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:

What is the length of a simple pendulum,which ticks seconds (in $m$)?

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ (in $g$)? ($g$ is acceleration due to gravity).

The linear displacement $x$ of the bob of a simple pendulum from its mean position varies as $x = a \sin \left(\frac{\pi}{\sqrt{2}} t\right)$,where $a$ is its amplitude expressed in meters and $t$ is in seconds. The length of the simple pendulum is (Take $g = \pi^{2} \ m/s^{2}$): (in $m$)

If a body dropped freely from a height of $20 \,m$ reaches the surface of a planet with a velocity of $31.4 \,ms^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is (in $\,m$)

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 ......... $s$.