Question

Define simple pendulum and derive it's expestion time period

Answer 1

You need to remember the formula of constant of simple pendulum such that:

k = L/T^2

You need to consider the value of constant equivalent to g/(4pi^2) (g expresses the gravity acceleration)

You need to set the equations g/(4pi^2) and L/T^2 equal such that:

 an ideal pendulum consisting of a point mass suspended by a weightless inextensible perfectly flexible thread and free to vibrate without friction

derivation==--

L/T^2 = g/(4pi^2)

You need to find time period such that:

g*T^2 = 4pi^2*L

T^2 = (4pi^2*L)/g => T = 2pi*sqrt(L/g)

Hence, evaluating the time period of simple pendulum under given conditions yields T = 2pi*sqrt(L/g).

Answer 2

To and Fro motion of the pendulum in two second is called Second pendulum