Question

The wavelength of the sound produce by a source is 0.8 m. If the source moves towards the stationary listener at $32m{s}^{-1}$, what will be apparent wavelength of the sound?
( velocity of sound is $320m{s}^{-1}$)

• A. 0.80 m
• B. 0.72 m
• C. 0.40 m
• D. 0.32 m

Answer 1

The correct option is B 0.72 m

When a source of wavelength $\lambda$ approaches to a stationary observer with a velocity $v_s$, apparent wavelength ${\lambda }^{\prime }$ of sound is given by ,

${\lambda }^{\prime }=\lambda \left(\frac{v-v_s}{v}\right)$ ,

where $v=320m/s$ (velocity of sound in air ) ,

given $\lambda =0.8m,v_s=32m/s$

therefore ${\lambda }^{\prime }=0.8\left(\frac{320-32}{320}\right)=0.8\times 288/320$ ,

or ${\lambda }^{\prime }=0.72m$