3.5: Nonlinear Pendulum

In this section we will introduce the nonlinear pendulum as our first example of periodic motion in a nonlinear system. Oscillations are important in many areas of physics. We have already seen the motion of a mass on a spring, leading to simple, damped, and forced harmonic motions. Later we will explore these effects on a simple nonlinear system. In this section we will introduce the nonlinear pendulum and determine its period of oscillation.

We begin by deriving the pendulum equation. The simple pendulum consists of a point mass $m$ hanging on a string of length $L$ from some support. [See Figure 3.11.] One pulls the mass back to some starting angle, $\theta_{0}$, and releases it. The goal is to find the angular position as a function of time, $\theta(t)$.

Figure 3.11. A simple pendulum consists of a point mass $m$ attached to a string of length $L$. It is released from an angle $\theta_{0}$.

There are a couple of derivations possible. We could either use Newton's Second Law of Motion, $F=m a$, or its rotational analogue in terms of torque. We will use the former only to limit the amount of physics background needed. There are two forces acting on the point mass, the weight and the tension in the string. The weight points downward and has a magnitude of $m g$, where $g$ is the standard symbol for the acceleration due to gravity. At the surface of the earth we can take this to be $9.8 \mathrm{~m} / \mathrm{s}^{2}$ or $32.2 \mathrm{ft} / \mathrm{s}^{2}$. In Figure 3.12 we show both the weight and the tension acting on the mass. The net force is also shown.

The tension balances the projection of the weight vector, leaving an unbalanced component of the weight in the direction of the motion. Thus, the magnitude of the sum of the forces is easily found from this unbalanced component as $F=m g \sin \theta$.

Newton's Second Law of Motion tells us that the net force is the mass times the acceleration. So, we can write

$$m \ddot{x}=-m g \sin \theta \nonumber$$

Next, we need to relate $x$ and $\theta$. $x$ is the distance traveled, which is the length of the arc traced out by our point mass. The arclength is related to the angle, provided the angle is measured in radians. Namely, $x=r \theta$ for $r=L$. Thus, we can write

Figure 3.12. There are two forces acting on the mass, the weight $mg$ and the tension $T$. The magnitude of the net force is found to be $F = mg \sin \theta$.

$$m L \ddot{\theta}=-m g \sin \theta \nonumber$$

Canceling the masses, leads to the nonlinear pendulum equation

$$L \ddot{\theta}+g \sin \theta=0 \label{3.8}$$

There are several variations of Equation (3.8) which will be used in this text. The first one is the linear pendulum. This is obtained by making a small angle approximation. For small angles we know that $\sin \theta \approx \theta$. Under this approximation (3.8) becomes

$$L \ddot{\theta}+g \theta=0 \label{3.9}$$

We can also make the system more realistic by adding damping. This could be due to energy loss in the way the string is attached to the support or due to the drag on the mass, etc. Assuming that the damping is proportional to the angular velocity, we have equations for the damped nonlinear and damped linear pendula:

$$L \ddot{\theta}+b \dot{\theta}+g \sin \theta=0 \label{3.10}$$

$$L \ddot{\theta}+b \dot{\theta}+g \theta=0 \label{3.11}$$

Finally, we can add forcing. Imagine that the support is attached to a device to make the system oscillate horizontally at some frequency. Then we could have equations such as

$$L \ddot{\theta}+b \dot{\theta}+g \sin \theta=F \cos \omega t \label{3.12}$$

We will look at these and other oscillation problems later in the exercises. These are summarized in the table below.

Equations for Pendulum Motion

1. Nonlinear Pendulum: $L \ddot{\theta}+g \sin \theta=0$.
2. Damped Nonlinear Pendulum: $L \ddot{\theta}+b \dot{\theta}+g \sin \theta=0$.
3. Linear Pendulum: $L \ddot{\theta}+g \theta=0$.
4. Damped Linear Pendulum: $L \ddot{\theta}+b \dot{\theta}+g \theta=0$.
5. Forced Damped Nonlinear Pendulum: $L \ddot{\theta}+b \dot{\theta}+g \sin \theta= F \cos \omega t$.
6. Forced Damped Linear Pendulum: $L \ddot{\theta}+b \dot{\theta}+g \theta=F \cos \omega t$.

