A source emitting sound of frequency 288 ,Hz | vedclass

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Question

(B) The radius of the circular path is $r = 100 \,cm = 1 \,m$. The angular velocity is $\omega = 20 \,rad/s$. The linear velocity of the source is $v_

Vedclass · Accepted Answer

$272 \,Hz$ to $306 \,Hz$

Vedclass · Answer

(B) The radius of the circular path is $r = 100 \,cm = 1 \,m$. The angular velocity is $\omega = 20 \,rad/s$. The linear velocity of the source is $v_s = r\omega = 1 	imes 20 = 20 \,m/s$.The observer is at a distance of $5 \,m$ from the center of the circle. Since the observer is far from the source compared to the radius of the circle, the maximum and minimum frequencies heard are determined by the Doppler effect formula: $f' = f \left( \frac{v}{v \mp v_s} ight)$.For the maximum frequency (source moving towards the observer): $f_{max} = 288 	imes \left( \frac{340}{340 - 20} ight) = 288 	imes \left( \frac{340}{320} ight) = 288 	imes 1.0625 = 306 \,Hz$.For the minimum frequency (source moving away from the observer): $f_{min} = 288 	imes \left( \frac{340}{340 + 20} ight) = 288 	imes \left( \frac{340}{360} ight) = 288 	imes 0.9444 = 272 \,Hz$.Thus, the range of frequencies heard is $272 \,Hz$ to $306 \,Hz$.

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