A particle executing SHM has total energy of 20 joules. If the PE at the mean position is 5J, the average K.E is

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15J

10J

7.5J

answer is D.

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Detailed Solution

$K E_{a v g} = \frac{K E_{\min} + K E_{\max}}{2}$
$K E_{\max} = T E - P E$
$\begin{aligned} = 20 - 5 \\ = 15 J \\ K E_{min} = 0 \\ K E_{avg} = \frac{0 + 15}{2} = 7.5 J \end{aligned}$

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