An engine moving towards a wall with a velocity 50 m/s emits a note of 1.2 kHz. Speed of sound in air is 350 m/s. The frequency of the note after reflection from the wall as heard by the driver of the engine is :

2.4 kHz

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0.24 kHz

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1.6 kHz

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1.2 kHz

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Solution

The correct option is B 1.6 kHz
From Doppler's Effect:
$f_{h e a r d} = f_{e m i t t e d} (\frac{v_{o} + v_{r e c e i v e r}}{v_{o} - v_{e m i t t e r}})$ .

The direction of sound is taken as positive and opposite is taken to be negative.

Initially let there be a stationary receiver at the wall:
$f_{1} = f_{0} (\frac{v_{o}}{v_{o} - 50}) H z$

After reflection let there be stationary source at the wall:
$f_{2} = f_{1} (\frac{v_{o} + 50}{v_{0}}) = f_{0} (\frac{v_{o} + 50}{v_{o} - 50})$

On solving, final sound frequency $= 1.6 k H z$

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