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Question

(A) The displacement of a particle performing $SHM$ starting from the mean position is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$

Vedclass · Accepted Answer

$1: 3$

Vedclass · Answer

(A) The displacement of a particle performing $SHM$ starting from the mean position is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.At $t = \frac{T}{12}$,the displacement is $x = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = A \sin\left(\frac{\pi}{6}\right) = \frac{A}{2}$.The potential energy $U$ is given by $U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2\left(\frac{A}{2}\right)^2 = \frac{1}{8}m\omega^2A^2$.The kinetic energy $K$ is given by $K = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}m\omega^2\left(\frac{3A^2}{4}\right) = \frac{3}{8}m\omega^2A^2$.The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8}m\omega^2A^2}{\frac{3}{8}m\omega^2A^2} = \frac{1}{3}$.Thus,the ratio is $1: 3$.

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t=\frac{T}{12}$,the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

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