$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ (in $g$)? ($g$ is acceleration due to gravity).

The bob of a pendulum of mass $m$, suspended by an inextensible string of length $L$ as shown in the figure, carries a small charge $q$. An infinite horizontal plane conductor with uniform surface charge density $\sigma$ is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1\,m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be:

At which position in the string of a simple pendulum is the tension maximum?

The metallic bob of a simple pendulum has the relative density $\rho$. The time period of this pendulum is $T$. If the metallic bob is immersed in water,then the new time period is given by