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Question

An observer moves towards a stationary source of sound with a speed 14th of the speed of sound. The wavelength and frequency of the source emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively.

A
1.25f, 1.25λ
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B
1.25f, λ
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C
f, 1.25λ
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D
0.75f, 0.75λ
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Solution

The correct option is B 1.25f, λ
Given that velocity of source,
vs = 0 (because it is stationary).

Velocity of observer vo=(1/4)v=0.25v
(where v is the velocity of sound).

Actual frequency of source is f and actual wavelength of sound emitted by the source is λ. We know from the equation of Doppler's effect that the apparent frequency recorded, when the observer is moving towards the stationary source, is given by
f=(v+vovvs)× f
=(v+0.25vv0)× f=1.25vv× f=1.25f

Since the source is stationary, therefore the apparent wavelength remains unchanged (since distance between successive wave peaks remains the same)
i.e. λ=λ

Hence the correct anser is option (b).

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