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(C) The potential energy $(U)$ of a particle in simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t)$.The total ener

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(C) The potential energy $(U)$ of a particle in simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t)$.The total energy $(E)$ is given by $E = \frac{1}{2} k A^2$.We are given that $U = \frac{1}{2} E$.Substituting the expressions,we get $\frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.This simplifies to $\sin^2(\omega t) = \frac{1}{2}$,which means $\sin(\omega t) = \frac{1}{\sqrt{2}}$.Thus,$\omega t = \frac{\pi}{4}$.Given the period $T = 8 \ s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.Substituting $\omega$ into the equation: $(\frac{\pi}{4}) t = \frac{\pi}{4}$.Therefore,$t = 1 \ s$.

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)

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