An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. The percentage increase in the apparent frequency is

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20%

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zero

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0.5%

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Solution

The correct option is C 20%
Apparent frequent, $f^{'} = (\frac{v + v_{o}}{v}) f$ $= \frac{(v + \frac{v}{5})}{v} f$ $= \frac{6 f}{5}$
Percentage increase $= \frac{f^{1} - f}{f} \times 100$ $= \frac{1}{5} \times 100 = 20$ %

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An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. The percentage increase in the apparent frequency is

The correct option is C 20%Apparent frequent, f′=(v+vov)f =(v+v5)vf =6f5Percentage increase =f1−ff×100 =15×100=20 %

The correct option is C 20%
Apparent frequent, $f^{'} = (\frac{v + v_{o}}{v}) f$ $= \frac{(v + \frac{v}{5})}{v} f$ $= \frac{6 f}{5}$
Percentage increase $= \frac{f^{1} - f}{f} \times 100$ $= \frac{1}{5} \times 100 = 20$ %