Draw a neat labelled diagram of variation of kinetic energy and potential energy with displacement in S.H.M.

Answer: [Figure: energy on the vertical axis and displacement \(x\) from \(-A\) to \(+A\) on the horizontal axis. Three graphs are drawn: (1) the total energy \(E = \frac{1}{2}kA^2\) as a horizontal straight line; (2) the potential energy \(\frac{1}{2}kx^2\) as an upward-opening parabola, zero at \(x = 0\) and equal to \(E\) at \(x = \pm A\); (3) the kinetic energy \(\frac{1}{2}k(A^2 - x^2)\) as a downward-opening parabola, equal to \(E\) at \(x = 0\) and zero at \(x = \pm A\). The K.E. and P.E. curves cross at \(x = \pm A/\sqrt{2}\).]

\(\text{K.E.} = \frac{1}{2}k(A^2 - x^2)\), \(\text{P.E.} = \frac{1}{2}kx^2\) and \(\text{K.E.} + \text{P.E.} = \frac{1}{2}kA^2\) = constant at every displacement, where \(k = m\omega^2\).