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(B) The displacement of a particle starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $t = \frac{T}{6}$ and $\omega = \fr

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$1:3$

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(B) The displacement of a particle starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $t = \frac{T}{6}$ and $\omega = \frac{2\pi}{T}$,we have $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{6}) = A \sin(\frac{\pi}{3}) = A \frac{\sqrt{3}}{2}$.The kinetic energy $(KE)$ is given by $KE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}k(A^2 - \frac{3A^2}{4}) = \frac{1}{2}k(\frac{A^2}{4})$.The potential energy $(PE)$ is given by $PE = \frac{1}{2}kx^2 = \frac{1}{2}k(\frac{3A^2}{4})$.Therefore,the ratio $\frac{KE}{PE} = \frac{\frac{1}{2}k(A^2/4)}{\frac{1}{2}k(3A^2/4)} = \frac{1}{3}$.

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

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