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Answer: \(x = 5\) cm from the mean position.

For S.H.M. of amplitude \(A\) and force constant \(k\), at displacement \(x\):

\[\text{K.E.} = \frac{1}{2}k(A^2 - x^2),\qquad \text{P.E.} = \frac{1}{2}kx^2\]

Given K.E. = 3 × P.E.:

\[\frac{1}{2}k(A^2 - x^2) = 3\times\frac{1}{2}kx^2 \Rightarrow A^2 = 4x^2 \Rightarrow x = \frac{A}{2} = \frac{10}{2} = 5\ \mathrm{cm}\]
Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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