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,

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given that velocity of source v_S = 0(because it is stationary). Velocity of observer $v_0=\left(\frac{1}{5}\right)v=0.2v$ (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
$n^{\prime }=\left(\frac{v+v_0}{v-v_s}\right)\times n=\left(\frac{v+0.2v}{v-0}\right)\times n=\frac{1.2v}{v}\times n=1.2n=1.2f$
Since the source is stationary, therefore the apparent wavelength remains unchanged, i.e., $\lambda$