17.2: Two Coupled Pendulums

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We’ll take two equal pendulums, coupled by a light spring. We take the spring restoring force to be directly proportional to the angular difference between the pendulums. (This turns out to be a good approximation.)

For small angles of oscillation, we take the Lagrangian to be

\begin{equation}
L=\frac{1}{2} m \ell^{2} \dot{\theta}_{1}^{2}+\frac{1}{2} m \ell^{2} \dot{\theta}_{2}^{2}-\frac{1}{2} m g \ell \theta_{1}^{2}-\frac{1}{2} m g \ell \theta_{2}^{2}-\frac{1}{2} C\left(\theta_{1}-\theta_{2}\right)^{2}
\end{equation}

Denoting the single pendulum frequency by \(\omega_{0}\), the equations of motion are (writing \(\omega_{0}^{2}=g / \ell, k=C / m \ell^{2}\), so \(\left.[k]=T^{-2}\right)\)

\begin{equation}
\begin{array}{l}
\ddot{\theta}_{1}=-\omega_{0}^{2} \theta_{1}-k\left(\theta_{1}-\theta_{2}\right) \\
\ddot{\theta}_{2}=-\omega_{0}^{2} \theta_{2}-k\left(\theta_{2}-\theta_{1}\right)
\end{array}
\end{equation}

We look for a periodic solution, writing

\begin{equation}
\theta_{1}(t)=A_{1} e^{i \omega t}, \quad \theta_{2}(t)=A_{2} e^{i \omega t}
\end{equation}

(The final physical angle solutions will be the real part.)

The equations become (in matrix notation):

\begin{equation}
\omega^{2}\left(\begin{array}{c}
A_{1} \\
A_{2}
\end{array}\right)=\left(\begin{array}{cc}
\omega_{0}^{2}+k & -k \\
-k & \omega_{0}^{2}+k
\end{array}\right)\left(\begin{array}{c}
A_{1} \\
A_{2}
\end{array}\right)
\end{equation}

Denoting the \(2 \times 2 \text { matrix by } \mathbf{M}\)

\begin{equation}
\mathbf{M} \vec{A}=\omega^{2} \vec{A}, \quad \vec{A}=\left(\begin{array}{l}
A_{1} \\
A_{2}
\end{array}\right)
\end{equation}

This is an eigenvector equation, with \(\omega^{2}\) the eigenvalue, found by the standard procedure:

\begin{equation}
\operatorname{det}\left(\mathbf{M}-\omega^{2} \mathbf{I}\right)=\left|\begin{array}{cc}
\omega_{0}^{2}+k-\omega^{2} & -k \\
-k & \omega_{0}^{2}+k-\omega^{2}
\end{array}\right|=0
\end{equation}

Solving, \(\omega^{2}=\omega_{0}^{2}+k \pm k\), that is

\begin{equation}
\omega^{2}=\omega_{0}^{2}, \quad \omega^{2}=\omega_{0}^{2}+2 k
\end{equation}

The corresponding eigenvectors are (1,1) and (1,−1).