Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$,then the ratio of lengths of the two pendulums will be

$A$ pendulum of length $l = 1\,m$ is released from $\theta_0 = 60^\circ$. The rate of change of speed of the bob at $\theta = 30^\circ$ is ........ $m/s^2$ $(g = 10\,m/s^2)$.

At what position is the speed of a simple pendulum maximum?

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

The ratio of frequencies of two oscillating pendulums are $3: 2$. Their lengths are in the ratio:

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is: