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(D) Given: Angular speed $\omega = 40 \ rad \ s^{-1}$, radius $r = 50 \ cm = 0.5 \ m$. The linear speed of the source is $v_s = r \omega = 0.5 	imes

Vedclass · Accepted Answer

$9: 8$

Vedclass · Answer

(D) Given: Angular speed $\omega = 40 \ rad \ s^{-1}$, radius $r = 50 \ cm = 0.5 \ m$. The linear speed of the source is $v_s = r \omega = 0.5 	imes 40 = 20 \ m \ s^{-1}$. The Doppler effect formula for maximum frequency (when the source moves towards the observer) is $n_{\max} = n \left( \frac{v}{v - v_s} ight)$. The Doppler effect formula for minimum frequency (when the source moves away from the observer) is $n_{\min} = n \left( \frac{v}{v + v_s} ight)$. The ratio of maximum to minimum frequency is $\frac{n_{\max}}{n_{\min}} = \frac{v + v_s}{v - v_s}$. Substituting the values $v = 340 \ m \ s^{-1}$ and $v_s = 20 \ m \ s^{-1}$: $\frac{n_{\max}}{n_{\min}} = \frac{340 + 20}{340 - 20} = \frac{360}{320} = \frac{9}{8}$. Thus, the ratio is $9: 8$.

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