## What is called a Compound Pendulum?

*A compound pendulum is a rigid body pegged and hanged freely to oscillate about a smooth horizontal axis passing through the peg.*

Consider about a rigid body ( compound pendulum ) as shown in figure. The rigid body is hanged through a peg at point ( O ) so that, it can oscillate freely about an horizontal axis passing through the peg ( O ) . The centre point of axis through the point ( O ) is called the **centre of suspension.**

In the figure, the point ( G ) is the *centre of gravity* of body. Now let, ( OG = h ) and ( \angle {AOG} = \theta )

Then *moment* of weight of body about point ( O ) will be –

M = W × OG \sin \theta = W h \sin \theta = W h \theta ( Because ( \theta ) is very small )

From definition of a *torque* –

\text {Moment or Torque} = \text {Moment of inertia} \times \text {Angular acceleration}

Therefore, moment of weight of the body will be –

M = \left ( \frac {W}{g} \right ) {k_0}^2 \times \left ( \frac {d^2 \theta}{dt^2} \right ) . Here, ( k_0 ) is the *radius of gyration* of body about ( O ) .

Hence, \quad \left ( \frac {W}{g} \right ) {k_0}^2 \times \left ( \frac {d^2 \ \theta}{dt^2} \right ) = - W h \theta

So, \quad \left ( \frac {d^2 \ \theta}{dt^2} \right ) = - \left ( \frac {g h}{k_0^2} \right ) \theta

Therefore, angular acceleration of a compound pendulum is –

\alpha = - \left ( \frac {g h}{k_0^2} \right ) \theta

But, for a given body forming the compound pendulum, the terms ( g ), \ ( h ) \ \text {and} \ ( k ) are constants. Therefore, \quad \alpha \propto \theta

*Hence, the angular acceleration of body is proportional to angular displacement. Therefore, motion of body is Simple Harmonic Motion.*

### Time Period of Compound Pendulum

Time period of a *Simple Harmonic Motion* is given by –

T = 2 \pi \left ( \frac {\text {Displacement}}{\text {Acceleration}} \right )

Angular acceleration of a compound pendulum is given by –

\alpha = \left ( \frac {g h}{k_0^2} \right ) \theta

Therefore, *Time period* of oscillation of compound pendulum will be –

T = 2 \pi \sqrt {\left ( \frac {k_0^2}{gh} \right )} ……. (1)

*A simple pendulum whose time period of oscillation is the same as that of a compound pendulum, is called its simple equivalent pendulum.*

Let, ( l ) is the length of the simple equivalent pendulum, then its period of oscillation will be –

T = 2 \pi \sqrt {\left ( \frac {l}{g} \right )} ……. (2)

Hence, equating equations (1) and (2), we get –

2 \pi \sqrt {\left ( \frac {k_0^2}{gh} \right )} = 2 \pi \sqrt {\left ( \frac {l}{g} \right )}

Or, \quad l = \left ( \frac {k_0^2}{h} \right )

If, ( k ) is the radius of gyrations about a parallel axis passing through ( G ) . Then –

k_0^2 = k^2 + h^2

So, \quad l = \left ( \frac {k^2 + h^2}{h} \right )

= h + \left ( \frac {k^2}{h} \right )

### Centre of Oscillation

A point ( C ) is taken on produced part of line ( OG ) , such that ( OC = l )

*Then point ( C ) is called the centre of oscillation or percussion.*

Therefore, \quad OC = h + \frac {k_G^2}{h}

But, \quad k_0^2 = k^2 + h^2 . Therefore, \quad k^2 = k_0^2 - h^2 = l h - h^2 = h ( l - h )

= OG ( OC - OG ) = OG \times GC

*Thus, symmetry of the result shows that, if the body were suspended from a parallel axis through ( C ) , then ( O ) will become the centre of oscillation or percussion.*

Hence, centre of oscillation ( C ) and point of suspension ( O ) are inter-changeable.

