A source of sound is moving along a circular $F$ orbit of radius 3 metres with an angular velocity o

Question

A source of sound is moving along a circular $F$ orbit of radius 3 metres with an angular velocity of $10 \mathrm{rad} / \mathrm{s}$. A sound detector located far away from the source is executing linear simple harmonic motion along the line $B D$ with an amplitude $B C=C D=6$ metres. The frequency of oscillation of the detector is 5 per second. The source is at the point A when the detector is at the point $B$. If the source emits a continuous sound wave of frequency 340 Hz ,  (Speed of sound $=340 \mathrm{~m} / \mathrm{s}$ )\

Professor PC Thomas Classes · Answer

The sound-source is moving along a circular orbit. A sound-detector is executing linear simple harmonic motion, far away from the source.\ The source is at $A$ when detector is at $B$. When the detectror is at $C$, moving towards $D$, the source is at $A_{1}$ moving leftwards. In such a situation, the frequency detected is minimum. When the detector is at $C$, moving towards $B$, the source is at $A_{3}$ moving rightwards. In this situation, the frequency detected is maximum.\ For detector, angular frequency $=\omega$\ $\omega=2 \pi f \quad 	herefore \omega=2 \pi 	imes \frac{5}{\pi}=10 \frac{\mathrm{rad}}{\mathrm{sec}}$\ $\omega$ of detector and $\omega$ of source of sound are equal.\ Velocity of source $=R \omega$\ $	herefore \quad v_{S}=(3)(10)$ or $v_{S}=30 \mathrm{~m} / \mathrm{s}$\ Velocity of observer/detector\ $=(\omega) 	imes($ amplitude $A)$\ $	herefore \quad v_{O}=10 	imes 6=60 \mathrm{~m} / \mathrm{s}$\ Since the detector/observer is located far away from the source, Doppler's formula can be applied.\ $	herefore f_{	ext {max }}=f\left(\frac{v+v_{O}}{v-v_{S}}ight)$\ or $f_{\max }=340\left(\frac{340+60}{340-30}ight)$\ or $f_{	ext {max }}=\frac{340 	imes 400}{310}$ or $f_{	ext {max }}=438.7 \mathrm{~Hz}$\ Similarly, $f_{\min }=f\left(\frac{v-v_{O}}{v+v_{S}}ight)$\ or $f_{	ext {min }}=340\left(\frac{340-60}{340+30}ight)$\ or $f_{\min }=\frac{340 	imes 280}{370}$ or $f_{\min }=257.3 \mathrm{~Hz}$.\