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(B) Let the original frequency of the whistle be $f_0$.When the train approaches the observer,the observed frequency is $f_1 = f_0 \left( \frac{v}{v -

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(B) Let the original frequency of the whistle be $f_0$.When the train approaches the observer,the observed frequency is $f_1 = f_0 \left( \frac{v}{v - v_s} ight)$.When the train moves away from the observer,the observed frequency is $f_2 = f_0 \left( \frac{v}{v + v_s} ight)$.However,the problem states the frequency drops by $140 \, Hz$ as the train passes. This implies the change is from the frequency when approaching to the frequency when receding: $f_1 - f_2 = 140 \, Hz$.Given $v = 330 \, m/s$ and $v_s = 70 \, m/s$:$f_0 \left( \frac{330}{330 - 70} - \frac{330}{330 + 70} ight) = 140$$f_0 \left( \frac{330}{260} - \frac{330}{400} ight) = 140$$f_0 \left( 1.269 - 0.825 ight) = 140$$f_0 (0.444) = 140 \implies f_0 \approx 315 \, Hz$. Wait,re-evaluating the standard interpretation: If the train is moving away,the drop is $f_0 - f_2 = 140$.$f_0 \left( 1 - \frac{330}{330 + 70} ight) = 140$$f_0 \left( 1 - \frac{330}{400} ight) = 140$$f_0 \left( \frac{70}{400} ight) = 140$$f_0 = 140 	imes \frac{400}{70} = 800 \, Hz$.

$A$ man standing on a platform observes that the frequency of the sound of a whistle emitted by a train drops by $140 \, Hz$. If the velocity of sound in air is $330 \, m/s$ and the speed of the train is $70 \, m/s$,the frequency of the whistle is .......... $Hz$.

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