82. A wave travels a distance of one wavelength during one oscillation period. Therefore, a time corresponding to two oscillation
periods is required for a wave to travel two wavelengths.
83.
Picture the Problem: Light is a periodic wave that travels at a constant speed, and the different colors of light have different
wavelengths.
Strategy: Recall the relationships among wavelength, frequency, and speed of a periodic wave to answer the conceptual
question.
Solution: The product of the wavelength and frequency is equal to the wave speed, v   f . Because red light has
the longer wavelength, we conclude that it also has the lower frequency, so that the product of its wavelength and
its frequency is equal to its speed as well as the speed of the blue light. Hence blue light has the greater frequency.
Insight: The speed of all electromagnetic waves in vacuum is exactly the same, 3.00×108 m/s, for all wavelengths,
from the longest radio waves, through the infrared and visible waves, and all the way down to x-rays and gamma rays.
84.
Picture the Problem: Fans at a stadium often create a transverse “wave” by standing and sitting in synchronized fashion. In this
problem we consider what motion would be required to make a longitudinal wave.
Strategy: The fans must move in a manner that simulates the motions of the masses involved in the propagation of a
longitudinal wave.
Solution: The fans would have to move closer together and spread further apart in synchronized fashion in order to
demonstrate a longitudinal wave. The horizontal oscillatory motion of each student would be parallel to the horizontal direction
of wave propagation.
Insight: Each fan oscillates about his or her original position, demonstrating that waves transport energy but they do not
transport mass (or in this case, people).
85.
Picture the Problem: In a classic TV commercial, a group of cats feed from
bowls of cat food that are lined up side by side. Initially there is one cat for each
bowl. When an additional cat is added to the scene, it runs to a bowl at the end of
the line and begins to eat. The cat that was there originally moves to the next
bowl, displacing that cat, which moves to the next bowl, and so on down the line.
Strategy: Consider the relative motions of the individual cats and the
propagation of the “wave.” If the motions are perpendicular, it is a transverse
wave, but if they are parallel, it is a longitudinal wave.
Solution: The cats move horizontally, and so does the “wave”. Because the motion of an individual cat is parallel to the
direction of wave propagation, we conclude this is a longitudinal wave.
Insight: A key difference between this motion and an actual wave is that each cat does not oscillate about an equilibrium
position, but moves to the next bowl of food. In this way mass is transported along the line of cats, and not energy, as is the case
for a wave.
86.
Picture the Problem: The three waves, A, B, and C, shown in the
figure at right propagate on strings with equal tensions and equal
mass per length.
Strategy: Use v   f and the fact that each wave propagates
with the same speed to determine the requested rankings.
Solution: 1. (a) The waves on the three strings have the same speed, v   f . The wave with the longest wavelength therefore
has the lowest frequency. The ranking for frequency is B < C < A.
2. (b) Wavelength is simply the distance from crest to crest. Therefore the ranking for wavelength is A < C < B.
Insight: If these were sound waves, the ranking for sound pitch would be the same as for frequency: B < C < A.
87.
Picture the Problem: Ripples on a pond expand outward from where you are dipping your
finger in a periodic fashion.
Strategy: Use v   f and the given frequency and wavelength to find the various values.
Solution: 1. Each time you dip your finger you create a wave. Therefore the frequency of the
wave is f  2 dips second  2 Hz
2. The period is the inverse of the frequency, T  1 2 Hz   0.50 s
3. The wavelength is the distance between adjacent crests, which is given in the problem
statement as   0.18 m
4. The speed of the waves is v  f   2 Hz 0.18 m  0.36 m/s
Insight: Dipping your finger even more frequently will not make the waves go any faster; it will
only shorten the wavelength. This is because the speed of any wave is determined by the
properties of the medium, not by wavelength or frequency.
88.
Picture the Problem: The dimensions of a wave are labeled in the
image shown at the right.
Strategy: Set the wavelength equal to the horizontal crest-to-crest
distance, or double the horizontal crest-to-trough distance. Set the
amplitude equal to the vertical crest-to-midline distance, or half the
vertical crest-to-trough distance.
Solution: 1. (a) Double the horizontal crest-to-trough distance:
  2 28 cm  56 cm
2. (b) Halve the vertical crest-to-trough distance:
A
1
2
13 cm 
6.5 cm
Insight: Note the difference between how wavelength and amplitude are measured. The wavelength is the entire distance from
crest to crest, but the amplitude is measured from the equilibrium level to the crest.
90.
Picture the Problem: The speed and wavelength of a tsunami (tidal wave) are measured.
Strategy: Use the relationship between frequency, wavelength, and speed to find the frequency.
Solution: Calculate the frequency:
f 
v


750 km/h  
1h 
4

  6.7  10 Hz
310 km 3600 s 
Insight: Although the tsunami has a very high speed, the long wavelength gives it a low frequency.
93.
Picture the Problem: Several waves are depicted in the provided figure.
The waves have different wavelengths and amplitudes.
Strategy: To determine which wave is the sum of the others, recall that
wave amplitudes add together. At some point two crests will add together
and make a wave amplitude that is larger than either of the other amplitudes.
Identify the resultant wave as the one that has an amplitude that is larger than
the others.
Solution: The blue wave has an amplitude that is greater than the amplitude
of either the red wave or the green wave, so we conclude that it is the
resultant wave.:
Insight: There is an exception to the strategy outlined above. If the red wave and the green wave had the same wavelength and
were out of phase with each other (crests overlap troughs) the resultant wave would be zero everywhere. That would be an
example in which a resultant wave does not have an amplitude that is larger than
either of the two waves that are summed. If the waves have different wavelengths the above strategy always works.
94.
Picture the Problem: Two waves (depicted with a brown color) are
added together.
Strategy: The resultant wave has an amplitude that is the sum of the
amplitudes of the two brown waves.
Solution: Where the two brown waves are both zero, the sum of their
amplitudes is also zero. For the leftmost pulses, the positive pulse from
the upper brown wave is cancelled by the negative pulse of the lower;
the resultant is zero. For the rightmost pulses, the negative pulse of the
lower brown wave half cancels the double-amplitude of the upper
brown wave. But the lower brown wave amplitude goes to zero
halfway through the upper brown wave pulse, so the remainder of the
resultant wave mimics the upper brown wave. The resultant of the two
brown waves is the lower wave drawn at the right in blue.
Insight: One of the characteristics that distinguish waves from particles is that waves pass through each other but particles
collide. When waves pass through each other their amplitudes add by the principle of wave superposition.
99.
Picture the Problem: Two wave pulses travel toward each
other at time t = 0 with a propagation speed of
1.0 m/s. We will make sketches of the wave pulses at times t =
1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s.
Strategy: Move the pulses one meter forward for each second. When the pulses overlap, add their amplitudes.
Solution: Sketches for times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s are shown below.
t = 1.0 s:
t = 2.0 s:
Add the amplitudes when the waves overlap at t = 2.5 s:
t = 3.0 s:
t = 4.0 s:
Insight: The pulses pass right through each other and continue their motion unchanged.