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(B) Let the position of the train be $T$,the crossing be $C$,and the observer be $O$. The distance $TC = 288 \ m$ and $CO = 384 \ m$. The distance $TO

Vedclass · Accepted Answer

$612$

Vedclass · Answer

(B) Let the position of the train be $T$,the crossing be $C$,and the observer be $O$. The distance $TC = 288 \ m$ and $CO = 384 \ m$. The distance $TO = \sqrt{288^2 + 384^2} = \sqrt{82944 + 147456} = \sqrt{230400} = 480 \ m$.The component of the train's velocity along the line joining the train and the observer is the effective velocity of the source moving towards the observer.Let $v_s$ be the velocity of the train $(120 \ km/h = 120 	imes \frac{5}{18} \ m/s = \frac{100}{3} \ m/s)$.The angle $	heta$ is between the track and the line $TO$. From the geometry,$\cos 	heta = \frac{TC}{TO} = \frac{288}{480} = \frac{3}{5} = 0.6$.The velocity of the source towards the observer is $u = v_s \cos 	heta = \frac{100}{3} 	imes 0.6 = 20 \ m/s$.Using the Doppler effect formula for a source moving towards a stationary observer:$f' = f \left( \frac{v}{v - u} ight)$Where $f = 576 \ Hz$,$v = 340 \ m/s$,and $u = 20 \ m/s$.$f' = 576 \left( \frac{340}{340 - 20} ight) = 576 \left( \frac{340}{320} ight) = 576 	imes 1.0625 = 612 \ Hz$.

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