$A$ simple pendulum,suspended from the ceiling of a lift,has a period of oscillation $T$,when the lift is at rest. If the lift starts moving upwards with an acceleration $a = 3g$,then the new period will be

In an experiment for determining the gravitational acceleration $g$ of a place with the help of a simple pendulum,the measured time period square $(T^2)$ is plotted against the string length $(L)$ of the pendulum as shown in the figure. What is the value of $g$ at the place in $m/s^2$?

The ratio of the frequencies of two simple pendulums is $4: 3$ at the same place. The ratio of their respective lengths is

$A$ simple pendulum is vibrating in an evacuated chamber. It will oscillate with:

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?

Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed a number of oscillations equal to