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(D) The position of the source is given by $x = 3 \sin \omega t$.The instantaneous velocity of the source is $v_s = \frac{dx}{dt} = 3 \omega \cos \ome

Vedclass · Accepted Answer

$10 \,rad \,s^{-1}$

Vedclass · Answer

(D) The position of the source is given by $x = 3 \sin \omega t$.The instantaneous velocity of the source is $v_s = \frac{dx}{dt} = 3 \omega \cos \omega t$.The maximum velocity of the source is $v_{s, \max} = 3 \omega$.According to the Doppler effect,the observed frequency $f'$ is given by $f' = f \left( \frac{v}{v \mp v_s} ight)$,where $v = 330 \,ms^{-1}$ is the speed of sound.The maximum frequency is $f_{\max} = f \left( \frac{v}{v - v_{s, \max}} ight)$ and the minimum frequency is $f_{\min} = f \left( \frac{v}{v + v_{s, \max}} ight)$.Given $f_{\max} - f_{\min} = 22 \,Hz$,we have:$120 \left( \frac{330}{330 - 3 \omega} - \frac{330}{330 + 3 \omega} ight) = 22$$120 	imes 330 \left( \frac{(330 + 3 \omega) - (330 - 3 \omega)}{330^2 - (3 \omega)^2} ight) = 22$$120 	imes 330 \left( \frac{6 \omega}{108900 - 9 \omega^2} ight) = 22$Since $v_s \ll v$,we can approximate $330^2 - (3 \omega)^2 \approx 330^2$:$120 	imes 330 	imes \frac{6 \omega}{330 	imes 330} = 22$$120 	imes \frac{6 \omega}{330} = 22$$120 	imes \frac{2 \omega}{110} = 22$$240 \omega = 2420 \Rightarrow \omega \approx 10 \,rad \,s^{-1}$.

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