Question

A source of sound is traveling at $\frac{100}{3}m{s}^{-1}$ along a road, towards a point A. When the source is 3 m away from A, a person standing at a point O on a road perpendicular to the track hears a sound of frequency v'. The distance of O from A at that time is 4 m . If the original frequency is 640 Hz, then the value of v' is :

(Given velocity of sound = $340m{s}^{-1}$)

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

Answer 1

The correct option is B 680 Hz

Given : Original frequency of the source $\nu =640$ Hz

Velocity of source towards A $v_s=\frac{100}{3}$ $m{s}^{-1}$

In right triangle $OAS$, $O{S}^2=3^2+4^2=25$

$⟹$ $OS=5$ m

There is no Doppler effect in perpendicular direction.

Velocity of source in parallel direction i.e $\overrightarrow{SO}$, $v_s^{\prime }=\frac{100}{3}cos\theta$

$\therefore$ $v_s^{\prime }=\frac{100}{3}\times \frac{3}{5}=20$ $m{s}^{-1}$

Let the apparent frequency heard by the observer be ${\nu }^{\prime }$.

Using Doppler effect when source moves towards the stationary observer :

${\nu }^{\prime }=\nu [\frac{v_{sound}}{v_{sound}-v_s^{\prime }}]$ $=640[\frac{340}{340-20}]=680$ Hz