Question

A source of sound of frequency 640 Hz is moving at a velocity of $\frac{100}{3}$ m/s along a road, and is at an instant $30m$ away from a point A on the road (as shown in figure). A person standing at O, $40m$ away from the road hears sound of apparent frequency ${\nu }^{\prime }$. The value of ${\nu }^{\prime }$ is:
(velocity of sound = $340m/s$)

• A. $620Hz$
• B. $680Hz$
• C. $720Hz$
• D. $840Hz$

Answer 1

The correct option is B $680Hz$

This is the case of doppler effect with source approaching the observer. The velocity of approach of source towards observer is = $vcos\theta$ = $\frac{100}{3}cos\theta =\frac{100}{3}\times \frac{3}{5}=20m/s$.

Hence ${\nu }^{\prime }=\nu \left(\frac{v_a}{v_a-v_s}\right)=640\left(\frac{340}{340-20}\right)=680Hz$