Question 495: Waves

Question

An audio transmitter $(T)$ and a receiver $(R)$ are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the $X$ axis. They are pulled from the equilibrium position in opposite directions along the $X$ axis by a small angular amplitude $	heta_0=\cos ^{-1}(0.9)$ and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, the maximum variation in the frequency (in Hz ) as measured by the receiver (Take the acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{s}^2$ ) is _____________.

Accepted Answer

$\begin{aligned} & \cos 	heta_0=1-\frac{	heta_0^2}{2}=0.9 \ & \frac{	heta_0^2}{2}=0.1 \Rightarrow 	heta_0=10.2=\frac{1}{\sqrt{5}}\end{aligned}$
$$ \begin{aligned} & f_{\max }=\frac{v+v^{\prime}}{v-v^{\prime}} f \ & f_{\min }=\frac{v-v^{\prime}}{v+v^{\prime}} f \ & \Delta f_{\max }=f_{\max }-f_{\min }=\frac{v+v^{\prime}}{v-v^{\prime}} f-\frac{v-v^{\prime}}{v+v^{\prime}} f \ & =\frac{\left(v+v^{\prime}ight)^2-\left(v-v^{\prime}ight)^2}{v^2-v^{\prime 2}} f \ & \Delta f_{\max }=\frac{4 v v^{\prime}}{v^2-v^{\prime 2}} f ..........(i) \end{aligned} $$
Here, $\mathrm{v}^{\prime}=\ell \Omega_{\max }$
$$ \begin{aligned} & =\ell \cdot 	heta_0 \cdot \omega \quad(\omega=	ext { angular frequency }) \ & =\ell 	heta_0 \sqrt{\frac{\mathrm{~g}}{\ell}} \ & \mathrm{v}^{\prime}=	heta_0 \sqrt{\mathrm{~g} \ell} \ & \mathrm{v}^{\prime}=\frac{1}{\sqrt{5}} \sqrt{10 	imes 8} \ & \mathrm{v}^{\prime}=4 \end{aligned} $$
Put in equation (i)
$$ \begin{aligned} & \Delta f_{\max }=\frac{4 	imes 330 	imes 4 	imes 660}{330^2-4^2} \ & \approx \frac{16 	imes 330 	imes 660}{330} \approx 32 \end{aligned} $$

Waves

Integer / Numeric Answer Type