1. Simple and Compound Pendulum
Prepared by;
Dr. Rajesh Sharma
Assistant Professor
Dept of Physics
P.G.G.C-11, Chandigarh

2. Simple Pendulum
Let the bob of the pendulum be displaced through a linear displacement ‘x’ and angular displacement  from its equilibruim position O to A.
When displacement x is taken:
At position A, the forces acting on the bob after resolving mg, are:
(i) mg cos, the horizontal component
(ii) mg sin, the vertical component
Since mg cos equals and opposite to the tension T, hence they cancel each other.
The net force acting on the bob towards its equilibrium = - mg cos = F.
This force is called Restoring force
If  is very small,
mgcos
mgsin mg  l O A x

3. Hence,
But, from the figure we have,
Or, (1)
Therefore, (2)
Or,
Or, (3)
Now, Acceleration
or,
We know, the time period of pendulum executing SHM is given by

4. Frequency
Now,
Or, (4)
II. When the displacement  is taken:
The torque acting on the pendulum
Or,
If, be the angular acceleration produced, then or
Where I = ml2, is the moment of inertia of the bob about horizontal axis passing through the centre of point of suspension

5. We have
Or,
If  is very small then
We have,
Or, (5)
Thus, the motion of the bob is SHM.
The time period is given by
But,
Or, (6)
Which is same as that given in eq.(4).

6. Compound Pendulum
Definition: A rigid body of any shape, capable of oscillating about a horizontal axis passing through it in a vertical plane is called a Compound Pendulum.
Centre of Suspension: The point through which the vertical plane passing through the centre of gravity of the pendulum meets the axis of rotation.
The distance between the point of suspension and the C.G. of the pendulum is called the length of the pendulum.
Equation of motion of the compound pendulum:
Let an arbitrary shaped rigid body of mass m is capable of oscillating freely about a horizontal axis passing through it perpendicular to its plane. mg G
l Sin l S  G’

7. Suppose S be the point of suspension of the body and let G be the C. G
Suppose S be the point of suspension of the body and let G be the C.G. of the body at a distance l from the point S. let  be the small angular displacement of the body.
Let G’ be the new C.G. of the body at this displaced position.
The following forces are acting on the body in displaced position:
(i) Weight of the body mg acting vertically downwards at G’.
(ii) The force of reaction equal to mg acting vertically upwards at S.
These forces constitute a Couple. This couple tends to bring the pendulum back to its mean position.
So, Moment of Restoring Couple
(1)
The –ve sign shows that torque is directed opposite to the increasing direction of displacement .
If be the angular acceleration and I be the moment of inertia of the pendulum about the axis passing through the point of suspension S.

8. Then the restoring couple is given by
(2)
Now comparing Eq.(1) and (2), we get
Or,
If the amplitude  is very small and hence
Therefore, (3)
The Eq.(3) is the Differential equation of the SHM.
Since, is constant so,
i.e. The Angular acceleration is proportional to angular displacement.
Hence, the motion of the pendulum is SHM.

9. Time Period