A particle undergoing simple harmonic motion has time dependent displacement give by The ratio of kinetic to potential energy of this particle at t = 210 s will be

Engineering

Physics

Basics of Simple Harmonic Motion

Question

A particle undergoing simple harmonic motion has time dependent displacement give by $x (t) = A sin \frac{πt}{90} .$ The ratio of kinetic to potential energy of this particle at t = 210 s will be

$\frac{1}{9}$

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$K = \frac{1}{2} {mv}^{2}; U = \frac{1}{2} {kx}^{2} = \frac{1}{2} m^{2} x^{2}$

$∴ \frac{k}{U} = \frac{v^{2}}{ω^{2} x^{2}} = {(\frac{\cos (wt)}{\sin (wt)})}^{2}$

$= \cot^{2} (\frac{π}{90} \times 210)$

$= \cot^{2} (2 π + \frac{π}{3})$

$= {(\frac{1}{\sqrt{3}})}^{2} = \frac{1}{3},$