A particle starts oscillating simple harmonically from its equilibrium position with time period $T$. The ratio of KE and PE of other particle at the $t = \frac{T}{12}$ is

Question

Dr. Abubacker · Accepted Answer

(c) When $t = \frac{T}{12},\ then\ x = A\sin{\frac{2\pi}{T} 	imes \frac{T}{12} = \frac{A}{2}}$
$KE = \frac{1}{2}\ mv^{2} = \frac{1}{2}\ m\omega^{2}(r^{2} - x^{2})$
$= \ \ \frac{1}{2}\ m\omega^{2}\left( A^{2} - \frac{A^{2}}{4} ight)$
$= \ \ \frac{3}{4}\left( \frac{1}{2}\ m\omega^{2}A^{2} ight)$
$PE = \frac{1}{2}\ m\omega^{2}x^{2} = \frac{1}{4}\left( \frac{1}{2}\ m\omega^{2}A^{2} ight)$
$\frac{KE}{PE} = \frac{3}{1}$

Q. A particle starts oscillating simple harmonically from its equilibrium position with time period $T$. The ratio of KE and PE of other particle at the $t = \frac{T}{12}$ is

Explanation

Reviewed by Dr. Abubacker

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