Two pendulums differ in lengths by $22 \ cm$. They oscillate at the same place such that one of them makes $15 \ oscillations$ and the other makes $18 \ oscillations$ during the same time. The lengths (in $cm$) of the pendulums are

The bob of a simple pendulum of mass $m$ and total energy $E$ will have maximum linear momentum equal to

$A$ uniform rod of length $1.8 \,m$ suspended by an end is made to undergo small oscillations. Find the length of the simple pendulum having the mass and time period equal to that of the rod. (in $\,m$)

$A$ simple pendulum of length $1 \ m$ and having a bob of mass $100 \ g$ is suspended in a car,moving on a circular track of radius $100 \ m$ with uniform speed $10 \ m/s$. If the pendulum makes small oscillations in a radial direction about its equilibrium position,then its time period is given by $T = 2\pi / \alpha^{1/4}$. The value of $\alpha$ is (Take $g = 10 \ m/s^2$)

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1 \ m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be (take $g = 10 \ m/s^2$):

$A$ second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket