A source of sound is travelling at frac{100}{ | vedclass

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(B) The source is moving towards point $A$ with velocity $v_s = \frac{100}{3} \, m/s$. The observer is at point $O$. The distance $AO = 4 \, m$ and th

Vedclass · Accepted Answer

$680$

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(B) The source is moving towards point $A$ with velocity $v_s = \frac{100}{3} \, m/s$. The observer is at point $O$. The distance $AO = 4 \, m$ and the distance of the source from $A$ is $3 \, m$. The distance of the source from the observer $O$ is $r = \sqrt{3^2 + 4^2} = 5 \, m$.The component of the source's velocity along the line joining the source and the observer is the effective velocity $v_{s, 	ext{eff}} = v_s \cos 	heta$,where $	heta$ is the angle between the velocity vector and the line $OS$.From the geometry,$\cos 	heta = \frac{3}{5}$.Therefore,$v_{s, 	ext{eff}} = \frac{100}{3} 	imes \frac{3}{5} = 20 \, m/s$.Using the Doppler effect formula for a moving source and stationary observer: $n' = n \left( \frac{v}{v - v_{s, 	ext{eff}}} ight)$.Substituting the values: $n' = 640 	imes \left( \frac{340}{340 - 20} ight) = 640 	imes \left( \frac{340}{320} ight) = 640 	imes 1.0625 = 680 \, Hz$.

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