The time period of a simple pendulum will be doubled if we:

$A$ piece of wood has length,breadth and height,'$a$','$b$' and '$c$' respectively. Its relative density is '$d$'. It is floating in water such that the side '$a$' is vertical. It is pushed down a little and released. The time period of $S.H.M.$ executed by it is ($g=$ acceleration due to gravity).

State the laws of a simple pendulum.

$A$ tunnel has been dug through the centre of the Earth and a ball is released in it. It will reach the other end of the tunnel after:

The bob of a pendulum of mass $m$, suspended by an inextensible string of length $L$ as shown in the figure, carries a small charge $q$. An infinite horizontal plane conductor with uniform surface charge density $\sigma$ is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

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?