For a particle executing SHM. The KE 'K' is given by $K = K_{0} c o s^{2}$ wt. The maximum value of PE is

K_{0}

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zero

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\frac{K_{0}}{2}

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\frac{K_{0}}{4}

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Solution

The correct option is C $K_{0}$
In S.H.M, The kinetic energy is given by,
$K = K_{0} c o s^{2} w t$
Kinetic energy is zero, then potential energy is maximum
The maximum potential energy is $K_{0}$
The correct option is A.

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