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## Problem

A man standing at a bus stop when two cyclists travelling in the same direction toward him are each blasting a radio with sound of a single frequency. Each cyclist is travelling at 5 m/s. As the first cyclist passes him but before the second cyclist does, the man hears a beat frequency of 5 Hz.

What is the frequency of the sound played by the cyclists?

Use the speed of sound of 343 m/s in air.

## Solution

This problem involves beat frequencies and Doppler shift. The formula for beat frequency is :${f}_{beat} = |f_1 - f_2|$.

Let the frequency from the second cyclist (as heard by the man) be $f_1$, and the frequency from the first cyclist (as heard by the man) be $f_2$.

Since the man is not moving, the beat frequency must be

${f}_{beat} = f_0 \left(\frac{v}{v - v_s} \right) - f_0 \left(\frac{v}{v + v_s} \right)$.

Solving for the original frequency from the radio,

$f_0 = {f}_{beat} \left(\frac{v^2 - {v_s}^{2}}{2v{v}_{s}} \right)$.

Thus

$f_0 = 5 \left(\frac{343^2 - 5^2}{2(343)(5)} \right)$
$f_0 = 171.46 \: Hz$.
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