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

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 bob of a simple pendulum is a spherical hollow ball filled with water. $A$ plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation,till water is coming out,the time period of oscillation would:

$A$ simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in a horizontal position from the wall and is left free to move. The bob hits the wall,and the coefficient of restitution is $\frac{2}{\sqrt{5}}$. After how many collisions will the amplitude of vibration become less than $60^\circ$?

Two masses,both equal to $100\, g$,are suspended at the ends of identical light strings of length $\ell = 1.0\, m$,attached to the same point on the ceiling (see figure). At time $t = 0$,they are simultaneously released from rest,one at angle $\theta_1 = 1^o$,the other at angle $\theta_2 = 2^o$ from the vertical. The masses will collide

If the length of a clock pendulum increases by $0.2 \%$ due to atmospheric temperature rise,then the loss in time of the clock per day is ........... $s$.