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 2m/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 = 300m/s; g = 10m/s2)

Answer 1

after 1 sec, falling point source (with frequency 150 Hz) will attain velocity of

Vs = 10 ms⁻¹ , and observer sitting on the balloon is coming upward with the velocity

V₀ = 2 ms⁻¹, say frequencies observed

by observer before and after crossing are n₁ and n₂

respectively, then

n₁ = ( 300 + 2 / 300 - 10 ) * 150 = 156.20 Hz

n₂ = ( 300 - 2 /  300 + 10 ) * 150 = 144.19 Hz

n₁ - n₂ = 12 Hz ( approx )

Answer 2

Answer:

$f = fo( \frac{v \frac{ + }{ - } vo }{v \frac{ + }{ - } vs} )$

When approaching : fa =

$150( \frac{300 + 2}{300 - 10} )$

When receding: fr =

$150( \frac{300 - 2}{300 + 10} )$

$fa \: - fr \: = approx.12 \\ \\ so \: the \: answer \: is \: approx. \: 12$

Thanks !!