A particle executes S.H.M. and its position v | vedclass

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(B) The particle starts from the mean position $(x = 0)$ at $t = 0$.At the mean position,the phase $\phi_1 = 0$.At the mid-point between the mean and

Vedclass · Accepted Answer

$\frac{3 A \omega}{\pi}$

Vedclass · Answer

(B) The particle starts from the mean position $(x = 0)$ at $t = 0$.At the mean position,the phase $\phi_1 = 0$.At the mid-point between the mean and extreme position,the displacement is $x = \frac{A}{2}$.Using the equation $x = A \sin \omega t$,we have $\frac{A}{2} = A \sin(\omega t)$.This gives $\sin(\omega t) = \frac{1}{2}$,so $\omega t = \frac{\pi}{6}$,which means $t = \frac{\pi}{6 \omega}$.The total distance traveled is $d = \frac{A}{2} - 0 = \frac{A}{2}$.Average speed is defined as $	ext{Total distance} / 	ext{Total time}$.Average speed $= \frac{A/2}{\pi / (6 \omega)} = \frac{A}{2} 	imes \frac{6 \omega}{\pi} = \frac{3 A \omega}{\pi}$.

$A$ particle executes $S.H.M.$ and its position varies with time as $x = A \sin \omega t$. Its average speed during its motion from mean position to mid-point of mean and extreme position is

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