There is a simple pendulum hanging from the ceiling of a lift. When the lift is standing still,the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4$,then the new time period of the pendulum is: (in $T$)

At a place,the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping the amplitude the same. Then,the total energy will be:

$A$ child swinging on a swing in a sitting position stands up. What will happen to the time period of the swing?

The length of a seconds pendulum on the surface of the Earth is $1\, m$. The length of a seconds pendulum on the surface of the Moon,where $g$ is $1/6$th the value of $g$ on the surface of the Earth,is:

$A$ simple pendulum of length $L$ has mass $M$ and it oscillates freely with amplitude $A$. At the extreme position,its potential energy is $(g =$ acceleration due to gravity$)$

$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.