A particle is executing linear SHM. The average kinetic energy and average potential energy, over a period of oscillation, respectively are $K_{av} and U_{av}$ . Then,

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$K_{a v} = U_{a v}$

$U_{av} = \frac{K_{av}}{3}$

$K_{av} = \frac{U_{av}}{2}$

$U_{av} = \frac{K_{av}}{2}$

answer is C.

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Detailed Solution

$K = \frac{1}{2} {mv}^{2} = \frac{1}{2} {mA}^{2} ω^{2} \cos^{2} ωt$

Average value of $\cos^{2} ωt is \frac{1}{2} over one cycle$

$∴ K_{av} = \frac{1}{4} {mω}^{2} A^{2}$ ------------(i)

$U = \frac{1}{2} {mω}^{2} x^{2} = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$

Average value of $\sin^{2} ωt is \frac{1}{2} over one cycle .$

$∴ U_{av} = \frac{1}{4} {mω}^{2} A^{2}$ ---------(iii)

From Eqs. (i) and (ii), we get $K_{av} = U_{av}$

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