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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 emited are `lamda` and f, respectively. The apparent frequency and wavelength recorded by the observer are, respectively-
(a) `f,1.2 lambda` (b) `0.8f,0.8lambda` (c) `1.2f,1.2lambda` (d) `1.2f,lambda`

A

`1.2f` and `lamda`

B

`f` and `1.2lamda`

C

`0.8` and `0.8lamda`

D

`1.2f` and `1.2lamda`

Text Solution

Verified by Experts

The correct Answer is:
A

Given that velocity of source `v_s=0` (because it is stationary). Velocity of observer `v_0=((1)/(5))=0.2v` (where v is the velocity of sound.) Actual frequency of source is f and actual wavelength of source is `lamda`. 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+v_0)/(v-v_S))xxn`
`=((v+0.2v)/(v-0))xxn=(1.2v)/(v)xxn=1.2n=1.2f`
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