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

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

$A$ coin is placed on a horizontal plate. The plate performs $S.H.M.$ vertically with an angular frequency $\omega$. The amplitude $A$ of oscillations is gradually increased. The coin will lose contact with the plate for the first time when the amplitude is ($g =$ acceleration due to gravity).

Two pendulums have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum will they be again in the same phase?

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:

The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$,then its new time period of oscillation will be ......... $s$.