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=10 m / s 2 )

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=10 m / s 2 ) (A) 12 (B) 6 (C) 8 (D) 4

Tardigrade Admin · Accepted Answer

f=f0((v ± v0/v ± vs)) When approaching: fa=150[(300+2/300-10)] When receding: fr=150[(300-2/300+10)] ⇒ fa=fr=12

12

6

8

4

f=f0​(v±vs​v±v0​​) When approaching: fa​=150[300−10300+2​] When receding: fr​=150[300+10300−2​] ⇒fa​=fr​=12

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