A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $\mathrm{t}=\frac{\mathrm{T}}{6}$, the ratio of the potential energy to kinetic energy of the particle is

$$ \left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right] $$

Two particles ' $A$ ' and ' $B$ ' execute SHMs of periods ' $T$ ' and $\frac{3 T}{2}$. If they start from the mean position then the phase difference between them, when the particle ' $A$ ' completes two oscillations will be

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere is (acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

The displacement of particle in S.H.M. is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6)$. Its speed will be maximum at time $\left(\sin 90^{\circ}=1\right)$