A source of sound of frequency 640 ,Hz is mov | vedclass

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Question

(B) The apparent frequency $v^{\prime}$ heard by a stationary observer when the source is moving at an angle $	heta$ with the line joining the source

Vedclass · Accepted Answer

$680$

Vedclass · Answer

(B) The apparent frequency $v^{\prime}$ heard by a stationary observer when the source is moving at an angle $	heta$ with the line joining the source and the observer is given by the Doppler effect formula:$v^{\prime} = v \left( \frac{V}{V - v_s \cos 	heta} ight)$Where $V = 340 \,m/s$ is the speed of sound, $v_s = \frac{100}{3} \,m/s$ is the speed of the source, and $v = 640 \,Hz$ is the source frequency.From the geometry of the right-angled triangle formed by the source, point $A$, and the observer $O$, the distance between the source and $O$ is $\sqrt{30^2 + 40^2} = 50 \,m$.Thus, $\cos 	heta = \frac{30}{50} = \frac{3}{5}$.Substituting the values:$v^{\prime} = 640 \left( \frac{340}{340 - (\frac{100}{3}) 	imes (\frac{3}{5})} ight)$$v^{\prime} = 640 \left( \frac{340}{340 - 20} ight)$$v^{\prime} = 640 	imes \frac{340}{320}$$v^{\prime} = 640 	imes \frac{34}{32} = 20 	imes 34 = 680 \,Hz$.

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