An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

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0 %

20 %

5 %

10 %

answer is A.

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Detailed Solution

$f_{0} = (\frac{v + v_{0}}{v}) f_{s} f_{0} = (\frac{v + \frac{v}{5}}{v}) f_{s} f_{0} = \frac{6}{5} f_{s} % change = \frac{f_{0} - f_{s}}{f_{s}} \times 100 = \frac{1}{5} \times 100 = 20 %$

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