An observer moves towards a stationary source of sound of frequency n. The apparent frequency heard by him is 2n. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is [MP PET 1990]

166 m/sec

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664 m/sec

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332 m/sec

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1328 m/sec

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Solution

The correct option is D 332 m/sec
$n^{'} = n (\frac{v + v_{0}}{v}) \Rightarrow= 2 n = n (\frac{v + v_{0}}{v}) \Rightarrow \frac{v + v_{0}}{v} = 2$
$\Rightarrow v_{0} = v = 332$ m/sec

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An observer moves towards a stationary source of sound of frequency n. The apparent frequency heard by him is 2n. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is [MP PET 1990]

The correct option is D 332 m/secn′=n(v+v0v)⇒=2n=n(v+v0v)⇒v+v0v=2⇒v0=v=332 m/sec

The correct option is D 332 m/sec
$n^{'} = n (\frac{v + v_{0}}{v}) \Rightarrow= 2 n = n (\frac{v + v_{0}}{v}) \Rightarrow \frac{v + v_{0}}{v} = 2$
$\Rightarrow v_{0} = v = 332$ m/sec