$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ (in $g$)? ($g$ is acceleration due to gravity).

The time period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration $g/3$,the new time period is:

Two simple pendulums of equal length cross each other at the mean position. What is their phase difference?

$A$ pendulum suspended from the ceiling of a train has a period $T$ when the train is at rest. When the train travels the same distance per unit time,the period of oscillation is

Answer the following questions:
$(a)$ The time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle: $T=2 \pi \sqrt{\frac{m}{k}}$. $A$ simple pendulum executes $SHM$ approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation,a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$. Think of a qualitative argument to appreciate this result.
$(c)$ $A$ man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).