3.10: Appendix - Period of the Nonlinear Pendulum

In Section 3.5.1 we saw that the solution of the nonlinear pendulum problem can be found up to quadrature. In fact, the integral in Equation (3.19) can be transformed into what is know as an elliptic integral of the first kind. We will rewrite our result and then use it to obtain an approximation to the period of oscillation of our nonlinear pendulum, leading to corrections to the linear result found earlier.

We will first rewrite the constant found in (3.18). This requires a little physics. The swinging of a mass on a string, assuming no energy loss at the pivot point, is a conservative process. Namely, the total mechanical energy is conserved. Thus, the total of the kinetic and gravitational potential energies is a constant. Noting that \(v = L \dot{\theta}\), the kinetic energy of the mass on the string is given as

\[T = \dfrac{1}{2}mv^2 = \dfrac{1}{2}mL^2 \dot{\theta} {}^2. \nonumber \]

The potential energy is the gravitational potential energy. If we set the potential energy to zero at the bottom of the swing, then the potential energy is \(U=m g h\), where \(h\) is the height that the mass is from the bottom of the swing. A little trigonometry gives that \(h=L(1-\cos \theta)\). This gives the potential energy as

\[U=m g L(1-\cos \theta). \nonumber \]

So, the total mechanical energy is

\[E=\dfrac{1}{2} m L^{2} \theta^{2}+m g L(1-\cos \theta) \label{3.56} \]

We note that a little rearranging shows that we can relate this to Equation (3.18):

\[\dfrac{1}{2}\left(\theta^{\prime}\right)^{2}-\omega^{2} \cos \theta=\dfrac{1}{m L^{2}} E-\omega^{2}=c. \nonumber \]

We can use Equation (3.56) to get a value for the total energy. At the top of the swing the mass is not moving, if only for a moment. Thus, the kinetic energy is zero and the total energy is pure potential energy. Letting \(\theta_{0}\) denote the angle at the highest position, we have that

\[E=m g L\left(1-\cos \theta_{0}\right)=m L^{2} \omega^{2}\left(1-\cos \theta_{0}\right). \nonumber \]

Here we have used the relation \(g=L \omega^{2}\).
Therefore, we have found that

\[\dfrac{1}{2} \dot{\theta}^{2}-\omega^{2} \cos \theta=\omega^{2}\left(1-\cos \theta_{0}\right) \label{3.57} \]

Using the half angle formula,

\[\sin ^{2} \dfrac{\theta}{2}=\dfrac{1}{2}(1-\cos \theta) \nonumber \]

we can rewrite Equation (3.57) as

\[\dfrac{1}{2} \dot{\theta}^{2}=2 \omega^{2}\left[\sin ^{2} \dfrac{\theta_{0}}{2}-\sin ^{2} \dfrac{\theta}{2}\right] \label{3.58} \]

Solving for \(\theta^{\prime}\), we have

\[\dfrac{d \theta}{d t}=2 \omega\left[\sin ^{2} \dfrac{\theta_{0}}{2}-\sin ^{2} \dfrac{\theta}{2}\right]^{1 / 2} \label{3.59} \]

One can now apply separation of variables and obtain an integral similar to the solution we had obtained previously. Noting that a motion from $\theta=0$ to \(\theta=\theta_{0}\) is a quarter of a cycle, then we have that

\[T=\dfrac{2}{\omega} \int_{0}^{\theta_{0}} \dfrac{d \phi}{\sqrt{\sin ^{2} \dfrac{\theta_{0}}{2}-\sin ^{2} \dfrac{\theta}{2}}} \label{3.60} \]

This result is not much different than our previous result, but we can now easily transform the integral into an elliptic integral. We define

\[z=\dfrac{\sin \dfrac{\theta}{2}}{\sin \dfrac{\theta_{0}}{2}} \nonumber \]

and

\[k=\sin \dfrac{\theta_{0}}{2}. \nonumber \]

Then Equation (3.60) becomes

\[T=\dfrac{4}{\omega} \int_{0}^{1} \dfrac{d z}{\sqrt{\left(1-z^{2}\right)\left(1-k^{2} z^{2}\right)}} \label{3.61} \]

This is done by noting that \(d z=\dfrac{1}{2 k} \cos \dfrac{\theta}{2} d \theta=\dfrac{1}{2 k}\left(1-k^{2} z^{2}\right)^{1 / 2} d \theta\) and that \(\sin ^{2} \dfrac{\theta_{0}}{2}-\sin ^{2} \dfrac{\theta}{2}=k^{2}\left(1-z^{2}\right)\). The integral in this result is an elliptic integral of the first kind. In particular, the elliptic integral of the first kind is defined as

\[F(\phi, k) \equiv=\int_{0}^{\phi} \dfrac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=\int_{0}^{\sin \phi} \dfrac{d z}{\sqrt{\left(1-z^{2}\right)\left(1-k^{2} z^{2}\right)}} \nonumber \]

In some contexts, this is known as the incomplete elliptic integral of the first kind and \(K(k)=F\left(\dfrac{\pi}{2}, k\right)\) is called the complete integral of the first kind.
There are tables of values for elliptic integrals. Historically, that is how one found values of elliptic integrals. However, we now have access to computer algebra systems which can be used to compute values of such integrals. For small angles, we have that \(k\) is small. So, we can develop a series expansion for the period, \(T\), for small \(k\). This is done by first expanding

\[\left(1-k^{2} z^{2}\right)^{-1 / 2}=1+\dfrac{1}{2} k^{2} z^{2}+\dfrac{3}{8} k^{2} z^{4}+O\left((k z)^{6}\right) \nonumber \]

Substituting this in the integrand and integrating term by term, one finds that

\[T=2 \pi \sqrt{\dfrac{L}{g}}\left[1+\dfrac{1}{4} k^{2}+\dfrac{9}{64} k^{4}+\ldots\right] \label{3.62} \]

This expression gives further corrections to the linear result, which only provides the first term. In Figure 3.37 we show the relative errors incurred when keeping the \(k^{2}\) and \(k^{4}\) terms versus not keeping them. The reader is asked to explore this further in Problem 3.8.

Figure 3.37. The relative error in percent when approximating the exact period of a nonlinear pendulum with one, two, or three terms in Equation (3.62).