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The frequency changes by 10% as a sound source approaches a stationary observer with constant speed Vs. What would be percentage change in the frequency as the source recedes the observer with same speed (V_s < V) ?

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1.5 %

10.5 %

4.5 %

8.5 %

answer is B.

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Detailed Solution

$\begin{aligned} \frac{v^{'}}{v} = \frac{V}{V - V_{s}} = 1.1 \\ V_{S} = \frac{1}{11} \times V \end{aligned}$

$\begin{aligned} \frac{v^{'}}{v} = \frac{V}{V + V_{s}} = \frac{V}{V + (1 / 11) \times V} = \frac{11}{12} \\ \frac{v^{'}}{v} \times 100 = \frac{11}{12} \times 100 = 91.66 % \end{aligned}$

hence percentage change= 8.5%

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