A particle starts oscillating simple harmonically from its mean position with time period '$$T$$'. At time $$t=\frac{T}{12}$$, the ratio of the potential energy to kinetic energy of the particle is $$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$$

The displacement of a particle performing S.H.M. is given by $$x=5 \sin (3 t+3)$$, where $$x$$ is in $$\mathrm{cm}$$ and $$t$$ is in second. The maximum acceleration of the particle will be

Two identical springs of constant '$$\mathrm{K}$$' are connected in series and parallel in shown in figure. A mass '$$\mathrm{M}$$' is suspended from them. The ratio of their frequencies is series to parallel combination will be

A particle performing linear S.H.M. of amplitude $$0.1 \mathrm{~m}$$ has displacement $$0.02 \mathrm{~m}$$ and acceleration $$0.5 \mathrm{~m} / \mathrm{s}^2$$. The maximum velocity of the particle in $$\mathrm{m} / \mathrm{s}$$ is