The ratio of frequencies of two pendulums are $2 : 3$, then their length are in ratio
$\sqrt {2/3} $
$\sqrt {3/2} $
$4/9$
$9/4$
The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with acceleration of $g/3,$ then the time period of the pendulum is
The length of a second's pendulum on the surface of earth is $1\, m$. What will be the length of a second's pendulum on the moon ?
The length of a seconds pendulum at a height $h=2 R$ from earth surface will be.(Given: $R =$ Radius of earth and acceleration due to gravity at the surface of earth $g =\pi^{2}\,m / s ^{-2}$ )
A simple pendulum of length $L$ and mass (bob) $M$ is oscillating in a plane about a vertical line between angular limits $ - \varphi $ and $ + \varphi $. For an angular displacement $\theta (|\theta | < \varphi )$, the tension in the string and the velocity of the bob are $T$ and $ v$ respectively. The following relations hold good under the above conditions
A solid cylinder of density $\rho_0$, cross-section area $A$ and length $l$ floats in a liquid of density $\rho\left( >\rho_0\right)$ with its axis vertical, as shown. If it is slightly displaced downward and released, the time period will be .......