If the time period of a $2 \ m$ long simple pendulum is $2 \ s$,the acceleration due to gravity at the place where the pendulum is executing $S.H.M.$ is

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

$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

The ratio of the period of oscillation of a conical pendulum to that of a simple pendulum is: (Assume the strings are of the same length in both cases and $\theta$ is the angle made by the string with the vertical in the case of the conical pendulum.)

$A$ simple pendulum oscillates in a vertical plane. When it passes through the mean position,the tension in the string is $3$ times the weight of the pendulum bob. What is the maximum displacement of the pendulum string with respect to the vertical in degrees?

The bob of a simple pendulum executes simple harmonic motion in water with a period $t$,while the period of oscillation of the bob is ${t_0}$ in air. Neglecting the frictional force of water and given that the density of the bob is $(4/3) \times 1000 \ kg/m^3$. What relationship between $t$ and ${t_0}$ is true?