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(B) Since the particle starts from the equilibrium position,the displacement is given by $x = a \sin(\omega t)$.At $t = T/12$,the displacement is $x =

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$3 : 1$

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(B) Since the particle starts from the equilibrium position,the displacement is given by $x = a \sin(\omega t)$.At $t = T/12$,the displacement is $x = a \sin(\frac{2\pi}{T} 	imes \frac{T}{12}) = a \sin(\frac{\pi}{6}) = \frac{a}{2}$.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 $\frac{K.E.}{P.E.} = \frac{a^2 - (a/2)^2}{(a/2)^2} = \frac{a^2 - a^2/4}{a^2/4} = \frac{3a^2/4}{a^2/4} = \frac{3}{1}$.

$A$ particle starts oscillating simple harmonically from its equilibrium position. The ratio of kinetic energy and potential energy of the particle at time $t = T/12$ is: ($T =$ time period)

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