Discussion
If the time period of a particle executing SHM is 8 s, then the time period of the potential energy of this particle is
Answer: B
The period of the potential energy is 4 s.
In SHM \(x = A\sin\omega t\), and the potential energy is \(U = \frac{1}{2}kx^{2} = \frac{1}{2}kA^{2}\sin^{2}\omega t\).
\[ \sin^{2}\omega t = \frac{1 - \cos 2\omega t}{2} \]
The energy therefore oscillates at twice the angular frequency, with half the period: \(\frac{8}{2} = 4\) s. It peaks at both extremes of each oscillation, x = +A and x = -A. 8 s is the period of the displacement itself, and 16 s doubles instead of halving.
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