## Conical Pendulum

*One end of a string ( AO ) is attached to a fixed point ( O ) and a weight ( W ) is tied to the other end ( A ) . If the weight is made to describe a horizontal circle so that the string describes a right circular cone, whose axis is a vertical through ( O ) , then system is called a conical pendulum.*

Consider about a conical pendulum as shown in figure. Let –

- ( N ) is the centre of horizontal circle of radius ( r ) described by weight.
- ( \omega ) is the angular velocity of the weight.
- ( T ) is the
*tension*of the string. - ( l ) is the length of the string and ( \angle {AON} = \theta )

Then \quad r = l \sin \theta

Resolving tension of the string, we will get –

- Vertical component is \quad T \cos \theta = W …… (1)
- Horizontal component is \quad T \sin \theta = F_C …… (2)

The horizontal component of tension provides the *centripetal force* ( F_C ) on the weight which is necessary to keep it moving in circle.

Then from definition of centripetal force, we have –

F_C = \left ( \frac {W}{g} \right ) \omega^2 r

Therefore, \quad T \sin \theta = \left ( \frac {W}{g} \right ) \omega^2 r = \left ( \frac {W}{g} \right ) \omega^2 l \sin \theta

So, \quad T = \left ( \frac {W}{g} \right ) \omega^2 l ……. (3)

Now dividing equation (1) by equation (3), we will get –

\cos \theta = \left ( \frac {g}{\omega^2 l} \right )

Here, ( \theta ) will be real if ( \theta < 1 )

This means that –

\left ( \frac {g}{\omega^2 l} \right ) < 1

So, \quad \omega^2 l > g

Or, \quad \omega > \sqrt {\left (\frac {g}{l} \right )}

This expression provides a necessary condition for angular velocity of a conical pendulum. It states that –

*The minimum angular velocity must be \left [ \sqrt {\left ( \frac {g}{l} \right )} \right ] so that the weight may revolve in a conical pendulum. If ( \omega ) is less than this value, the string will hang vertically.*

### Use of Conical Pendulum

From the geometry of the figure –

ON = l \cos \theta = l \left ( \frac {g}{\omega^2 l} \right )

= l \times \left ( \frac {g}{\omega^2 l} \right ) = \left ( \frac {g}{\omega^2} \right )

So, \quad ON \propto \left ( \frac {1}{\omega^2} \right )

*Therefore, the depth of weight below ( O ) is inversely proportional to the square of the angular velocity and is independent of the length of the string. Hence, it is used as a model to analyze the motion of planets.*

__Oscillations of a liquid column__

*When the surface of a liquid column kept in a U tube is displaced and released, the liquid column starts oscillating with Simple Harmonic Motion.*

Consider about a U tube as shown in figure. Let –

- Cross section area of tube is ( A ) .
- Density of liquid in tube is ( \rho ) .

Height of liquid in both arms of tube is ( h ) . Mass of liquid column in the U tube will be-

m = A \times 2 h \times \rho

Let, the liquid in one of the arm is depressed by a distance ( y ) . Then rise of the liquid in the other arm will be the same. Therefore, total difference in height of liquid column in both arms will become ( 2y ) .

*Restoring force* acting on the liquid column will be –

F = Weight of unbalanced liquid column.

= - A \times 2y \times \rho \times g = - 2 A \rho g y

= - \left ( 2 A \rho g \right ) y

But, ( A ), \ ( \rho ) \ \text {and} \ ( g ) are constants. Therefore, \quad F \propto y

*The force on the liquid column is proportional to the displacement and acts in opposite direction. Hence, the liquid in the tube executes Simple Harmonic Motion.*

### Force Constant of liquid column

Restoring force in a simple harmonic motion is given by –

F = - k y

Restoring force in a liquid column is given by –

F = - \left ( 2 A \rho g \right ) y

Comparing the equations, *force constant* for liquid column will be –

k = \left ( 2 A \rho g \right )

### Time Period of liquid column

Time period of oscillations of liquid column will be –

T = 2 \pi \sqrt { \frac { m }{ k }}

= 2 \pi \sqrt { \frac { A \times 2 h \times \rho }{ 2 A \rho g }}

= 2 \pi \sqrt { \frac { h }{ g }}

Let, ( l ) * *is the total length of the liquid column. Then –

l = 2 h

Therefore, \quad T = 2 \pi \sqrt { \frac { l }{ 2 g }}