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Find the frequency of Sue's scream in the cave and calculate distance to the pool

John and Sue are descending on ropes into a cave. Sue's rope breaks and she falls screaming past John. Apply doppler effect.

John, hanging stationary on a rope in an underground cave, hears Sue screaming with frequency f as she falls past him with constant terminal speed S= 53 m/sec. While she is above him, he receives her scream as frequency fA, and after she passes him he receives frequency fB, which is 96 cy/sec less than fA. Exactly t= 3.6 seconds after she passes him he hears the sound of her arrival as she splashes into a pool.

The velocity of sound is 345 m/sec.
PART a. Find the frequency f, of her scream.
PART b. How far below John is the pool?

Solution Preview

The doppler effect is expressed as: (1) f' = f (v + vs) / (v - vo) in which
f' is the frequency received by the observer
f is the frequency emitted by the source
v is the velocity of sound
vs is the velocity of the source, (+ if toward, - if away from the observer)
vo is the velocity of the observer, (+ if toward, - if away from the ...

Solution Summary

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