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(B) The displacement of a body starting from the origin in $SHM$ is given by $x = A \sin(\omega t)$.Velocity is $v = \frac{dx}{dt} = A \omega \cos(\om

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$\frac{1}{6} \ s$

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(B) The displacement of a body starting from the origin in $SHM$ is given by $x = A \sin(\omega t)$.Velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.Kinetic Energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.Total Energy $(T.E.)$ is given by $T.E. = \frac{1}{2} m A^2 \omega^2$.Given that $K.E. = 75\% 	ext{ of } T.E. = 0.75 \ T.E.$Substituting the expressions: $\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = 0.75 	imes \frac{1}{2} m A^2 \omega^2$.$\cos^2(\omega t) = 0.75 = \frac{3}{4}$.Taking the square root,$\cos(\omega t) = \frac{\sqrt{3}}{2}$.This implies $\omega t = \frac{\pi}{6}$.Since $\omega = \frac{2\pi}{T}$ and $T = 2 \ s$,we have $\omega = \frac{2\pi}{2} = \pi \ rad/s$.Substituting $\omega$: $\pi t = \frac{\pi}{6}$.Therefore,$t = \frac{1}{6} \ s$.

Starting from the origin,a body oscillates simple harmonically with a period of $2 \ s$. After what time will its kinetic energy be $75\%$ of the total energy?

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