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 $a$ . The frequency registered by the receiver at a time $t=10s$ is $250Hz$ . What is the time (in seconds) at which the corresponding registered frequency of $250Hz$ was emitted by the source?
[Given that $\frac{m}{K}=\frac{25}{{\pi }^2}$ and amplitude of oscillation $=\frac{100}{\pi }m$ , $v_{sound}=320m{s}^{-1}$ ]

Answer 1

Answer: 8

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
$v_R=A\omega =\frac{100}{\pi }\times \frac{\pi }{5}=20m{s}^{-1}$
Using doppler’s formula
$f_{\text{app}}=f_0\left(\frac{v-v_R}{v+v_S}\right)$
$250=400\left(\frac{320-20}{320+v_S}\right)\Rightarrow v_S=160m{s}^{-1}$
So the velocity of the source is $v_S=160m{s}^{-1}$ 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 $\frac{1}{2}a{t}^2$ . This distance is travelled by the wave in a time $10-T$ . So,
$\frac{1}{2}a{T}^2=v\left(10-T\right)$
$v_s=aT$
$\Rightarrow \frac{1}{2}\frac{v_s}{T}{T}^2=v\left(10-T\right)$
$\Rightarrow \frac{1}{2}\times 160\times T=320\times \left(10-T\right)$
$\Rightarrow T=40-4T$
$\Rightarrow T=8s$