20.3: Example- Pendulum Driven at near Double the Natural Frequency

A simple pendulum of length \(ℓ\), mass \(m\) is attached to a point which oscillates vertically \(y=a \cos \Omega t\). Measuring \(y\) downwards, the pendulum position is

\begin{equation}
x=\ell \sin \phi, y=a \cos \Omega t+\ell \cos \phi
\end{equation}

The Lagrangian

\begin{equation}
\begin{array}{c}
L=\frac{1}{2} m\left(\dot{x}^{2}+\dot{y}^{2}\right)+m g \ell \cos \phi \\
=\frac{1}{2} m\left(\ell^{2} \cos ^{2} \phi\right) \dot{\phi}^{2}+\frac{1}{2} m(a \Omega \sin \Omega t+\ell \dot{\phi} \sin \phi)^{2}+m g \ell \cos \phi \\
=\frac{1}{2} m \ell^{2} \dot{\phi}^{2}-m a \ell \Omega \sin \Omega t \frac{d}{d t} \cos \phi+a^{2} \Omega^{2} \sin ^{2} \Omega t+m g \ell \cos \phi
\end{array}
\end{equation}

The purely time-dependent term will not affect the equations of motion, so we drop it, and since the equations are not affected by adding a total derivative to the Lagrangian, we can integrate the second term by parts (meaning we’re dropping a term \(\begin{equation}
\left.\frac{d}{d t}(\operatorname{mal} \Omega \sin \Omega t \cos \phi)\right)
\end{equation}\) to get

\begin{equation}
L=\frac{1}{2} m \ell^{2} \dot{\phi}^{2}+m a \ell \Omega^{2} \cos \Omega t \cos \phi+m g \ell \cos \phi
\end{equation}

(We’ve also dropped the term \(\begin{equation}
m g a \cos \Omega t
\end{equation}\) from the potential energy term—it has no \(\phi \text { or } \dot{\phi}\) dependence, so will not affect the equations of motion.)

The equation for small oscillations is

\begin{equation}
\ddot{\phi}+\omega_{0}^{2}\left[1+(4 a / \ell) \cos \left(2 \omega_{0}+\varepsilon\right) t\right] \phi=0, \quad \omega_{0}^{2}=g / \ell
\end{equation}

Comparing this with

\begin{equation}
\ddot{x}+\omega_{0}^{2}\left[1+h \cos \left(2 \omega_{0}+\varepsilon\right) t\right] x=0
\end{equation}

we see that \(4 a / \ell \equiv h\), so the parametric resonance range around \(2 \omega_{0}=2 \sqrt{g / \ell} \text { is of width } \frac{1}{2} h \omega_{0}=2 a \sqrt{g / \ell^{3}}\).