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

A mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period $T$. If the mass is increased by $m_0$, the periodic time becomes $\frac{5 \mathrm{~T}}{4}$. The ratio $\frac{\mathrm{m}_0}{\mathrm{M}}$

A simple pendulum has time period ' $\mathrm{T}_1$ '. The point of suspension is now moved upward according to equation $\mathrm{y}=\mathrm{kt}^2$ where $\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^2$. If new time period is ' $\mathrm{T}_2$ ' then $\mathrm{T}_1^2 / \mathrm{T}_2^2$ will be ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )