A particle undergoing simple harmonic motion | vedclass

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(D) The displacement is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{\pi}{90} \ rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \c

Vedclass · Accepted Answer

$1/3$

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(D) The displacement is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{\pi}{90} \ rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)$.Kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.Potential energy $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$.The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\cos^2(\omega t)}{\sin^2(\omega t)} = \cot^2(\omega t)$.At $t = 210 \ s$,the phase is $\omega t = \frac{\pi}{90} 	imes 210 = \frac{21\pi}{9} = \frac{7\pi}{3} = 2\pi + \frac{\pi}{3}$.Therefore,$\frac{K}{U} = \cot^2\left( 2\pi + \frac{\pi}{3} ight) = \cot^2\left( \frac{\pi}{3} ight) = \left( \frac{1}{\sqrt{3}} ight)^2 = \frac{1}{3}$.

$A$ particle undergoing simple harmonic motion has time-dependent displacement given by $x(t) = A \sin \left( \frac{\pi t}{90} \right)$. The ratio of kinetic energy to potential energy of the particle at $t = 210 \ s$ will be:

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