= Solution
Subtracting the two linearized equations eliminates their acceleration terms:
$$
\omega_0^2\theta=\omega_1^2\eta.
$$
For $g,k>0$, the assumed nontrivial <simple harmonic motion> therefore has
$$
\boxed{\eta(t)=\frac{\omega_0^2}{\omega_1^2}\theta_0\sin(\omega t)
=\frac{mg}{k\ell}\theta_0\sin(\omega t).}
$$
The support displacement and pendulum angle oscillate in phase. Inserting this relation into either <Euler-Lagrange equation> yields
$$
\omega^2\left(1+\frac{\omega_0^2}{\omega_1^2}\right)=\omega_0^2,
$$
so the one positive finite <angular frequency> is
$$
\boxed{\omega=\frac{\omega_0\omega_1}{\sqrt{\omega_0^2+\omega_1^2}}
=\sqrt{\frac{gk}{mg+k\ell}},\qquad \frac1{\omega^2}=\frac\ell g+\frac mk.}
$$
The negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos(\omega t)+B\sin(\omega t)$.
There is no missing second finite branch. The quadratic <Lagrangian> is
$$
L_2=\frac12(\dot\eta+\dot\theta)^2-\frac12\omega_1^2\eta^2-\frac12\omega_0^2\theta^2,
$$
whose kinetic matrix is $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$ and has rank one. Its characteristic <determinant> is $\omega_0^2\omega_1^2-(\omega_0^2+\omega_1^2)\omega^2$, not a quadratic polynomial in $\omega^2$. This <singular kinetic matrix in small oscillations> results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.
For a stiff spring,
$$
\boxed{k\to\infty:\qquad\eta\to0,\qquad\omega\to\sqrt{g/\ell}.}
$$
The hinge becomes fixed, recovering the <simple pendulum> frequency, as expected for the <pendulum with a massless spring-mounted support>. At finite $k$, horizontal support motion makes the frequency smaller.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018-q2-massless-support-frequency.png]
{title=A massless spring-mounted support lowers the pendulum's one finite frequency}
{description=Original curve of $\omega/\omega_0=r/\sqrt{1+r^2}$ against $r=\omega_1/\omega_0$. The small-$r$ spring-controlled limit and the large-$r$ fixed-support limit are marked.}
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