If the length of an oscillating simple pendulum is made $\frac{1}{3}$ times at a place while keeping the amplitude the same,then its total energy $(E)$ will be: (in $E$)

For a simple pendulum, the graph of time period $(T)$ versus length $(l)$ is depicted by:

$A$ simple pendulum of length $1\,m$ is allowed to oscillate with an amplitude of $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be: (use $g = \pi^2$)

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

The ratio of frequencies of oscillations of two simple pendulums is $3: 4$,then their lengths are in the ratio