A body performs linear S.H.M. with amplitude | vedclass

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(D) Let $a$ be the amplitude of the $S$.$H$.$M$.Given that the particle is at a distance $\frac{a}{3}$ from the extreme position,its displacement $x$

Vedclass · Accepted Answer

$\frac{4 \pi}{3 \sqrt{5}} 	ext{ s}$

Vedclass · Answer

(D) Let $a$ be the amplitude of the $S$.$H$.$M$.Given that the particle is at a distance $\frac{a}{3}$ from the extreme position,its displacement $x$ from the mean position is $x = a - \frac{a}{3} = \frac{2a}{3}$.The magnitude of acceleration is $a_p = \omega^2 x = \omega^2 \left( \frac{2a}{3} ight)$.The magnitude of velocity is $v_p = \omega \sqrt{a^2 - x^2} = \omega \sqrt{a^2 - \left( \frac{2a}{3} ight)^2} = \omega \sqrt{a^2 - \frac{4a^2}{9}} = \omega \sqrt{\frac{5a^2}{9}} = \frac{\omega a \sqrt{5}}{3}$.According to the problem,$v_p = \frac{1}{3} a_p$.Substituting the expressions for $v_p$ and $a_p$:$\frac{\omega a \sqrt{5}}{3} = \frac{1}{3} \left( \omega^2 \frac{2a}{3} ight)$.$\frac{\omega a \sqrt{5}}{3} = \frac{2 \omega^2 a}{9}$.Dividing both sides by $\omega a$ (assuming $\omega, a 
eq 0$):$\frac{\sqrt{5}}{3} = \frac{2 \omega}{9}$.$\omega = \frac{9 \sqrt{5}}{3 	imes 2} = \frac{3 \sqrt{5}}{2}$.Since the period $T = \frac{2 \pi}{\omega}$,we have:$T = \frac{2 \pi}{\frac{3 \sqrt{5}}{2}} = \frac{4 \pi}{3 \sqrt{5}} 	ext{ s}$.

$A$ body performs linear $S$.$H$.$M$. with amplitude $a$. When it is at a distance $\frac{a}{3}$ from the extreme position,the magnitude of velocity is $\frac{1}{3}$ times the magnitude of acceleration. The period of $S$.$H$.$M$. is:

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