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A sound source S, emitting a sound of fr...

A sound source S, emitting a sound of frequency 400 Hz and a receiver R of mass m are at the same point. R is performing SHM with the help of a spring of force constant K. At a time t = 0, R is at the mean position and moving towards the right, as shown. At the same time, the source starts moving away from R with some acceleration a. The frequency registered by the receiver at a time t = 10 s is 250 Hz. What is the time (in seconds) at which the corresponding registered frequency of 250 Hz was emitted by the source? [Given that `(m)/(K)=(25)/(pi^(2))` and amplitude of oscillation = `(100)/(pi) m, V_("sound")` = 320 m s^(-1)]`

Text Solution

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The correct Answer is:
8
The time period of oscillation
`T=2pi sqrt((m)/(K))=10 s`
So at t= 10 s, the receiver passes through mean position towards the right with a speed
`v_(R)=A omega=(100)/(pi xx(pi)/(5)=20 ms^(-1)`
`f_("app")=f_(0) ((v-v_(R))/(v+v_(s)))`
`250=400 ((320-20)/(320+v_(s))) rArr v_(S)=160 m`
So the velocity of the source is `v_(S)=160 ms^(-1)` at the time of emission of the wave which was received by the receiver att = 10 s
Let.s say the time instant when the wave was emitted is `(1)/(2)at^(2)` then the distance of the scource from the receiver is at. This distance is travelled by the wave in a time 10-T. So,
`(1)/(2)aT^(2)=v(10-T)`
`v_(s)=aT`
`rArr (1)/(2)(v_(s))/(T)T^(2)=v(10-T)`
`rArr (1)/(2)xx160xxT=320xx(10-T)`
`rArr T=40-4T`
`rArr T=8s`
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