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

The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be

Two simple harmonic motions of angular frequency $300 \mathrm{rad} / \mathrm{s}$ and $3000 \mathrm{rad} / \mathrm{s}$ have same amplitude. The ratio of their maximum accelerations is