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Question

(B) The sound from the source $P$ reaches the observer at $Q$ along the line joining them. The source $P$ is moving towards the crossing with a veloci

Vedclass · Accepted Answer

$717$

Vedclass · Answer

(B) The sound from the source $P$ reaches the observer at $Q$ along the line joining them. The source $P$ is moving towards the crossing with a velocity $v_s = 10\, m/s$.The component of the velocity of the source $P$ along the line $PQ$ (the direction of sound propagation) is the effective velocity of approach,given by $v_{eff} = v_s \cos 	heta$.From the geometry of the problem,the distance of $P$ from the crossing is $40\, m$ and the distance of $Q$ from the crossing is $30\, m$. The distance $PQ$ is $\sqrt{40^2 + 30^2} = 50\, m$.Thus,$\cos 	heta = \frac{40}{50} = \frac{4}{5}$.The apparent frequency $f'$ heard by the stationary observer $Q$ is given by the Doppler effect formula:$f' = f \left( \frac{v}{v - v_s \cos 	heta} ight)$Substituting the given values:$f' = 700 	imes \left( \frac{340}{340 - 10 	imes \frac{4}{5}} ight)$$f' = 700 	imes \left( \frac{340}{340 - 8} ight)$$f' = 700 	imes \left( \frac{340}{332} ight)$$f' \approx 716.86\, Hz \approx 717\, Hz$.

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