= Solution
For two bars, after subtracting the gravitational <equilibrium>,
$$
K=\begin{pmatrix}2&-1\\-1&2\end{pmatrix},\qquad M=\begin{pmatrix}1&0\\0&m_2\end{pmatrix}.
$$
A <normal mode> $\boldsymbol\eta=\mathbf v e^{i\omega t}$ requires the <generalized eigenvalue problem> $(K-\omega^2M)\mathbf v=0$. With $z=\omega^2$,
$$
0=(2-z)(2-m_2z)-1=m_2z^2-2(m_2+1)z+3.
$$
Thus, for $m_2>0$,
$$
\boxed{\omega_{1,2}=\sqrt{\frac{m_2+1\mp\sqrt{m_2^2-m_2+1}}{m_2}},\qquad\omega_1<\omega_2.}
$$
Both squared <angular frequencies> are positive. A corresponding <eigenvector> is $(1,2-\omega_j^2)^T$, from the first row of the <matrix> equation.
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