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)
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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 )
Answered by
31
Answer:
When approaching : fa =
When receding: fr =
Hz
Thanks !!
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