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When a source moves away from a stationary observer, the frequency is 6/7 times the original frequency. Given: speed of sound = 330 m/s. The speed of the source is

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40 m/s

55 m/s

330 m/s

165 m/s

answer is B.

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Detailed Solution

$\begin{array}{l} f^{'} = \frac{v}{v + v_{s}} f \\ or \frac{6}{7} f = \frac{330}{330 + v} f or 6 \times 330 + 6 v = 7 \times 330 \\ or 6 v = 330 or v = 55 m / s \end{array}$

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