The particle executing simple harmonic motion has a kinetic energy $K_0 \cos ^2$ $\omega t$. The maximum…

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.
  1. $K_0 / 2$ and $K_0$
  2. $K_0$ and $2 K_0$
  3. $K_0$ and $K_0$
  4. 0 and $2 K_0$.

Solution

$\because \mathrm{K}$. E. $=K_0 \cos ^2 \omega t$ $\therefore$ Maximum P. E. = Maximum K. E. $=$ Total energy $=\mathrm{K}_0$ /

Asked in: NEET 2007

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