A sound source S, emitting a sound of frequency 400Hz 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 $α$ . The frequency registered by the receiver at a time t = 10s 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 $\frac{m}{k} = \frac{25}{π^{2}}$ and amplitude of oscillation $= \frac{100}{π} m, v_{sound} = 320 {ms}^{- 1}]$

Question

A sound source S, emitting a sound of frequency 400Hz 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 $α$ . The frequency registered by the receiver at a time t = 10s 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 $\frac{m}{k} = \frac{25}{π^{2}}$ and amplitude of oscillation $= \frac{100}{π} m, v_{sound} = 320 {ms}^{- 1}]$

Infinity Learn · Accepted Answer

The time period of oscillation T=2πmk=10sSo at t = 10s, the receiver passes through mean position towards the right with a speedvR=Aω=100π×π5=20ms−1Using doppler’s formulafapp=f0v−vRv+vs250=400320−20320+vs⇒vs=160ms−1So the velocity of the source is vs=160ms−1 at the time of emission of the wave which was received by the receiver at t=10s.Let’s say the value of time at the instant when the wave was emitted is T, then the distance of the sourcefrom the receiver is 12at2. This wave travels in a time 10 – T to receiver.So, for sound wave12αT2=320(10−T) andvs=αT=160After solving we get, T=8s

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