An audio transmitter ( $T$ ) and a receiver ( $R$ ) are hung vertically from two identical massless

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\

Professor PC Thomas Classes · Answer

(32): Given, length of string, $l=8 \mathrm{~m}$ Angular amplitude, $	heta_{0}=\cos ^{-1}(0.9)$\ Natural frequency, $f_{0}=660 \mathrm{~Hz}$\ According to energy conservation, $$ \begin{aligned} & \frac{1}{2} m v^{2}=m g l(1-\cos 	heta) \ & v=\sqrt{2 g l(1-\cos 	heta)}=\sqrt{2 	imes 10 	imes 8 	imes(1-0.9)} \ & =4 \mathrm{~m} / \mathrm{sec} \end{aligned} $$ Due to motion of transmitter and receiver natural frequency of receiver will change.\ Maximum frequency received by receiver,\ $f_{\max }=f_{0}\left(\frac{v_{s}+v}{v_{s}-v}ight)=660\left(\frac{330+4}{330-4}ight)=676.196 \mathrm{~Hz}$\ Minimum frequency,\ $f_{\min }=f_{0}\left(\frac{v_{s}-v}{v_{s}+v}ight)=660\left(\frac{330-4}{330+4}ight)=644.191 \mathrm{~Hz}$\ Maximum variation, $\Delta f$\ $\Delta f=f_{	ext {max }}-f_{	ext {min }}=676.196-645.191 \approx 32 \mathrm{~Hz}$