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(B) The displacement of a particle starting from the equilibrium position is given by $x = a \sin(\omega t)$.Given $t = \frac{T}{12}$ and $\omega = \f

Vedclass · Accepted Answer

$3: 1$

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(B) The displacement of a particle starting from the equilibrium position is given by $x = a \sin(\omega t)$.Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,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 kinetic energy $(K.E.)$ is given by $\frac{1}{2}k(a^2 - x^2)$ and the potential energy $(P.E.)$ is given by $\frac{1}{2}kx^2$.The ratio of kinetic energy to potential energy is $\frac{K.E.}{P.E.} = \frac{a^2 - x^2}{x^2}$.Substituting $x = \frac{a}{2}$,we get $\frac{K.E.}{P.E.} = \frac{a^2 - (a/2)^2}{(a/2)^2} = \frac{a^2 - a^2/4}{a^2/4} = \frac{3a^2/4}{a^2/4} = \frac{3}{1}$.Thus,the ratio is $3: 1$.

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