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 sound emitted are $\lambda$ and $f$ respectively. The apparent frequency and wavelength recorded by the observer are, respectively:

• A. $1.2f$ and $\lambda$
• B. $f$ and $1.2\lambda$
• C. $0.8f$ and $0.8\lambda$
• D. $1.2f$ and $1.2\lambda$

Answer 1

The correct option is A $1.2f$ and $\lambda$

Given that velocity of source,
$v_s=0$ (because it is stationary).

Velocity of observer $v_o=\left(1/5\right)v=0.2v$
(where $v$ is the velocity of sound).

Actual frequency of source is $f$ and actual wavelength of sound emitted by the source is $\lambda$. 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
$f^{\prime }=\left(\frac{v+v_o}{v-v_s}\right)\times f$
$=\left(\frac{v+0.2v}{v-0}\right)\times f=\frac{1.2v}{v}\times f=1.2f$

Since the source is stationary, therefore the apparent wavelength remains unchanged (since distance between successive wave peaks remains the same)
i.e. ${\lambda }^{\prime }=\lambda$
Option (a) is the correct answer.