Question

Derive an expression for time period of a simple pendulum.

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:

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

$\textbf{Simple Pendulum :}$

It is a arrangement of a heavy point mass suspended by a weightless, inextensible and perfectly flexible spring from a rigid support about which it is free to oscillate..