Question

Derive expression for time period of simple pendulum

Answer 1

Answer:

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

– ω0^2 θ = -(mgLθ)/I

ω0^2 = (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)

⚡Hope it works.⚡

Answer 2

Answer:

In harmonic motion there is always a restorative force, which acts in the opposite direction of the velocity. The restorative force changes during oscillation and depends on the position of the object. In a pendulum it is the component of gravity along the path of motion. The force on the oscillating object is directly opposite that of the direction of velocity.

For pendulums, the period gets larger as the length of the pendulum increases.

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