The amplitude of an oscillating simple pendulum is $10 \ cm$ and its period is $4 \ s$. Its speed $1 \ s$ after it passes its equilibrium position is ... $m/s$.

The time period of a simple pendulum is $T$. The angular amplitude is $\beta$. How much time will the bob of the pendulum take to move from the equilibrium position $O$ to position $A$,where the string makes an angle $\alpha$ with the vertical?

The time period of a simple pendulum when it is made to oscillate on the surface of the moon:

In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be:

$A$ cylindrical piece of cork of density $\rho$, base area $A$, and height $h$ floats in a liquid of density $\rho_{l}$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period $T=2 \pi \sqrt{\frac{h \rho}{\rho_{l} g}}$. (Ignore damping due to viscosity of the liquid).

The period of a simple pendulum gets doubled when