17.2: Two Coupled Pendulums

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

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

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).