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Question

(A) The displacement of a particle performing $S.H.M.$ starting from the mean position is given by $x = A \sin(\omega t)$.The velocity of the particle

Vedclass · Accepted Answer

$0.75$

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(A) The displacement of a particle performing $S.H.M.$ starting from the mean position is given by $x = A \sin(\omega t)$.The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.The total energy $(T.E.)$ is $T.E. = \frac{1}{2} m A^2 \omega^2$.We are given that $K.E. = 50\% 	ext{ of } T.E.$, so $\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = 0.5 	imes \frac{1}{2} m A^2 \omega^2$.This simplifies to $\cos^2(\omega t) = 0.5$, which means $\cos(\omega t) = \frac{1}{\sqrt{2}}$.Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, we have $\omega t = 45^\circ = \frac{\pi}{4} 	ext{ radians}$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} 	ext{ rad/s}$.Substituting the values: $(\frac{\pi}{3}) t = \frac{\pi}{4}$.Solving for $t$: $t = \frac{3}{4} = 0.75 	ext{ s}$.

$A$ particle starts from mean position and performs $S.H.M.$ with period $T = 6 \text{ s}$. At what time is its kinetic energy $50\%$ of its total energy (in $\text{ s}$)? (Given: $\cos 45^\circ = 1/\sqrt{2}$)

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