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For a particle executing SHM. The KE ‘K’ is given by K = K₀ cos² wt. The maximum value of PE is

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a

K₀

b

zero

c

\frac{K_{0}}{2}

d

\frac{K_{0}}{4}

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detailed solution

Correct option is A

$K . E = K_{0} \cos^{2} ω t$
But $K . E = \frac{1}{2} {mv}^{2} x = Asin ωt$
$\begin{array}{l} V = Aωcos ωt; K . E = \frac{1}{2} {mA}^{2} ω^{2} \cos^{2} ωt \\ compare with K . E = K_{0} \cos^{2} ωt \\ K_{0} = \frac{1}{2} {mω}^{2} A^{2} \to max value of P.E \\ P.E E_{max} = \frac{1}{2} m^{2} A^{2} = k_{0} \end{array}$

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