A car P approaching a crossing at a speed of | vedclass

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(B) The sound from the source $P$ reaches the observer at $Q$ along the line joining them. The distance between $P$ and $Q$ is $\sqrt{40^2 + 30^2} = 5

Vedclass · Accepted Answer

$717$

Vedclass · Answer

(B) The sound from the source $P$ reaches the observer at $Q$ along the line joining them. The distance between $P$ and $Q$ is $\sqrt{40^2 + 30^2} = 50 \, m$.Let $	heta$ be the angle between the velocity vector of car $P$ and the line $PQ$. From the geometry,$\cos 	heta = \frac{40}{50} = \frac{4}{5}$.The component of the velocity of the source $P$ along the line $PQ$ (the direction of sound propagation) is $v_s \cos 	heta = 10 	imes \frac{4}{5} = 8 \, m/s$.Since the source is approaching the observer,the apparent frequency $f'$ is given by the Doppler effect formula:$f' = f \left( \frac{v}{v - v_s \cos 	heta} ight)$Substituting the values $f = 700 \, Hz$,$v = 340 \, m/s$,and $v_s \cos 	heta = 8 \, m/s$:$f' = 700 \left( \frac{340}{340 - 8} ight) = 700 \left( \frac{340}{332} ight)$$f' = 700 	imes 1.0241 \approx 716.87 \, Hz$.Rounding to the nearest integer,the apparent frequency is $717 \, Hz$.

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