Question

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 $\frac{m}{K}=\frac{25}{\pi ^{2}}$ and amplitude of oscillation $=\frac{100}{\pi } \, m$ , $v_{s o u n d}=320\,ms^{- 1}$ ]

Answer 1

The time period of oscillation $T=2\pi \sqrt{\frac{m}{K}}=10s$
So at t = 10 s, the receiver passes through mean position towards the right with a speed
$v_{R}=A\omega =\frac{100}{\pi }\times \frac{\pi }{5}=20 \, ms^{- 1}$
Using doppler’s formula
$f_{\text{app}}=f_{0}\left(\frac{v - v_{R}}{v + v_{S}}\right)$
$250=400\left(\frac{320 - 20}{320 + v_{S}}\right)\Rightarrow v_{S}=160 \, ms^{- 1}$
So the velocity of the source is $v_{S}=160ms^{- 1}$ at the time of emission of the wave which was received by the receiver at $t=10s$
Let's say the time instant when the wave was emitted is $T$ , then the distance of the source from the receiver is $\frac{1}{2} \, at^{2}$ . This distance is travelled by the wave in a time $10-T$ . So,
$\frac{1}{2}aT^{2}=v\left(10 - T\right)$
$v_{s}=aT$
$\Rightarrow \frac{1}{2}\frac{v_{s}}{T}T^{2}=v\left(10 - T\right)$
$\Rightarrow \frac{1}{2}\times 160\times T=320\times \left(10 - T\right)$
$\Rightarrow T=40-4T$
$\Rightarrow T=8s$