Question

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

Answer 1

Answer: (A)

9 : 8

Answer 2

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 : $\frac{n_{a}}{n_{r}} = \frac{v + v_{S}}{v - v_{S}} = \frac{340 + 20}{340 - 20} = \frac{9}{8}.$