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The particle executing simple harmonic motion has a kinetic energy $K_0 \cos ^2$ $\omega t$. The maximum values of the potential energy and the total energy are respectively.
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The correct answer is:
$K_0$ and $K_0$
$\because \mathrm{K}$. E. $=K_0 \cos ^2 \omega t$
$\therefore$ Maximum P. E. = Maximum K. E. $=$ Total energy $=\mathrm{K}_0$
$\therefore$ Maximum P. E. = Maximum K. E. $=$ Total energy $=\mathrm{K}_0$
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