Question
An observer moves towards a stationary source of sound with a speed (1/5)th 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,
Answer: Option A
:
A
Given that velocity of source vS = 0(because it is stationary). Velocity of observer v0=(15)v=0.2v (where v is the velocity of sound). Actual frequency of source is f and actual wavelength of source is λ. We know from the Doppler’s effect that the apparent frequency recorded, when the observer is moving towards the stationary source, is given by
n′=(v+v0v−vs)×n=(v+0.2vv−0)×n=1.2vv×n=1.2n=1.2f
Since the source is stationary, therefore the apparent wavelength remains unchanged, i.e., λ
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:
A
Given that velocity of source vS = 0(because it is stationary). Velocity of observer v0=(15)v=0.2v (where v is the velocity of sound). Actual frequency of source is f and actual wavelength of source is λ. We know from the Doppler’s effect that the apparent frequency recorded, when the observer is moving towards the stationary source, is given by
n′=(v+v0v−vs)×n=(v+0.2vv−0)×n=1.2vv×n=1.2n=1.2f
Since the source is stationary, therefore the apparent wavelength remains unchanged, i.e., λ
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