Modelling of a simple pendulum

Introduction

A simple pendulum is a weight suspended by a non-deformable wire. In this post, we'll write the equations of the system. We'll then solve the equations to calculate the oscillation frequency.

In this post, we will assume the following hypotheses:

• The weight can swing freely.
• $$L$$ is the wire’s lenght.
• $$m$$ is the mass of the pendulum’s body.
• The mass of the wire is neglected.
• The frictions are neglected.

Equation of the system

To find the equation of the system, we will apply the fundamental principle of dynamics:

$$m.\vec{a} = \sum{\vec{F}}$$

Forces acting on the system are the gravity force $$\vec{F_g}$$ and the wire tension force $$\vec{T}$$. The body's trajectory describes an arc of circle. The body's acceleration is tangent to this trajectory. The wire tension force is perpendicular to the body's trajectory. Its projection along the motion axis is null. Only one force have to be considered in the following : the gravity force.

$$m.a = \vec{F_m} = m.g.sin(\Theta)$$

The relation between angular and linear velocity is given by:

$$v=L\frac{d \Theta }{dt}$$

By deriving the previous equation, we get:

$$a=\dfrac{dv}{dt}=L. \dfrac{d^2 \Theta }{dt^2}$$

The equation calculated from the fundamental principle of dynamics can be updated:

$$m.L. \frac{d^2 \Theta }{dt^2} = m.g.sin(\Theta)$$

Dividing by $$m$$:

$$L. \frac{d^2 \Theta }{dt^2} = g.sin(\Theta)$$

The differential equation of the system is:

$$\frac{d^2 \Theta }{dt^2} = \frac{g}{L}.sin(\Theta)$$

Solving the equation

The previous differential equation is non linear. A commonly used approximation is to approximate $$sin(\Theta)$$ by $$\Theta$$ for small values of $$\Theta$$. The equation becomes linear:

$$\frac{d^2 \Theta }{dt^2} = \frac{g}{L}.\Theta$$

The general solution to the previous differential equation is:

$$\Theta = A.cos(\omega.t + \phi)$$

with:

$$\omega^2=\frac{g}{L}$$

Given the initial conditions $$\Theta=\Theta_0$$ (initial angle of the pendulum), and assuming that the initial speed is null:

$$A.cos(\phi)=\Theta_0$$ $$-A.\omega.sin(\phi)=0$$

The previous system can be solved:

$$A=\Theta_0$$ $$\phi=0$$

The general solution of the system is :

$$\Theta(t)=\Theta_0.cos(\omega.t)$$

Where $$\omega = \sqrt { \frac {g}{L} }$$. The period of the motion is given by:

$$T=2.\pi.\sqrt { \frac {g}{L}}$$

Acknowledgements

I want to thank my coworker and friend Jean-Pierre Pecqueur from the University of Angers for his help.