Equation of motion of compound pendulum:

Let us consider a compound pendulum is suspended in a vertical plane, from a horizontal axis through the point O as shown in the Fig.1. Let G be the centre of mass of the compound pendulum and \( l \) be the distance of centre of gravity from the axis through the point O.

When the body is at the rest position, the line OG is vertical. If the body is displaced through an angle \( \theta \) then the centre of gravity of the body shifts to the new position G’. If \( M \) be the mass of the compound pendulum then a couple of moment \( Mgl\sin\theta \) acts on the body due to its weight \( Mg \), where \( g \) is the acceleration due to gravity. The upward reaction at O tends to restore the body to its original position. So the equation of motion of the body is given by,

\( I\frac{d^2\theta}{dt^2}=-Mgl\sin\theta \)
[where, \( I \) is the moment of inertia of the body about the axis of rotation through the point O]

Since \( \theta \) is very small, so \( \sin\theta=\theta \)

Therefore, \( I\frac{d^2\theta}{dt^2}=-Mgl\theta \)

or, \( I\frac{d^2\theta}{dt^2}+Mgl\theta=0 \)

or, \( \displaystyle{\frac{d^2\theta}{dt^2}+\frac{Mgl}{I}\theta=0} \)

This is the equation of motion of compound pendulum.

Time period of compound pendulum:

If \( MK^2 \) be the moment of inertia of the body about an axis through the point \( G \) and parallel to the axis of rotation through the point \( O \), then from theorem of parallel axes we can write

\( I=MK^2+Ml^2 \)

or, \( I=M(K^2+l^2) \)

where K is the radius of gyration.

Since the equation of motion of the of the compound pendulum is

\( \displaystyle{\frac{d^2\theta}{dt^2}+\frac{Mgl}{I}\theta=0} \)

So the time period of oscillation of the compound pendulum is

\( \displaystyle{T=2\pi\sqrt{\frac{I}{Mgl}}} \)

Putting the value of \( I \) we get,

\( \displaystyle{T=2\pi\sqrt{\frac{M(K^2+l^2)}{Mgl}}} \)

or, \( \displaystyle{T=2\pi\sqrt{\frac{K^2+l^2}{gl}}} \)