Question

Derive the equation for period of oscillation of a simple petulant. The period of oscillation (T) depends on
(1)Length of the pendulum(l)
(2)Mass of the Bob(m)
(3)Acceleration due to gravity(g)
​

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

– 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)

Answer 2

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