Question

Write an expression for the time period of simple pendulum.​

Answer 1

A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º. The period of a simple pendulum is T=2π√Lg T = 2 π L g , where L is the length of the string and g is the acceleration due to gravity.

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Answer 2

Answer:

2π × √(L/g)

Explanation:

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 ≈ θ,

Therefore, Iα = -mgLθ

α = -(mgLθ)/I

– ω02 θ = -(mgLθ)/I

ω02 = (mgL)/I

ω0 = √(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)

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