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A person B accidentally slips from a lar...

A person B accidentally slips from a large building and screams with a sound of constant frequency v as he falls as shown in the figure. What will be apparent frequency of the second of the scream as heard by the person A, at the top of the building as a function of time?

A

`v'=v(v/(v-"gt"))`

B

`v'=v(v/(v+"gt"))`

C

`v'=v((2v)/(2v-"gt"^(2)))`

D

`v'=v((2v)/(2v-"gt"^(2)))`

Text Solution

Verified by Experts

The correct Answer is:
B

The person B moves away from person A
or the source o f sound moves away from A.
Hence the apparent frequency is
`v.=v_(0)(v/(v+v_(s)))v_(0)`
Where `v_(B)` is the speed of sound and `v_(B)` is the speed of the person `B` (source) since (from `v=u+at` and u=0)
`v_(B)="gt"`
we get `v.=v_(0)(v/(v+"gt"))`
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