d BC  1.15 m , we have
2
2
d  dAB
 d BC
 d BC 
 2.50 m 2  1.15 m 2  1.15 m  1.60 m
Since 1.60 m  2  0.800 m   2 , it follows that the interference is destructive (the
speakers vibrate out of phase).
b. When d BC  2.00 m , we have
2
2
d  dAB
 dBC
 d BC 
 2.50 m 2   2.00 m 2  2.00 m  1.20 m
   , it follows that the interference is
Since 1.20 m  1.5  0.800 m   1
1
2
constructive (the
speakers vibrate out of phase).
8.
REASONING The fact that a loud sound is heard implies constructive interference, which
occurs when the difference in path lengths is an integer number 1, 2, 3,  of
wavelengths. This difference is 50.5 m  26.0 m = 24.5 m. Therefore, constructive
interference occurs when 24.5 m = n where n = 1, 2, 3, …. The wavelength is equal to the
speed v of sound divided by the frequency f;  = v/f (Equation 16.1). Substituting this
relation for  into 24.5 m = n, and solving for the frequency gives
nv

24.5 m

This relation will allow us to find the two lowest frequencies that the listener perceives as
being loud due to constructive interference.
f 
SOLUTION The lowest frequency occurs when n = 1:
1 343 m/s   14 Hz 
nv

24.5 m
24.5 m


This frequency lies below 20 Hz, so it cannot be heard by the listener. For n = 2 and n =
3, the frequencies are 28 and 42 Hz , which are the two lowest frequencies that the listener
perceives as being loud.
f 
12. REASONING Equation 17.1 specifies the diffraction angle  according to sin    / D ,
where  is the wavelength of the sound and D is the width of the opening. The wavelength
depends on the speed and frequency of the sound. Since the frequency is the same in winter
and summer, only the speed changes with the temperature. We can account for the effect of
the temperature on the speed by assuming that the air behaves as an ideal gas, for which the
speed of sound is proportional to the square root of the Kelvin temperature.
SOLUTION Equation 17.1 indicates that
sin  

D
Into this equation, we substitute   v / f (Equation 16.1), where v is the speed of sound and
f is the frequency:
 v/ f
sin   
D
D
Assuming that air behaves as an ideal gas, we can use v   kT / m (Equation 16.5), where
 is the ratio of the specific heat capacities at constant pressure and constant volume, k is
Boltzmann’s constant, T is the Kelvin temperature, and m is the average mass of the
molecules and atoms of which the air is composed:
sin  
v
1  kT

fD fD m
Applying this result for each temperature gives
sin summer 
1  kTsummer
fD
m
and
sin  winter 
1  kTwinter
fD
m
Dividing the summer-equation by the winter-equation, we find
sin summer
sin  winter
1  kTsummer
T
fD
m

 summer
Twinter
1  kTwinter
fD
m
Thus, it follows that
sin summer  sin  winter
Tsummer
Twinter
 sin 15.0
311 K
 0.276
273 K
or
summer  sin 1  0.276   16.0
54. REASONING AND SOLUTION The speed of the speakers is
vs = 2 r/t = 2 (9.01 m)/(20.0 s) = 2.83 m/s
The sound that an observer hears coming from the right speaker is Doppler shifted to a new
frequency given by Equation 16.11 as
f OR 
fs
1–vs /v

100.0 Hz
 100.83 Hz
1–  2.83 m/s  /  343.00 m/s  
The sound that an observer hears coming from the left speaker is shifted to a new frequency
given by Equation 16.12 as
f OL 
fs
1  vs /v

100.0 Hz
 99.18 Hz
1   2.83 m/s  /  343.00 m/s  
The beat frequency heard by the observer is then
100.83 Hz  99.18 Hz = 1.7 Hz