Answer:
Time period for Simple pendulum, 
Explanation:
The Simple Pendulum
Consider a small bob of mass
is tied to extensible string of length
that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let
is the angle made by string with vertical during oscillation.
Vertical component of the force on bob,
Negative sign shows that its opposing the motion of bob.
Taking
as very small angle then, 
Let
is the displacement made by bob from its mean position ,
then, 
so,
........(1)
Since, pendulum is in hormonic motion,
as we know, 
where
is the constant and 
.........(2)
From equation (1) and (2)


Since, 

