As a body performs S.H.M.,its potential energ | vedclass

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(B) The potential energy $U$ of a particle performing $S.H.M.$ is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t + \phi)$.Substituting $x

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(B) The potential energy $U$ of a particle performing $S.H.M.$ is given by $U = \frac{1}{2} k x^2$,where $x = A \sin(\omega t + \phi)$.Substituting $x$,we get $U = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) = \frac{1}{4} k A^2 (1 - \cos(2\omega t + 2\phi))$.This expression shows that $U$ is always non-negative $(U \ge 0)$ and varies sinusoidally with a frequency double that of the displacement $(2\omega)$.Looking at the options,graph $B$ represents a function that is always non-negative and has a frequency double that of the standard sine wave,starting from zero at $t=0$ (assuming $\phi = 0$).Therefore,graph $B$ correctly represents the variation of potential energy with time.

As a body performs $S.H.M.$,its potential energy $U$ varies with time $t$ as indicated in:

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