Embibe Experts Solutions for Chapter: Simple Harmonic Motion, Exercise 1: AMU-AT (B.Tech.) 2017

The time-period of a physical pendulum is $2\pi \sqrt{\frac{I}{mgd}},$ where $I$ is the moment of inertia of the pendulum about the axis of rotation and $d$ is perpendicular distance between the axis of rotation and the centre of mass of the pendulum. A circular ring hangs from a nail on a wall. The mass of the ring is $3 \mathrm{kg}$ and its radius is $20 \mathrm{cm}$. If the ring is slightly displaced, the time of resulting oscillations will be

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to the relation $y=k{t}^2,\left(k=1 \mathrm{m} {\mathrm{s}}^{-2}\right)$  where $y$ is the vertical displacement. The time
period now becomes $T_2$.

The ratio of $\frac{T_1^2}{T_2^2}\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$ is