A simple pendulum of length $1\, m$ is oscillating with an angular frequency $10\, rad/s$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1\, rad/s$ and an amplitude of $10^{-2}\, m$. The relative change in the angular frequency of the pendulum is best given by

The period of a simple pendulum is doubled, when

To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that

A pendulum clock that keeps correct time on the earth is taken to the moon it will run (it is given that $g_{Moon} = g_{Earth}/6$ )

A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A.$ Its speed as it passes through the equilibrium position is $V.$ If extended $2A$ and released, the speed of the mass passing through the equilibrium position will be

The acceleration due to gravity on the surface of moon is $1.7 \;ms ^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period (in $sec$) on the surface of earth is $3.5\; s ?( g$ on the surface of earth is $9.8\; ms ^{-2} )$