The variation of displacement with time of a | vedclass

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(D) The displacement of a particle in simple harmonic motion is given by $x(t) = A \sin(\omega t)$.The potential energy $U$ is given by $U = \frac{1}{

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(D) The displacement of a particle in simple harmonic motion is given by $x(t) = A \sin(\omega t)$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.Using the trigonometric identity $\sin^2(	heta) = \frac{1 - \cos(2	heta)}{2}$,we get $U = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.This shows that the potential energy $U$ is always non-negative and varies with a frequency double that of the displacement.At $t = 0$,$x = 0$,so $U = 0$.At the extreme positions,$x = \pm A$,so $U$ is maximum.Looking at the options,figure $D$ represents a graph where $U$ is zero at $t = 0$,reaches a maximum at the extreme positions,and has a frequency twice that of the displacement graph. Thus,figure $D$ is the correct representation.

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U(x)$ versus time $t$ plot of the particle is correctly shown in figure:

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