= Solution
Set $q=\ell_1/\ell_2$ and use <normal mode> amplitudes $\phi_j=A_je^{i\omega t}$, with real physical displacements obtained at the end. In units of $\omega_0^2$, the stiffness matrix is
$$
\mathsf K=\begin{pmatrix}1+q&-q\\-q&q\end{pmatrix}.
$$
Writing $\lambda=\omega^2/\omega_0^2$, the <eigenvalue> condition is
$$
\det(\mathsf K-\lambda\mathsf 1)=(1+q-\lambda)(q-\lambda)-q^2
=\lambda^2-(1+2q)\lambda+q=0.
$$
Therefore
$$
\boxed{\omega_\pm^2=\omega_0^2\left[\frac12+\frac{\ell_1}{\ell_2}
\pm\sqrt{\frac14+\left(\frac{\ell_1}{\ell_2}\right)^2}\right].}
$$
For finite positive fibre lengths both <eigenvalues> are positive. The corresponding amplitude ratio is
$$
\frac{A_2}{A_1}=\frac q{q-\lambda}=\frac1{1-R\lambda},\qquad R=\frac{\ell_2}{\ell_1}.
$$
The lower <normal mode> has the spheres twisting in the same sense, and the higher one has opposite signs of angular displacement.
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