Q. A source of sound is moving with constant velocity of $20\ m/s$ emitting a note of frequency $1000\ Hz.$ The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (Speed of sound $v = 340\ m/s$)
A) 9 : 8
B) 8 : 9
C) 1 : 1
D) 9 : 10
Explanation
(a) When source is approaching the observer, the frequency heard
$n_{a} = \left( \frac{v}{v - v_{S}} \right) \times n = \left( \frac{340}{340 - 20} \right) \times 1000 = 1063Hz$
When source is receding, the frequency heard
$n_{r} = \left( \frac{v}{v + v_{S}} \right) \times n = \frac{340}{340 + 20} \times 1000 = 944$
$\Rightarrow n_{a}:n_{r} = 9\ :8$
Short tricks : $\frac{n_{a}}{n_{r}} = \frac{v + v_{S}}{v - v_{S}} = \frac{340 + 20}{340 - 20} = \frac{9}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.