$A$ simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$. The time period is given by $T = 2\pi \sqrt{\frac{l}{g'}}$,where $g'$ is equal to:

$A$ light balloon filled with helium of density $\rho_{He}$ is tied to a light string of length $L.$ The string is tied to the ground forming an "inverted" simple pendulum (figure). If the balloon is displaced slightly from equilibrium as in figure and released, the period of the motion is. Take the density of air to be $\rho_{air}$. Assume the air applies a buoyant force on the balloon but does not otherwise affect its motion.

If a simple pendulum is released from the given position,find the velocity of the bob when it reaches the lowest position. (Take $g = 10 \ m/s^2$) (in $m/s$)

The angular amplitude of a simple pendulum is $\theta_0$. The maximum tension in its string will be

$A$ string of a pendulum of length $l$ is displaced through $90^{\circ}$ from its vertical and released. Then,the minimum strength of the string needed to withstand the tension as the pendulum passes through its mean position is

$A$ coin is placed on a horizontal plate. The plate performs $S.H.M.$ vertically with an angular frequency $\omega$. The amplitude $A$ of oscillations is gradually increased. The coin will lose contact with the plate for the first time when the amplitude is ($g =$ acceleration due to gravity).