Normal modes of coupled oscillators

Question

For two pendulums of mass $m_1$ and $m_2$, coupled by a spring of constant k, both suspended by strings of length $l$, the following matrix equality results from their equations of motion:

$$ \omega^2 \begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} g/l+k/m1 & -k/m_1\\ -k/m_2 & g/l + k/m_2\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $$

But I have stumbled onto the weirdest thing. In theory, to find the normal mode frequencies, you would have to subtract $\omega^2$ from the main diagonal of the matrix, and find when that determinant equals zero. However, while messing around with this expression, I decided to see what would happen if the determinant of the matrix itself (without subtracting the eigenvalue) equals zero, and I found that the determinant equals zero if:

$ \frac{g}{l} = 0 $ , or, $ \frac{g}{l}+k(\frac{1}{m_1}+\frac{1}{m_2})=0 $

Doing this was much faster than dealing with the trinomial that would have formed if I had gone and actually found when the determinant of:

$$ \begin{pmatrix} g/l+k/m1-\omega^2 & -k/m_1\\ -k/m_2 & g/l + k/m_2-\omega^2\end{pmatrix} $$

Equals zero. And nevertheless, the values stated above turn out to be the actual eigenvalues of the matrix in question, which speed up the process of finding normal modes immensely. What happened? Why did I find the correct eigenvalues by finding under which conditions the original matrix was zero? Was this purely coincidental, or is this an easier, faster method of finding eigenvalues for 2x2 matrices?

Answer 1

For a 2x2 invertible matrix A with eigenvalues $\lambda_1$ and $\lambda_2$, its determinant is the product $\lambda_1 \lambda_2$.

$\lambda_1$ and $\lambda_2$ are also the roots of $det(A-\lambda I)=0$.

With respect to your question, I would stick with the second way of finding the eigenvalues. This is because you found out what the product $\lambda_1 \lambda_2$ was but in general they won't factor nicely into $\lambda_1$ and $\lambda_2$ like they did here.