The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration of $g/3$,then the time period of the pendulum is

Two pendulums of length $1.21 \,m$ and $1.0 \,m$ start vibrating. At some instant,the two are in the mean position in the same phase. After how many vibrations of the longer pendulum will the two be in phase again?

The motion of a simple pendulum executing $S.H.M.$ is represented by the following equation:
$Y = A \sin (\pi t + \phi)$,where time is measured in $seconds$.
The length of the pendulum is .............$cm$.

The bob of a simple pendulum of length $L$ has a mass $m$ and charge $q$. The pendulum is suspended between the plates of a charged parallel plate capacitor. The direction of the electric field is shown in the figure. The period of oscillations of the simple pendulum is (acceleration due to gravity $g > qE/m$):

The time period of a simple pendulum is $2 \, sec$. If its length is increased $4$ times,then its period becomes ..... $\sec$.

The bob of a swinging seconds pendulum (one whose time period is $2 \,s$) has a small speed $v_{0}$ at its lowest point. Its height from this lowest point $2.25 \,s$ after passing through it is given by