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(C) The source is moving in a circular path. The linear speed of the source is given by $v_S = r\omega$.Given $r = 70 	ext{ cm} = 0.70 	ext{ m}$ and

Vedclass · Accepted Answer

$941$

Vedclass · Answer

(C) The source is moving in a circular path. The linear speed of the source is given by $v_S = r\omega$.Given $r = 70 	ext{ cm} = 0.70 	ext{ m}$ and frequency of revolution $f_{rev} = 5 	ext{ rev/s}$.Angular velocity $\omega = 2\pi f_{rev} = 2 	imes \pi 	imes 5 = 10\pi 	ext{ rad/s}$.Thus,$v_S = 0.70 	imes 10\pi = 7\pi \approx 7 	imes 3.1416 = 21.99 \approx 22 	ext{ m/s}$.The minimum frequency is heard when the source is moving directly away from the listener.Using the Doppler effect formula for a receding source: $n_{\min} = n \left( \frac{v}{v + v_S} ight)$.Substituting the values: $n_{\min} = 1000 	imes \left( \frac{352}{352 + 22} ight) = 1000 	imes \left( \frac{352}{374} ight) \approx 941.17 	ext{ Hz}$.Rounding to the nearest integer,the minimum frequency is $941 	ext{ Hz}$.

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