First slide

Doppler's effect

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)

Moderate

A

9 : 8
B

8 : 9
C

1 : 1
D

9 : 10

Solution

When source is approaching the observer, the frequency heard
$n_{a} = (\frac{v}{v - v_{S}}) \times n = (\frac{340}{340 - 20}) \times 1000 = 1063 Hz$
When source is receding, the frequency heard $n_{r} = (\frac{v}{v + v_{S}}) \times n = \frac{340}{340 + 20} \times 1000 = 944$
$\Rightarrow n_{a} : n_{r} = 9 : 8$
Short tricks : $\frac{n_{a}}{n_{r}} = \frac{v + v_{S}}{v - v_{S}} = \frac{340 + 20}{340 - 20} = \frac{9}{8} .$

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