39.
Picture the Problem: A crowd produces a “wave” at a sports stadium.
Strategy: Determine whether the wave is transverse or longitudinal by examining the relative directions of the oscillations and
of the wave propagation.
Solution: The motion of each fan in the crowd is vertical, perpendicular to the horizontal direction of wave propagation. We
conclude that the wave is transverse.
Insight: A longitudinal wave could propagate through the crowd, but it would cause many fans to bump into each other and
likely result in the spilling of many refreshments!
40.
Picture the Problem: A wave propagates through a chocolate candy bar and a chocolate cake.
Strategy: Note that the wave propagation speed depends upon the restoring forces that allow the masses in the medium to
oscillate about their equilibrium positions. Use this fact to predict the relative wave speeds.
Solution: The cake is a softer medium than a solid chocolate bar. A softer wave medium implies a weaker restoring force and a
smaller acceleration of the masses, which results in a longer oscillation period and a slower wave propa-gation speed. We
therefore expect the speed of sound to be greater in a chocolate candy bar than in a chocolate cake.
Insight: The speed of sound in steel (6100 m/s) is much faster than the speed of sound in hard wood (3960 m/s)
because steel is a much stiffer medium than wood. Likewise, each of these speeds is much greater than the speed of sound in air
(340 m/s).
42.
Picture the Problem: The frequency and speed of a wave are measured.
Strategy: Use the definition of frequency, period, and speed of a wave to answer the questions.
Solution: 1. (a) The frequency is the number of wave
cycles per second, which is also the number of oscillations
per second for individual masses:
f 
5.0 oscillations 5.0 wave cycles

 5.0 Hz
1 second
1 second
2. (b) The period is the inverse of the frequency:
T
1
1 second

 0.20 s
f 5.0 cycles

   vT  6.0 m s 0.20 s   1.2 m
T
Insight: A faster wave speed would correspond to a longer wavelength because the wave must travel one wavelength during one
oscillation period, and in this example the oscillation period is fixed.
3. (c) Find the wavelength from the speed and period:
43.
v
Picture the Problem: The frequency and wavelength of waves that pass a fishing boat are measured.
Strategy: Use the relationship between wave speed, frequency, and wavelength to answer the question.
Solution: 1. The frequency is the number of wave cycles
per second, which is also the number of oscillations per
second for individual masses:
f 
12 oscillations 12 wave cycles

 0.2667 Hz
45 seconds
45 seconds

  f  7.5 m0.2667 Hz   2.0 m s
T
Insight: If the wave crests were closer together but the frequency was the same, we would conclude that the wavelength was
shorter than 7.5 m and that the wave speed was slower than 2.0 m/s.
2. The speed of the wave is one wavelength per period:
44.
v
Picture the Problem: The speed and wavelength of a passing periodic wave are measured.
Strategy: Use the relationship among wave speed, frequency, and wavelength to answer the question.
Solution: 1. The wavelength is the distance between
adjacent crests, so the distance between a crest and an
adjacent trough is only half a wavelength:
2. The frequency of the wave is related to
wavelength and speed:
1
2
  2.4 m    4.8 m
vf
 f 
v


5.6 m/s
 1.2 Hz
4.8 m
Insight: In the laboratory it is easier to measure the period of a wave with a stopwatch than it is to measure the distance between
adjacent crests because the wave is continuously moving.