Physics - Sound Wave Question with Solution | TestHub

The time period of oscillation $T=2\pi \sqrt{\frac{m}{K}}=10 \mathrm{s}$

So at t = 10 s, the receiver passes through mean position towards the right with a speed

$v_R=A\omega =\frac{100}{\pi }\times \frac{\pi }{5}=20 \mathrm{m} {\mathrm{s}}^{-1}$

Using doppler’s formula

$f_{\text{app}}=f_0\left(\frac{v-v_R}{v+v_S}\right)$

$250=400\hspace{0.33em}\left(\frac{320-20}{320+v_S}\right)\hspace{0.33em}\Rightarrow \hspace{0.33em}{v}_S=160 \mathrm{m} {\mathrm{s}}^{-1}$

So the velocity of the source is $v_S=160 \mathrm{m} {\mathrm{s}}^{-1}$ at the time of emission of the wave which was received by the receiver at $t=10 \mathrm{s}$

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,

$\frac{1}{2}a{T}^2=v\left(10-T\right)$

$v_s=aT$

$\Rightarrow \frac{1}{2}\frac{v_s}{T}{T}^2=v\left(10-T\right)$

$\Rightarrow \frac{1}{2}\times 160\times T=320\times \left(10-T\right)$

$\Rightarrow T=40-4T$

$\Rightarrow T=8 \mathrm{s}$