= Solution
The <kinetic energy> is $m(\dot\eta_1^2+\dot\eta_2^2)/2$. The Euler–Lagrange equations give
$$
m\ddot\eta_1=-(j+k)\eta_1+k\eta_2,\qquad m\ddot\eta_2=k\eta_1-(\ell+k)\eta_2.
$$
For a <normal mode> $\eta=Ae^{i\omega t}$, the <stiffness matrix> is
$$
K=\begin{pmatrix}j+k&-k\\-k&\ell+k\end{pmatrix},\qquad \det(K-m\omega^2I)=0.
$$
Solving this quadratic yields
$$
\boxed{\omega_\pm^2=\frac{j+\ell+2k\pm\sqrt{(j-\ell)^2+4k^2}}{2m}.}
$$
Both are positive for positive spring constants. A corresponding amplitude ratio is $A_2/A_1=(j+k-m\omega_\pm^2)/k$; the lower-frequency mode has the masses moving in the same direction and the higher-frequency mode in opposite directions.
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