Q15.6. Reason: The speed of sound in a gas (vsound   RT /M ) 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 0C to 20 C, the speed of sound in air increases by 2 m/s.
Q15.9. Reason: Please refer to Figure Q15.9. Since the wave is traveling to the left, the snapshot has the same shape as the history graph.
To understand why, consider that the history graph tells us about the displacement at one point in space. As the wave moves to the left, that
point witnesses spots on the wave further to the right. Thus increasing t by t on the history graph has the same effect as increasing x by
v t on the snapshot graph, where v is the speed of the wave. Consequently the history graph and snapshot graph have the same shape. The
only difference is that in going from the history graph to the snapshot graph, the horizontal axis is scaled by a factor of v.
Assess: A similar argument can be used to show that if a wave is traveling to the right, the snapshot graph can be obtained from the history
graph by reflecting the graph about the y-axis and scaling the horizontal axis by a factor of v.
Q15.20. Reason: Knowing the frequency of the electromagnetic waves traveling from the station, we can determine their wavelength by
  c / f  (3.0  108 m/s)/(1.05  108 Hz)  2.85 m.
The station would be named KOOL 2.85. The correct choice is B.
Assess: The number obtained is in the radio wave range.
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 3.00  108 m/s) of a given wavelength (  400 nm).
f 
v


3.00  108 m/s
 7.50  1014 Hz
400  109 m
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  1014 Hz for visible light in your
brain for later recall.
This calculation is easily done in one’s head (keep track of the significant figures).
v 3.00  108 m/s 3  108 
f  
  7  Hz  0.75  1015 Hz  7.50  1014 Hz

400  10 9 m
4  10 
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  200(v / f )  200(343 m/s)/5  10 6 Hz  1.4 cm.
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.
Q15.24. 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 .
P15.1. Prepare: The wave is a traveling wave on a stretched string. We will use Equation 15.2 to find the wave speed.
Solve: The wave speed on a stretched string with linear density  is vstring  TS /  . The wave speed if the tension is doubled will be
 
vstring
2TS

 2 vstring  2 (200 m/s)  280 m/s