Physics 107 TUTORIAL ASSIGNMENT #7
Cutnell & Johnson, 7th edition
Chapter 16: Problems 51, 65, 79, 93
Chapter 17: Problems 7, 11, 29, 37, 48
Chapter 16
51 Suppose that sound is emitted uniformly in all directions by a public address system. The
intensity at a location 22 m away from the sound source is 3.0 x 10-4 W/m2. What is the intensity
at a spot that is 78 m away?
*65 Two identical rifles are shot at the same time, and the sound intensity level is 80.0 dB. What
would be the sound intensity level if only one rifle were shot? (Hint: The answer is not 40.0 dB.)
*79 ssm Two submarines are underwater and approaching each other head-on. Sub A has a
speed of 12 m/s and sub B has a speed of 8 m/s. Sub A sends out a 1550-Hz sonar wave that
travels at a speed of 1522 m/s. (a) What is the frequency detected by sub B? (b) Part of the sonar
wave is reflected from B and returns to A. What frequency does A detect for this reflected wave?
93 ssm A listener doubles his distance from a source that emits sound uniformly in all
directions. By how many decibels does the sound intensity level change?
Chapter 17
7 ssm The drawing shows a loudspeaker A and point C, where a listener is positioned. A second
loud-speaker B is located somewhere to the right of A. Both speakers vibrate in phase and are
playing a 68.6-Hz tone. The speed of sound is 343 m/s. What is the closest to speaker A that
speaker B can be located, so that the listener hears no sound?
11 ssm Consult Multiple-Concept Example 3 for background pertinent to this problem. A
speaker has a diameter of 0.30 m. (a) Assuming that the speed of sound is 343 m/s, find the
diffraction angle θ for a 2.0-kHz tone. (b) What speaker diameter D should be used to generate a
6.0-kHz tone whose diffraction angle is as wide as that for the 2.0-kHz tone in part (a)?
29 A string has a linear density of 8.5 × 10−3 kg/m and is under a tension of 280 N. The string is
1.8 m long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the
drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that
make up the standing wave.
37 Refer to Interactive Solution 17.37 to review a method by which this problem can be
solved. The fundamental frequencies of two air columns are the same. Column A is open at both
ends, while column B is open at only one end. The length of column A is 0.70 m. What is the
length of column B?
48 Two ultrasonic sound waves combine and form a beat frequency that is in the range of human
hearing. The frequency of one of the ultrasonic waves is 70 kHz. What is (a) the smallest
possible and (b) the largest possible value for the frequency of the other ultrasonic wave?
SOLUTIONS
Chapter 16
51. REASONING AND SOLUTION The intensity of the sound falls off according to the
square of the distance according to Equation 16.9. Therefore, the intensity at the new
location will be
2
I 78
 22 m 
 3.0  104 W/m2   2.4  10–5 W/m 2


 78 m 
65. REASONING AND SOLUTION Let 2 and I2 denote the intensity level and the intensity,
respectively, when two rifles are shot. Let 1 and I1 denote the intensity level and intensity,
respectively, when one rifle is shot. Then
 I2 
 2I 
  10 dB  log  1 
 I0 
 I0 
 2  10 dB  log 
where I1 is the intensity of a single rifle. Therefore,
2 I1
 10  2 /10 dB
I0
 I 0   2 / 10 dB  1.00 1012 W/m 2  8.00
I1   10
 
 5.00  105 W/m 2
10
2
 2


I 
 5.00 105 W/m 2 
 1  10 dB  log  1   10 dB  log 
  77.0 dB
12
I 
W/m 2 
 1.00 10
 0
79.
SSM REASONING
a. Since the two submarines are approaching each other head on, the frequency fo detected
by the observer (sub B) is related to the frequency fs emitted by the source (sub A) by
vo

1  v
f o  fs 
 1  vs
v






(16.15)
where vo and vs are the speed of the observer and source, respectively, and v is the speed of
the underwater sound
b. The sound reflected from submarine B has the same frequency that it detects, namely, fo.
Now sub B becomes the source of sound and sub A is the observer. We can still use
Equation 16.15 to find the frequency detected by sub A.
SOLUTION
a. The frequency fo detected by sub B is
vo

1  v
f o  fs 
 1  vs
v

8 m/s 


 1 + 1522 m/s 

  1570 Hz
  1550 Hz  
 1  12 m/s 


1522 m/s 


b. The sound reflected from submarine B has the same frequency that it detects, namely,
1570 Hz. Now sub B is the source of sound whose frequency is fs = 1570 Hz. The speed of
sub B is vs = 8 m/s. The frequency detected by sub A (whose speed is vo = 12 m/s) is
vo

1  v
f o  fs 
 1  vs
v

93.
12 m/s 


1+


1522 m/s 

  1590 Hz

1570
Hz

 
 1  8 m/s 


1522 m/s 


SSM REASONING According to Equation 16.10, the sound intensity level  in decibels
(dB) is related to the sound intensity I according to   10 dB log ( I / I0 ) , where the
quantity I0 is the reference intensity. Since the sound is emitted uniformly in all directions,
the intensity, or power per unit area, is given by I  P /(4 r 2 ) . Thus, the sound intensity at
position 1 can be written as I1  P /(4 r12 ) , while the sound intensity at position 2 can be
written as I 2  P /(4 r22 ) . We can obtain the sound intensity levels from Equation 16.10 for
these two positions by using these expressions for the intensities.
SOLUTION Using Equation 16.10 and the expressions for the intensities at the two
positions, we can write the difference in the sound intensity levels 21 between the two
positions as follows:
I 
I 
 21   2  1  10 dB  log  2   10 dB  log  1 
 I0 
 I0 
I /I 
I 
 10 dB  log  2 0   10 dB  log  2 
 I /I 
 I1 
 1 0
 P /(4 r22 ) 
 r12 
 r1
 21  10 dB  log 

