Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... 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:

$A$ ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

$A$ simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio $1: 2$, then the acceleration of the lift is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,ms^{-2}$)

$A$ simple pendulum has a time period $T$. The bob is given a negative charge and the surface below it is given a positive charge. The new time period will be

If a simple pendulum is released from the given position,find the velocity of the bob when it reaches the lowest position. (Take $g = 10 \ m/s^2$) (in $m/s$)