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

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?

Frequency $f$ of a simple pendulum depends on its length $\ell$ and acceleration $g$ due to gravity according to the following equation $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$. $A$ graph between which of the following quantities is a straight line?

The time period of a simple pendulum in a lift descending with constant acceleration $g$ is

$A$ simple pendulum of length $l$ has a brass bob attached at its lower end. Its period is $T$. $A$ steel bob of the same size,having density $x$ times that of brass,replaces the brass bob. Its length is then changed such that the period becomes $2T$. What is the new length?

In $S.H.M.$,a simple pendulum oscillates with frequency $f$. If the length of the pendulum is increased by three times its original length,then the frequency of oscillation of the pendulum will be