10
dB
log

10
dB
log








2
 2

 P /(4 r1 ) 
 r2 
 r2
r
  20 dB  log  1
r
 2




2

r 
   20 dB  log  1    20 dB  log 1/ 2   –6.0 dB

 2r 

 1
The negative sign indicates that the sound intensity level decreases.
Chapter 17
7.
SSM WWW REASONING The geometry of the positions of the loudspeakers and the
listener is shown in the following drawing.
The listener at C will hear either a loud sound or no sound, depending upon whether the
interference occurring at C is constructive or destructive. If the listener hears no sound,
destructive interference occurs, so
n
d 2  d1 
n  1, 3, 5,
(1)
2
SOLUTION Since v   f , according to Equation 16.1, the wavelength of the tone is
=
v 343 m/s
=
 5.00 m
f 68.6 Hz
Speaker B will be closest to Speaker A when n  1 in Equation (1) above, so
d2 
n
5.00 m
 d1 
 1.00 m  3.50 m
2
2
From the figure above we have that,
x1  (1.00 m) cos 60.0  0.500 m
y  (1.00 m) sin 60.0  0.866 m
Then
x22  y 2  d 22  (3.50 m)2
or
x2  (3.50 m)2  (0.866 m)2  3.39 m
Therefore, the closest that speaker A can be to speaker B so that the listener hears no
sound is x1  x2  0.500 m  3.39 m  3.89 m .
11. SSM REASONING The diffraction angle for the first minimum for a circular opening is
given by Equation 17.2: sin  1.22  / D , where D is the diameter of the opening.
SOLUTION
a. Using Equation 16.1, we must first find the wavelength of the 2.0-kHz tone:

v
343 m/s

 0.17 m
f 2.0  10 3 Hz
The diffraction angle for a 2.0-kHz tone is, therefore,


  sin –1 1.22 
0.17 m 
 
0.30 m 
44
b. The wavelength of a 6.0-kHz tone is

v
343 m/s

 0.057 m
f 6.0  103 Hz
Therefore, if we wish to generate a 6.0-kHz tone whose diffraction angle is as wide as that
for the 2.0-kHz tone in part (a), we will need a speaker of diameter D, where
D
1.22  (1.22)(0.057 m)

 0.10 m
sin 
sin 44
29. REASONING A standing wave is composed of two oppositely traveling waves. The speed
F
v of these waves is given by v 
(Equation 16.2), where F is the tension in the string
m/ L
and m/L is its linear density (mass per unit length). Both F and m/L are given in the
statement of the problem. The wavelength  of the waves can be obtained by visually
inspecting the standing wave pattern. The frequency of the waves is related to the speed of
the waves and their wavelength by f = v/ (Equation 16.1).
SOLUTION
a. The speed of the waves is
v
F
280 N

 180 m/s
m/ L
8.5 103 kg/m
1.8 m
b. Two loops of any standing wave comprise one
wavelength. Since the string is 1.8 m long and consists
of three loops (see the drawing), the wavelength is

2
3
1.8 m   1.2 m



c. The frequency of the waves is
f 
v


180 m/s
 150 Hz
1.2 m
37. REASONING The fundamental frequency f1A of air column A, which is open at both
 v 
ends, is given by Equation 17.4 with n = 1: f1A  1 
 , where v is the speed of sound
 2 LA 
in air and LA is the length of the air column. Similarly, the fundamental frequency f1B of air
column B, which is open at only one end, can be expressed using Equation 17.5 with n = 1:
 v 
f1B  1 
 4 L  . These two relations will allow us to determine the length of air column B.
 B
SOLUTION Since the fundamental frequencies of the two air columns are the same
f1A  f1B , so that
 v 
 v 
  1 

 2 LA 
 4 LB 
1 
f1A
f1B
or
LB  12 LA  12  0.70 m   0.35 m
48. REASONING AND SOLUTION Two ultrasonic sound waves combine and form a beat
frequency that is in the range of human hearing. We know that the frequency range of
human hearing is from 20 Hz to 20 kHz. The frequency of one of the ultrasonic waves is
70 kHz. The beat frequency is the difference between the two sound frequencies. The
smallest possible value for the ultrasonic frequency can be found by subtracting the upper
limit of human hearing from the value of 70 kHz. The largest possible value for the
ultrasonic frequency can be determined by adding the upper limit of human hearing to the
value of 70 kHz.
a. The smallest possible frequency of the other ultrasonic wave is
f  70 kHz – 20 kHz = 50 kHz
which results in a beat frequency of 70 kHz – 50 kHz = 20 kHz .
b. The largest possible frequency for the other wave is
f  70 kHz + 20 kHz = 90 kHz
which results in a beat frequency of 90 kHz – 70 kHz = 20 kHz .