Question

Derive an expression for the time period and frequency of a simple pendulum.​

Answer 1

Time Period of Simple Pendulum

A point mass M suspended from the end of a light inextensible string whose upper end is fixed to a rigid support. The mass displaced from its mean position.

Assumptions:

There is negligible friction from the air and the system

The arm of the pendulum does not bend or compress and is massless

The pendulum swings in a perfect plane

Gravity remains constant

Simple Pendulum

Time Period of Simple Pendulum Derivation

Using the equation of motion, T – mg cosθ = mv2L

The torque tending to bring the mass to its equilibrium position,

τ = mgL × sinθ = mgsinθ × L = I × α

For small angles of oscillations sin ≈ θ,

Therefore, Iα = -mgLθ

α = -(mgLθ)/I

– ω02 θ = -(mgLθ)/I

ω02 = (mgL)/I

ω20 = √(mgL/I)

Using I = ML2, [where I denote the moment of inertia of bob]

we get, ω0 = √(g/L)

Therefore, the time period of a simple pendulum is given by,

T = 2π/ω0 = 2π × √(L/g)