A spring force constant $180$ N/m is loaded with a mass $0.2$ kg. The amplitude of oscillations is $4$ cm. When mass comes to equilibrium position, its velocity is

A particle executing linear S.H.M. has velocities $V_1$ and $V_2$ at distance $x_1$ and $x_2$ respectively, from the mean position, its angular velocity is

A particle executes a simple harmonic motion with a periodic time $8$ second. At time $t = 0$, it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is
($\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\sin 90^\circ = \cos 0^\circ = 1$)