$A$ pendulum performs $S.H.M.$ with period $\sqrt{3} \ s$ in a stationary lift. If the lift moves up with acceleration $\frac{g}{3}$,the period of the pendulum is $[g=$ acceleration due to gravity $]$. (in $s$)

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 .............$cm$.

$A$ rubber ball filled with water,having a small hole,is used as the bob of a simple pendulum. The time period of such a pendulum

$A$ simple pendulum suspended from the ceiling of a stationary lift has period $T_0$. When the lift descends at steady speed,the period is $T_1$,and when it descends with constant downward acceleration,the period is $T_2$. Which one of the following is true?

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length,then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is

$A$ pendulum clock is placed inside an elevator. Under what condition does it show the correct time?