In S.H.M. the displacement of a particle at an instant is $Y=A \cos 30^{\circ}$, where $A=40 \mathrm{~cm}$ and kinetic energy is 200 J . If force constant is $1 \times 10^{\times} \mathrm{N} / \mathrm{m}$, then x will be $\left(\cos 30^{\circ}=\sqrt{3} / 2\right)$

A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is $12 \mathrm{~cm} / \mathrm{s}$. When the speed of the particle becomes $6 \mathrm{~cm} / \mathrm{s}$, the distance of the particle from mean position is

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is

A mass ' $m$ ' attached to a spring oscillates with a period of 3 second. If the mass is increased by 0.6 kg , the period increases by 3 second. The initial mass ' $m$ ' is equal to