The average kinetic energy of a simple harmonic oscillator with respect to mean position will be:

\frac{k a^{2}}{6}

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\frac{k a^{2}}{4}

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\frac{k a^{2}}{3}

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\frac{k a^{2}}{2}

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Solution

The correct option is B $\frac{k a^{2}}{3}$

The average K.E can be given by
$K . E = \int_{- a}^{a} \frac{1}{2} k (a^{2} - x^{2}) = \frac{k a^{3}}{3}$
Hence, average KE with respect to mean position will be
$\frac{\frac{k a^{3}}{3}}{a} = \frac{k a^{2}}{3}$ .
Option C is correct.

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