The position average kinetic energy per oscillation is

\frac{1}{4} k a^{2}

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\frac{1}{3} k a^{2}

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

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\frac{1}{6} k a^{2}

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Solution

The correct option is D $\frac{1}{3} k a^{2}$
$K E = \frac{1}{2} m ω^{2} (a^{2} - x^{2})$
Position average $= \frac{\int_{- a}^{a} \frac{1}{2} m ω^{2} (a^{2} - x^{2}) d x}{\int_{- a}^{a} d x}$
$= \frac{1}{3} m ω^{2} a^{2}$

$= \frac{1}{3} k a^{2}$

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The position average kinetic energy per oscillation is

The correct option is D 13ka2KE=12mω2(a2−x2)Position average=∫a−a12mω2(a2−x2)dx∫a−adx=13mω2a2=13ka2

The correct option is D $\frac{1}{3} k a^{2}$
$K E = \frac{1}{2} m ω^{2} (a^{2} - x^{2})$
Position average $= \frac{\int_{- a}^{a} \frac{1}{2} m ω^{2} (a^{2} - x^{2}) d x}{\int_{- a}^{a} d x}$
$= \frac{1}{3} m ω^{2} a^{2}$

$= \frac{1}{3} k a^{2}$