Q15.3.
Reason: Wave speed is independent of wave amplitude, v
1
  v
3
.
Wave speed for mechanical waves depends on the properties of the medium, not the amplitude of the vibration.
Assess: If the wave speed were dependent on the amplitude then it might be the case that a later shout could overtake an earlier whisper.
Q15.5.
Reason: Equation 15.2, v string
 T s
 gives the wave speed on a stretched string with tension T s and linear mass density   / .
We will investigate how T s and  are changed in each case below and how that affects the wave speed. Use a subscript 1 for the original string, and a subscript 2 for the altered string.
We are given that ( v )  200 cm/s.
(a) ( )  T s 1

2
 
1
( v
( v

T 
2 s 1

1

T
T


2
1
 s 1

1 s 1

1
 2
( v  2( v 
(b) ( )  T s 1 m
2
 4 m
1
 
2
 4 
1
( v
( v

T 
2
T 
1

T 
2
T 
1

T 
1
T 
1

1
2
( v 
1
2
( v 
1
2
(c) ( )  T s 1
L
2
 4 L
1
 
2
 1
4

1
( v
( v

T 
2
T 
1

T 
T 
1
2 
T
T
1
4

1

1  2
(d) ( )  T s 1 m
2
 4 m
1 and
( v )  2( v )  
L
2
 4 L
1
 
2
 
1
( v
( v

T 
2
T 
1

T 
2
T 
1

T 
1
T 
1
 1
( v )  ( v )  200 cm/s
Assess: Notice as in part (d) that if both the mass and length of the string are increased by the same factor, then speed is the same (with no change in tension).
 is not changed, so the
Q15.6.
Reason: The speed of sound in a gas ( v sound
  / ) depends on the temperature of the gas. Since air is made up of gases, the speed of sound in air must also be temperature-dependent.
Assess: The temperature dependence of the speed of sound in air is small but noticeable. For example, as shown in Table 15.1 when the temperature of air goes from   the speed of sound in air increases by 2 m/s.
Q15.11
.
Reason: The relationship between wavelength and frequency is   / .
This tells us that the wavelength is inversely proportional to the frequency—the greater the frequency then, the smaller the wavelength. Since f
1
 f
2
 f
3
, we can write  
2
 
3
.
Assess: Since the speed of sound is a constant for any given situation, the wavelength depends only on the frequency.
Q15.21.
Reason: The fundamental relationship for periodic waves is given in Equation 15.10, v =  f .
We are asked for the frequency of an electromagnetic wave (which travels at a speed of 8 of a given wavelength (   400 nm).

f  v


8
 9
 14
The correct choice is D.
Assess: Note the tremendously high frequency of visible light: seven and a half hundred trillion hertz! You may not remember the exact number (which, of course, varies for different colors), but it would be fun to tuck away the tidbit that f  for visible light in your brain for later recall.
This calculation is easily done in one’s head (keep track of the significant figures). f  v




8
 9


4 10
8
 7



 15   14
Q15.22.
Reason: If the depth of penetration is 200 times the wavelength, the depth for a 5.0 MHz signal may be obtained by d  200   v f   6 
The correct choice is B.
Assess: Considering the distance from skin surface to bone, this depth would allow the signal to impact most body tissue.
Q15.23.
Reason: The fundamental relationship for periodic waves is given in Equation 15.10, v   f .
We are told that   32 cm. We’ll compute the frequency from f  1/ T   v   f  (32 cm) (5.0 Hz)  160 cm/s.
The correct choice is D.
Assess: Equation 15.9 is even more directly applicable, but the customary form (that you should remember) is Equation 15.10. The amplitude was unneeded information.
Q15.24.
Reason: As the disturbance travels from string 1 to string 2, the frequency remains the same every disturbance in string 1 creates just one disturbance in string 2 and according to the figure, the wavelength decreases. Knowing the relationship between speed of disturbance, wavelength, and frequency ( v  f  ), we can establish what happens to the speed of the disturbance as it travels from string 1 to string 2. If the frequency is constant, as the wavelength decreases, the speed at which the disturbance travels decreases. As a result the disturbance will travel slower in string 2 than in string 1. The correct choice is C.
Assess: The answer is obtained by translating the meaning of the expression v f  .
P15.81.
Prepare: Knowing that that a bat is ready to detect its echo.
Solve: v

/ and that the distance d includes out and back, we can determine the time after a pulse is emitted t

/
  x v

2(1.0 m)/343 m/s

6.0 ms
The correct choice is D.
Assess: This time will allow the bat to keep track of insects flying at standard insect speeds.
P15.82.
Prepare: “Changes in frequency” makes us think of the Doppler effect.
Solve: The Doppler effect is able to tell the bat about the speed of its prey.
The correct answer is B.
Assess: If the bat also measures the time it takes for the reflected pulse to return it could determine distance; and if the bat also measures the intensity of the reflected pulse it might get some idea of size as well. But the changes in frequency indicate speed.
P15.83.
Prepare: Some bats emit ultrasonic pulses through their nostrils into a parabolic nose reflector that concentrates the pulse in the forward direction. As a result, the reflected signal from the prey (usually in the forward direction) is much stronger than the signal from other directions. Furthermore, since the pulse is concentrated in a small area, it is more intense, which makes it easier to detect prey. If we could hear the ultrasound produced by the bat, it would be as loud as a jet engine (100 dB), so the bat does need some ear protection from the ultrasound. However this is provided by another mechanism (see Problem P15.81).
Solve: Based on the above, answers A and B are correct.
Assess: We can learn a lot from bats. They might hold the key to detecting stealth bombers.
P15.84.
Prepare: We want to invoke the fundamental relationship for sinusoidal waves, v
  f .
Solve: The speed of sound in air is determined by the properties of the air (including its temperature), but the speed is not altered by a decreasing frequency sound wave. That is, v is constant for given air conditions.
So, in v
  f , if f is decreasing, then  must be increasing.
The correct answer is C.
Assess: Always remember that for mechanical waves and sound waves the speed is determined by the properties of the medium.