3.5.1 In Search of Solutions

Before returning to studying the equilibrium solutions of the nonlinear pendulum, we will look at how far we can get at obtaining analytical solutions. First, we investigate the simple linear pendulum.

The linear pendulum equation (3.9) is a constant coefficient second order linear differential equation. The roots of the characteristic equations are $r= \pm \sqrt{\dfrac{g}{L}} i$. Thus, the general solution takes the form

$$\theta(t)=c_{1} \cos \left(\sqrt{\dfrac{g}{L}} t\right)+c_{2} \sin \left(\sqrt{\dfrac{g}{L}} t\right) \label{3.13}$$

We note that this is usually simplified by introducing the angular frequency

$$\omega \equiv \sqrt{\dfrac{g}{L}} \label{3.14}$$

One consequence of this solution, which is used often in introductory physics, is an expression for the period of oscillation of a simple pendulum. REcall that the period is the time it takes to complete one cycle of the oscillation. The period is found to be

$$T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{L}{g}} \label{3.15}$$

This value for the period of a simple pendulum is based on the linear pendulum equation, which was derived assuming a small angle approximation. How good is this approximation? What is meant by a small angle? We recall the Taylor series approximation of $\sin \theta$ about $\theta=0$:

$$\sin \theta=\theta-\dfrac{\theta^{3}}{3 !}+\dfrac{\theta^{5}}{5 !}+\ldots \label{3.16}$$

One can obtain a bound on the error when truncating this series to one term after taking a numerical analysis course. But we can just simply plot the relative error, which is defined as

Relative Error $=\left|\dfrac{\sin \theta-\theta}{\sin \theta}\right| \times 100 \%$.

A plot of the relative error is given in Figure 3.13. We note that a one percent relative error corresponds to about 0.24 radians, which is less that fourteen degrees. Further discussion on this is provided at the end of this section.

Figure 3.13. The relative error in percent when approximating $\sin \theta$ by $\theta$.

We now turn to the nonlinear pendulum. We first rewrite Equation (3.8) in the simpler form

$$\ddot{\theta}+\omega^{2} \sin \theta=0 \label{3.17}$$

We next employ a technique that is useful for equations of the form

$$\ddot{\theta}+F(\theta)=0 \nonumber$$

when it is easy to integrate the function $F(\theta)$. Namely, we note that

$$\dfrac{d}{d t}\left[\dfrac{1}{2} \dot{\theta}^{2}+\int^{\theta(t)} F(\phi) d \phi\right]=[\ddot{\theta}+F(\theta)] \dot{\theta} \nonumber$$

For our problem, we multiply Equation (3.17) by $\dot{\theta}$,

$$\ddot{\theta} \dot{\theta} + w^2 \sin \theta \dot{\theta} = 0 \nonumber$$

and note that the left side of this equation is a perfect derivative. Thus,

$$\dfrac{d}{d t}\left[\dfrac{1}{2} \dot{\theta}^{2}-\omega^{2} \cos \theta\right]=0. \nonumber$$

Therefore, the quantity in the brackets is a constant. So, we can write

$$\dfrac{1}{2} \dot{\theta}^{2}-\omega^{2} \cos \theta=c \label{3.18}$$

Solving for $\dot{\theta}$, we obtain

$$\dfrac{d \theta}{d t}=\sqrt{2\left(c+\omega^{2} \cos \theta\right)} \nonumber$$

This equation is a separable first order equation and we can rearrange and integrate the terms to find that

$$t=\int d t=\int \dfrac{d \theta}{\sqrt{2\left(c+\omega^{2} \cos \theta\right)}} \label{3.19}$$

Of course, one needs to be able to do the integral. When one gets a solution in this implicit form, one says that the problem has been solved by quadratures. Namely, the solution is given in terms of some integral. In the appendix to this chapter we show that this solution can be written in terms of elliptic integrals and derive corrections to formula for the period of a pendulum.