Question

A source of sound is moving with constant velocity of 20 $m/s$ emitting a note of frequency $1000Hz$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (Velocity of sound $\left(v\right)=340m/s)$

• A. 9 : 8
• B. 8 : 9
• C. 1 : 1
• D. 9 : 10

Answer 1

The correct option is A 9 : 8

When car is approaching towards observer, then frequency observed by stationary observer will be given by
$f^{\prime }=\left(\frac{v}{v-v_s}\right)f.....\left(i\right)$
When car is approaching away from observer, then frequency observed by stationary observer will be given by
$f^{\prime \prime }=\frac{v}{v+v_s}f.....\left(ii\right)$
On comparing (1) and (2) equation
$\frac{f^{\prime }}{f^{\prime \prime }}=\frac{v+v_s}{v-v_s}\frac{f^{\prime }}{f^{\prime \prime }}=\frac{340+20}{340-20}=\frac{360}{320}=\frac{9}{8}$