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

The bob of a simple pendulum executes simple harmonic motion in water with a period $t$,while the period of oscillation of the bob is ${t_0}$ in air. Neglecting the frictional force of water and given that the density of the bob is $(4/3) \times 1000 \ kg/m^3$. What relationship between $t$ and ${t_0}$ is true?

$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

$A$ cylindrical block of wood (density $= 650 \ kg \ m^{-3}$),of base area $30 \ cm^2$ and height $54 \ cm$,floats in a liquid of density $900 \ kg \ m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... $cm$ (nearly).

$A$ 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

$A$ simple pendulum is made of a body which is a hollow sphere containing mercury, suspended by means of a wire. If a little mercury is drained off, the period of the pendulum will