Question

An observer moves towards a stationary source of sound with a speed ${\frac{1}{4}}^{th}$ of the speed of sound. The wavelength and frequency of the source emitted are $\lambda$ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively.

• A. $1.25f,1.25\lambda$
• B. $1.25f,\lambda$
• C. $f,1.25\lambda$
• D. $0.75f,0.75\lambda$

Answer 1

The correct option is B $1.25f,\lambda$

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

Velocity of observer $v_o=\left(1/4\right)v=0.25v$
(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
$f^{\prime }=\left(\frac{v+v_o}{v-v_s}\right)\times f$
$=\left(\frac{v+0.25v}{v-0}\right)\times f=\frac{1.25v}{v}\times f=1.25f$

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

Hence the correct anser is option (b).