$A$ simple pendulum hanging from the ceiling of a stationary lift has a time period $T_1$. When the lift moves downward with constant velocity,the time period is $T_2$,then

$A$ pendulum suspended from the ceiling of a train oscillates with a time period $2 \ s$,when the train is accelerating at $10 \ m/s^2$. What will be its time period when the train retards at $10 \ m/s^2$?

If the length of a seconds pendulum becomes $\frac{1}{3}$ of its original length,what will be its new periodic time?

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

$A$ trolley of mass $m_1$ is placed on a horizontal rigid pair of rails at the same height. $A$ mass $m_2$ is suspended from the trolley vertically by means of an ideal massless rope. The rope hangs between the rails without touching them. The trolley can move along smooth rails but cannot move in any other direction. The suspended mass is given a small oscillation and performs $SHM$ after being displaced slightly from its stable equilibrium position in two ways: first,perpendicular to the rails,and second,parallel to the rails. The ratio of the time periods of these two $SHM$s (second case to first case) is:

$A$ simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle $\theta$ with the vertical,the angle $\phi$ which the acceleration vector of the bob makes with the string is given by