A stationary source (see figure) emits sound waves of frequency $f$ towards a stationary wall. If an observer moving with speed $u$ in a direction perpendicular to the wall measures a frequency $f' = \frac{11}{8} f$ at the instant shown, then $u$ is related to the speed of sound $V_{s}$ as

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TestHub · Accepted Answer

This is the case of Doppler's effect in reflected sound.

The apparent frequency is given by

$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$

Given, $f^{'} = \frac{11}{8} f$

$v_{0} = u$

$v_{s} = 0$ (stationary source)

$v = V_{s}$

$∴ \frac{11}{8} f = (\frac{v_{s} + u \cos 60}{V_{s}})$

$\frac{11}{8} = 1 + \frac{u}{2} V_{s}$

$\frac{3}{8} = \frac{u}{2 V_{s}}$

$V_{s} = \frac{4}{3} u$

Physics - Sound Wave Question with Solution | TestHub

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