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

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(B) The apparent frequency $v^{\prime}$ heard by a stationary observer when the source is moving is given by the Doppler effect formula:$v^{\prime} =

Vedclass · Accepted Answer

$680$

Vedclass · Answer

(B) The apparent frequency $v^{\prime}$ heard by a stationary observer when the source is moving is given by the Doppler effect formula:$v^{\prime} = v \left[ \frac{V}{V - v_s \cos 	heta} ight]$Where $v = 640 ~Hz$ is the source frequency,$V = 340 ~m/s$ is the speed of sound,$v_s = \frac{100}{3} ~m/s$ is the speed of the source,and $	heta$ is the angle between the velocity vector of the source and the line joining the source to the observer.From the geometry of the right-angled triangle formed by the source,point $A$,and the observer $O$:The hypotenuse is $\sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = 50 ~m$.Thus,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{30}{50} = \frac{3}{5} = 0.6$.Substituting the values into the formula:$v^{\prime} = 640 \left[ \frac{340}{340 - (\frac{100}{3}) 	imes 0.6} ight]$$v^{\prime} = 640 \left[ \frac{340}{340 - 20} ight] = 640 	imes \frac{340}{320}$$v^{\prime} = 2 	imes 340 = 680 ~Hz$.

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