Q. A source of sound S is moving with a velocity of $50ms^{- 1}$ towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? The velocity of the sound in medium is $350m^{- 1}$.
A) 750 Hz
B) 857 Hz
C) 1143 Hz
D) 1333 Hz
Explanation
(a) Here, $u_{s} = 50ms^{- 1}$, $v_{L} = 0,\ v = 350ms^{- 1}$
When source is moving towards observer,
$v^{'} = 1000$
$v^{'} = \frac{u \times v}{u - u_{s}}$
$v = \frac{\left( u - u_{s} \right)v^{'}}{u}$
$= \frac{(350 - 50)1000}{350} = \frac{6000}{7}Hz$
When source is moving away from observer,
$v^{'} = \frac{u \times v}{u + v_{s}}$
$= \frac{350}{(350 + 50)} \times \frac{6000}{7}$
$= 750\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.