Question

A simple pendulum consists of a metallic bob suspended from a long straight thread whose one
end is fixed to a rigid support.

1. What is the time period of second’s pendulum?

2. Derive an expression for period of simple pendulum.

Answer 1

1. 2 sec

2.

Consider a mass m suspended from one end of a string of length L fixed at the other end as shown in figure. Suppose P is the instantaneous position of the pendulum. At this instant its string makes an angle θ with the vertical.

The forces acting on the bob are (1) weight of bob F_g(mg) acting vertically downward. (2) Tension T in the string.

The gravitational force F_g can be divided into a radial component F_gCosθ and tangential component

F_gSinθ. The tangential component F_gSinθ produces a restoring torque.

Restoring torque τ = – F_g sinθ . L

τ = – m_g sinθ . L _____(1)

-ve sign shown that the torque and angular displacement θ are oppositely directed. For rotational motion of bob,

τ = Iα ______(2)

Where I is moment of inertia about the point of suspension and α is angular acceleration. From eq (1) and eq (2).

Iα = – mgsinθ . L

If we assume that the displacement θ is small, sinθ ≈ θ

∴ Iα = -mgθ . L

Iα + mgθ L = 0

Compairing eq (3) with standard differential equation

∴ period of simple pendulum,

for simple pendulum I = mL²

Substituting I = mL² we get

The above equation gives that T is independent of mass.