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(A) Given,position of particle of mass $m = 3 \ kg$ is $x = 0.3 \cos (\omega t)$.Velocity of particle,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \ome

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(A) Given,position of particle of mass $m = 3 \ kg$ is $x = 0.3 \cos (\omega t)$.Velocity of particle,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \omega t) = -0.3 \omega \sin (\omega t)$.Kinetic energy $K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m (-0.3 \omega \sin \omega t)^2 = \frac{1}{2} m (0.09 \omega^2 \sin^2 \omega t)$.At $t_1 = \frac{\pi}{6 \omega}$,$K\left(\frac{\pi}{6 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{6}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{4}\right)$.At $t_2 = \frac{\pi}{3 \omega}$,$K\left(\frac{\pi}{3 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{3}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{3}{4}\right)$.Taking the ratio: $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)} = \frac{1/4}{3/4} = \frac{1}{3}$.

Position of a $3 \ kg$ mass moving along the $X$-axis is given by $x = 0.3 \cos (\omega t) \ m$. If $K(t)$ denotes the kinetic energy at time $t$,then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

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