Q. The source of sound generating a frequency of 3kHz reaches an observer with a speed of 0.5 times, the velocity of sound in air. The frequency heard by the observer is
A) 1 kHz
B) 2 kHz
C) 4 kHz
D) 6 kHz
Explanation
(d) From Doppler’s effect in sound,
$v^{'} = v_{o}\left( \frac{v \pm v_{o}}{v \pm v_{s}} \right)$
In the given case, $v_{s} = 0.5v,\ v_{o} = 0,\ v_{o} - 3kHz$
$\therefore v^{'} = 3 \times \frac{v}{v - 0.5v} = 6kHz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.