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Set 56 Problem number 4


Problem

If a wave with 5 meters between peaks passes at the velocity of 50 meters/second, then how many peaks pass in a second?

Solution

Visualize a 1-second section of the wave. It is 50 meters long, with peaks 5 meters apart. This 1-second section therefore contains 50 meters / ( 5 meters / peak) = 10 peaks.

In 1 second, therefore, these 10 peaks will pass.

We call the quantity 10 peaks / second the frequency of the wave.

Frequency is measured in peaks/second, or, since a peak occurs once per cycle of the wave, in cycles/second. The unit cycles/second is also called the 'Hertz', abbreviated Hz (for a guy named Hertz who had a lot to do with developing wave theories).

Generalized Solution

If a wave passes at velocity v with wavelength `lambda, then in time interval `dt, a wave segment of  length v `dt will pass.

With distance `lambda between peaks, the number of peaks passing in this time  interval is n = v `dt / `lambda.

The number of peaks per unit of time is thus

Explanation in terms of Figure(s), Extension

The figure below shows the distance moved in time `dt by a wave with velocity v, and the wavelength `lambda, or distance between peaks.   We see that the number of peaks in time `dt is equal to the distance traveled in this time, divided by the distance between consecutive peaks, or v `dt / `lambda.   The frequency is the number of peaks divided by the time `dt, or [ v `dt / lambda ] / `dt = v / `lambda.

Figure(s)

frequency_from_velocity_and_wavelength.gif (9488 bytes)

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