$A$ pendulum is suspended in a lift and its period of oscillation when the lift is stationary is $T_0$. What must be the acceleration of the lift for the period of oscillation of the pendulum to be $T_0/2$?

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 is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$. The time period is given by $T = 2\pi \sqrt{\frac{l}{g'}}$,where $g'$ is equal to:

The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).

The periodic time of a simple pendulum of length $1\, m$ and amplitude $2\, cm$ is $5\, s$. If the amplitude is made $4\, cm$,its periodic time in seconds will be

$A$ positively charged pendulum is oscillating in a uniform electric field pointing upwards. How does its time period compare to the time period when it oscillates without an electric field?