First slide

Doppler

Question

A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2 m/s one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be: (given that velocity of sound = 300 m/s; g = l0 $m / s^{2}$ )

Moderate

A

12
B

6
C

8
D

4

Solution

$f = f_{0} (\frac{v \pm v_{0}}{v \pm v_{s}})$
When approaching: $f_{a} = 150 [\frac{300 + 2}{300 - 10}]$
When receding: $f_{r} = 150 [\frac{300 - 2}{300 + 10}]$
$\Rightarrow f_{a} = f_{r} = 12$

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