A particle executes linear S.H.M. with amplit | vedclass

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(B) For a particle in $S.H.M.$, the velocity $v$ at position $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration $a$ at position $x$ is giv

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$\frac{6\pi}{\sqrt{7}} 	ext{ s}$

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(B) For a particle in $S.H.M.$, the velocity $v$ at position $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration $a$ at position $x$ is given by $a = \omega^2 x$.Given that the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.$\omega \sqrt{A^2 - x^2} = \omega^2 x$.$\sqrt{A^2 - x^2} = \omega x$.Given $A = 4 	ext{ cm}$ and $x = 3 	ext{ cm}$:$\sqrt{4^2 - 3^2} = \omega (3)$.$\sqrt{16 - 9} = 3\omega$.$\sqrt{7} = 3\omega \implies \omega = \frac{\sqrt{7}}{3} 	ext{ rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{(\sqrt{7}/3)} = \frac{6\pi}{\sqrt{7}} 	ext{ s}$.

$A$ particle executes linear $S.H.M.$ with amplitude $A = 4 \text{ cm}$. The magnitude of velocity and acceleration is equal when it is at a distance $x = 3 \text{ cm}$ from the mean position. The time period of the oscillation is:

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