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

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Question

(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 $\sqrt{40^2 + 30^2} = 50 \, m$.The source $P$ is moving towards the crossing with velocity $v_s = 10 \, m/s$. The component of this velocity along the line $PQ$ (the direction of sound propagation) is the effective velocity of the source towards the observer.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 effective velocity of the source towards the observer is $v_{s, 	ext{eff}} = v_s \cos 	heta = 10 	imes \frac{4}{5} = 8 \, m/s$.Using the Doppler effect formula for a moving source and stationary observer:$f' = f \left( \frac{v}{v - v_{s, 	ext{eff}}} ight)$Substituting the values:$f' = 700 	imes \left( \frac{340}{340 - 8} ight) = 700 	imes \left( \frac{340}{332} ight)$$f' = 700 	imes 1.0241 = 716.87 \, Hz \approx 717 \, Hz$.

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