= Solution
Set the support displacement to zero and seek a <normal mode> $\mathbf x=(A,B)^{\mathsf T}\cos(\omega t+\varphi)$. With $s=m\omega^2/k$, the <generalized eigenvalue problem> requires
$$
\det\begin{pmatrix}3-s&-2\\-2&2-2s\end{pmatrix}=2(s^2-4s+1)=0.
$$
Thus
$$
\boxed{\omega_\pm^2=\frac km(2\pm\sqrt3),\qquad f_\pm=\frac1{2\pi}\sqrt{\frac km(2\pm\sqrt3)}.}
$$
The second row gives $A=(1-s)B$. The signed displacement ratios are therefore
$$
\boxed{\left(\frac{x_1}{x_2}\right)_{\rm slow}=\sqrt3-1,\qquad\left(\frac{x_1}{x_2}\right)_{\rm fast}=-1-\sqrt3.}
$$
The slow mode is in phase and the fast mode out of phase. Keeping the unequal mass matrix is essential: replacing it by the identity changes the eigenfrequencies.
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