The motion of a pair of identical pendulums coupled with a spring is modeled by the system

${\mathbf{mx}}_{\mathbf{1}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\frac{\mathbf{mg}}{\mathbf{l}}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{k}\left({\mathbf{x}}_{\mathbf{1}}\mathbf{-}{\mathbf{x}}_{\mathbf{2}}\right){\mathbf{,}\mathbf{mx}}_{\mathbf{2}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\frac{\mathbf{mg}}{\mathbf{l}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{k}\left({\mathbf{x}}_{\mathbf{1}}\mathbf{-}{\mathbf{x}}_{\mathbf{2}}\right)$

for small displacements (see Figure 5.36). Determine the two normal frequencies for the system.

Let's rewrite this system in operator form;

$\begin{array}{rcl}\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_1\right]-\mathrm{k}\left[{\mathrm{x}}_2\right]& =& 0\\ -\mathrm{k}\left[{\mathrm{x}}_1\right]+\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_2\right]& =& 0\end{array}$

One will eliminate ${\mathrm{x}}_2$from the system by multiplying the first equation by${\mathrm{mD}}^2+(\mathrm{mg}+\mathrm{lk})/\mathrm{l}$ and the second by k and then adding those equations together:

$\begin{array}{rcl}{\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)}^2\left[{\mathrm{x}}_1\right]-\mathrm{k}\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_2\right]& =& 0\\ -{\mathrm{k}}^2\left[{\mathrm{x}}_1\right]+\mathrm{k}\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_2\right]& =& 0\\ \left({\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)}^2-{\mathrm{k}}^2\right)\left[{\mathrm{x}}_1\right]& =& 0\\ \left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}-\mathrm{k}\right)\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}+\mathrm{k}\right)\left[{\mathrm{x}}_1\right]& =& 0\\ \left({\mathrm{mD}}^2+\frac{\mathrm{mg}}{\mathrm{l}}\right)\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_1\right]& =& 0\end{array}$

The auxiliary equation is $\left({\mathrm{mr}}^2+\frac{\mathrm{mg}}{\mathrm{l}}\right)\left({\mathrm{mr}}^2+\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{l}}\right)=0$

Let's find the roots of the auxiliary equation:

$$\begin{gathered} {\mathrm{r}}^2=-\frac{\mathrm{g}}{\mathrm{l}}\Rightarrow {\mathrm{r}}_{1,2}=\pm \sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{i}, \\ {\mathrm{r}}^2=-\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}\Rightarrow {\mathrm{r}}_{3,4}=\sqrt{\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}}\mathrm{i} \end{gathered}$$

So, the general solution for${\mathrm{x}}_1$ is:

${\mathrm{x}}_1\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{t}+{\mathrm{c}}_2\sin \sqrt{\frac{\mathrm{g}}{\mathrm{l}}}\mathrm{t}+{\mathrm{c}}_3\cos \sqrt{\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}}\mathrm{t}+{\mathrm{c}}_4\sin \sqrt{\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}}\mathrm{t}$.

One can solve for ${\mathrm{x}}_2$by substituting solution for ${\mathrm{x}}_1$into the first equation of the given system which gives us that ${\mathrm{kx}}_2=\left({\mathrm{mD}}^2+\frac{\mathrm{mg}+\mathrm{lk}}{\mathrm{l}}\right)\left[{\mathrm{x}}_1\right]$but since ${\mathrm{D}}^2\left[{\mathrm{x}}_1\right]$contains the same trigonometric functions as ${\mathrm{x}}_1$then ${\mathrm{x}}_2$also contains the same trigonometric functions with the same frequencies, one doesn't need to solve for ${\mathrm{x}}_2$.

The normal angular frequencies are $\sqrt{\frac{\mathrm{g}}{\mathrm{l}}}$and $\sqrt{\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}}$so, the normal frequencies are $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{g}}{\mathrm{l}}}$ and $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{mg}+2\mathrm{lk}}{\mathrm{ml}}}$.