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 `lamda` and f, respectively. The apparent frequency and wavelength recorded by the observer are, respectively.

Options

A. `f`, `1.2lambda`
B. `0.8f,0.8lambda`
C. `1.2f,1.2lambda`
D. `1.2f, lambda`

Answer

D

Explanation

`f.=f(v+v_(0))/(v)` Given `v_(0)=(v)/(5)` Substituting in the relation for `f.` , we have
`f=(v+v//5)/(v)f=(6)/(5)f=1.2f`
Motion of observer does not affect the wavelength reaching the observer , hence wavelength remains `lambda`.