The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02 \text{ J}$. The speed of the simple pendulum bob at the equilibrium position is approximately: (Consider mass of the bob = $20 \text{ g}$) (in $\text{ m/s}$)

$A$ simple pendulum of length '$l$' and a bob of mass '$m$' is executing $S$.$H$.$M$. of small amplitude '$A$'. The maximum tension in the string will be ($g=$ acceleration due to gravity).

$A$ simple pendulum with a bob of mass $m$ and length $x$ is held in position at an angle $\theta_1$ and then at an angle $\theta_2$ with the vertical. When released from these positions,the speeds with which it passes the lowest position are $v_1$ and $v_2$ respectively. Then,the ratio $\frac{v_1}{v_2}$ is .............

The time period of a simple pendulum in air is $T$. If the pendulum is immersed in water and executes $SHM$,its time period is $t$. The value of $\frac{T}{t}$ is [density of the bob is $\frac{5000}{3} \ kg \ m^{-3}$ and density of water is $1000 \ kg \ m^{-3}$].

Two pendulums have time periods $T$ and $\frac{5T}{4}$. They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has completed one oscillation (in $^o$)?

The acceleration due to gravity at a place is $\pi^2 \, m/s^2$. Then the time period of a simple pendulum of length $1 \, m$ is: