The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

What is the number of degrees of freedom for an oscillating simple pendulum?

$A$ pendulum has a time period $T$. If it is taken to another planet having an acceleration due to gravity half that of the Earth and a mass $9$ times that of the Earth,then its time period on the other planet will be:

$A$ cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period $T$. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period $2T$. The value of $\rho_2/\rho_1$ is :

$A$ simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2$ $(l_1 > l_2)$ has time period $T_2$. Then the time period of the pendulum of length $(l_1 - l_2)$ will be

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \text{ m}$. The period of oscillation of the sphere is (Acceleration due to gravity $g = 10 \text{ m/s}^2$)