A particle executes S.H.M. starting from the mean position. Its amplitude is ' a ' and its periodic time is ' $T$ '. At a certain instant, its speed ' $u$ ' is half that of maximum speed $\mathrm{V}_{\text {max }}$. The displacement of the particle at that instant is

A particle performing linear S.H.M. has period 8 seconds. At time $\mathrm{t}=0$, it is in the mean position. The ratio of the distances travelled by the particle in the $1^{\text {st }}$ and $2^{\text {nd }}$ second is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}}{2} \mathrm{~A}$, where A is the amplitude of S.H.M. The value of ' $n$ ' is