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Set 56 Problem number 1


Problem

If a wave has the characteristic that 40 equally spaced peaks pass a given point every second, with the wave traveling at 160 meters/second, then what is the distance between the peaks?

Solution

If 40 peaks pass in a second, then the section of the wave that passes in a second must contain 40 peaks.

Since the wave travels at 160 meters/second, the section of the wave that passes in a second must be 160 meters long.

Therefore there are 40 peaks in 160 meters.

The distance between peaks must therefore be

We call this distance between peaks the wavelength1 of the wave.

Generalized Solution

If f peaks per second pass at velocity v, then in time interval `dt, the number of peaks will be f `dt and the distance moved will be v `dt.

If there are f `dt peaks spread over distance v `dt, then the distance between peaks is v `dt / f `dt = v / f.

Explanation in terms of Figure(s), Extension

The figure below depicts the segment of the wave that will pass in time interval `dt. This segment has length v `dt. Since its frequency is f, which represents the number of peaks passing per unit of time, we also seen that the number of peaks is f `dt.

The distance between peaks is easily found by dividing the number of peaks into the distance over which they are spread. In this case the distance is

This distance is called the wavelength of the wave.

Figure(s)

wavelength_from_frequency_and_velocity.gif (8856 bytes)

 

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