A first tour through examples - Chapter 5

# A new look at the pendulum Consider the motion of the following idealized pendulum: a bob of mass $m$ is attached to one end of a massless rigid rod. The other end of the rod is pivoted so that the mass may swing in a vertical plane. We neglect both the friction of the pivot and air drag.

The swinging of the pendulum is governed by

$$\ddot{\theta}=-\frac{g}{L} \sin \theta$$

where $\theta$ is the angle between the rod and the downward vertical. This equation is derived in all textbooks of classical mechanics.

Phase plane representation.
Let’s do the same trick as for the harmonic oscillator by representing the state of the pendulum as a point in the phase plane $x=\theta$, $y=\dot{\theta}$. This yields

$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2 \sin x \end{cases}$$

where $\omega^2=\frac{g}{L}$. (Notice that $\omega$ has the dimension of a frequency.)

Let’s draw the phase portrait!
We can now look at the trajectories of this system. In the following digital experiment, one can pull the mass away from its rest position, drop it (without speed) and see at once the resulting trajectory in the phase plane. One can also select a point in the phase plane and see at once the resulting motion of the pendulum. If one does not take a point on the $x$-axis, this means that one start with a nonzero velocity (we kick the mass).

Actually the equilibrium points are $(k\pi,0)$, where $k\in\mathbb{Z}$. The origin in the phase portrait corresponds to the stable equilibrium position of the pendulum hanging straight down. The points $(\pm k\pi,0)$, $k\in\mathbb{Z}\backslash \{0\}$, correspond to the unstable equilibrium position where the pendulum is straight up. Let’s restrict to the points $(-\pi,0)$, $(0,0)$, $(\pi,0)$, since the phase portrait is periodic in the $x$-direction (translating horizontally points of the phase portrait by a distance that is a multiple of $2\pi$ yields the same phase portrait).

We observe two generic behaviors which take place into two separate regions:

• closed trajectories encircling the origin, which describe the periodic motions of the pendulum swinging back and forth;
• non closed trajectories, which are periodic in the $x$-direction, and which describe motions where the mass whirls around the pivot endlessly.

These two kinds of trajectories are separated by a eye-like curve, called the separatrix. It can be (approximately) obtained by dropping the mass very close from its upside down rest position.

Notice that, mathematically, the true separatrix corresponds to a set of initial states that has area zero in the plane. It corresponds to exceptional motions such that the pendulum tends to the unstable equilibrium, which takes an infinite amount of time.

It is well-known that we have conservation of total energy (kinetic $+$ potential) because we neglect friction of the pivot and air resistance. As shown in any textbook of classical mechanics

$$\frac{\dot{\theta}^2}{2}-\omega^2(1-\cos\theta)=\text{const}$$

which means that each motion individually has its own energy which does not change in time. Translated into our phase plane language, this means that the trajectories of solutions in the $xy$-plane are level curves of the function

$$E(x,y)=\frac{y^2}{2}-\omega^2(1-\cos x).$$

The damped pendulum.
When we add to the system of the pendulum a perturbation term to represent air friction, a physically more realistic situation, energy is no more conserved : it decreases over time and we have damped oscillations. We assume that the friction is proportional to the velocity and call $\mu$ the friction coefficient. This gives

$$\ddot{\theta}=-\omega^2\sin \theta -\mu \dot{\theta}.$$

Passing to the phase plane representation we thus obtain

$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2\sin x -\mu y. \end{cases}$$

We now see that the equilibrium point $(0,0)$ is an attractor : all solutions converge toward this point. The corresponding trajectories have a spiraling behavior if $\mu$ is not too large. Of course we recover the phase portrait of the undamped pendulum when $\mu=0$.

Linear approximation.
Because of the sine function, the equation of the pendulum is nonlinear. However, if the angular deviation of the pendulum is very small, then it is tempting to approximate $\sin x$ by $x$, which leads to

$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2 x-\mu y. \end{cases}$$

We get the same equation as the one for the harmonic oscillator! This is in accordance with what can be seen when we compare the phase portrait of the harmonic oscillator to the one of the pendulum in a close neighborhood of $(0,0)$.
We have an instance of what a linear approximation can tell us about a nonlinear system, but only locally about an equilibrium point. We shall develop much further this approach later on.