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Set 56 Problem number 3


Problem

What is the velocity of wave propagaion  if the wave has frequency 30 cycles/sec, and if the wavelength  is 4 meters?

Solution

A section of the wave which passes in a second contains 30 peaks. Every peak is 4 meters from the next. Therefore this 1-second section of the wave is 30 peaks ( 4 meters / peak) = 120 meters long.

The wave thus moves 120 meters in a second, and its velocity is therefore 120 meters/second.

Generalized Solution

If f peaks/sec pass a given point, and there is distance `lambda meters between peaks, then a wave  segment of length f `lambda must pass every second. This means that the velocity is f `lambda  meters/second.

In symbols, we use `lambda for wavelength, f for frequency and v for wave velocity and we have

    v = f `lambda.

Explanation in terms of Figure(s), Extension

The figure below depicts the f peaks that pass in one second when a wave has frequency f. If the peaks have uniform separation `lambda, then the distance represented is f * `lambda. This distance represents one second's travel for the wave. Therefore the velocity of the wave is v = f * `lambda.

Figure(s)

velocity_from_frequency_and_wavelength.gif (8262 bytes)

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