At a place,the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping the amplitude the same. Then,the total energy will be:

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?

$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

The bob of a simple pendulum performs $SHM$ with period $T$ in air and with period $T_1$ in water. The relation between $T$ and $T_1$ is (neglect friction due to water,density of the material of the bob is $\frac{9}{8} \times 10^3 \ kg/m^3$,density of water is $1 \ g/cc$):

The length of the seconds pendulum is $1 \,m$ on Earth. If the mass and diameter of a planet are $1.5$ times that of the Earth, the length of the seconds pendulum on the planet will be nearly (in $\,m$)

Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.