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(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.Given

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$1: 3$

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(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{12}) = A \sin(\frac{\pi}{6}) = \frac{A}{2}$.The potential energy ($P$.$E$.) is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$.Substituting $x = \frac{A}{2}$,we get $U = \frac{1}{2} m \omega^2 (\frac{A^2}{4}) = \frac{1}{8} m \omega^2 A^2$.The kinetic energy ($K$.$E$.) is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Substituting $x = \frac{A}{2}$,we get $K = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3A^2}{4}) = \frac{3}{8} m \omega^2 A^2$.The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{3}{8} m \omega^2 A^2} = \frac{1}{3}$.

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

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