= Solution
For the <elastic pendulum>, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin\theta$ and $Z=-r\cos\theta$, with $Z$ upward. Differentiating gives $\dot X^2+\dot Z^2=\dot x^2+r^2\dot\theta^2$. Thus the <Lagrangian> is
$$
\boxed{L=\frac12m\left(\dot x^2+(l+x)^2\dot\theta^2\right)
-\frac12kx^2+mg(l+x)\cos\theta.}
$$
For $x$, the <Euler-Lagrange equation> gives
$$
\boxed{m\ddot x=m(l+x)\dot\theta^2-kx+mg\cos\theta.}
$$
For $\theta$, the <canonical momentum> is $p_\theta=m(l+x)^2\dot\theta$. Its <Euler-Lagrange equation> is
$$
\frac{d}{dt}\left[m(l+x)^2\dot\theta\right]+mg(l+x)\sin\theta=0,
$$
or, for positive length,
$$
\boxed{(l+x)\ddot\theta+2\dot x\dot\theta+g\sin\theta=0.}
$$
At an <equilibrium>, all time derivatives vanish, so $kx=mg\cos\theta$ and $\sin\theta=0$. Modulo full rotations, the two formal possibilities are
$$
\boxed{(x,\theta)=\left(\frac{mg}{k},0\right),\qquad
(x,\theta)=\left(-\frac{mg}{k},\pi\right).}
$$
The lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.
The <potential energy> is $V=kx^2/2-mg(l+x)\cos\theta$. At either formal stationary point its mixed second derivative vanishes, while
$$
V_{xx}=k,\qquad V_{\theta\theta}=mg(l+x)\cos\theta.
$$
Both curvatures are positive at the lower state, making it a <stable equilibrium>; the upper state, when it has positive length, has negative angular curvature and is unstable.
To linearize about the stable state, define the small displacement $\xi=x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as $\dot\theta^2$, $\dot\xi\dot\theta$ and $\xi\ddot\theta$ are second order, while $\cos\theta=1+O(\theta^2)$ and $\sin\theta=\theta+O(\theta^3)$. The first-order <equations of motion> therefore decouple:
$$
\boxed{\ddot\xi+\frac{k}{m}\xi=0,\qquad
\ddot\theta+\frac{g}{l+mg/k}\theta=0.}
$$
There are two independent <normal modes>. The radial one is vertical <simple harmonic motion> with <angular frequency> $\omega_r=\sqrt{k/m}$. The transverse one is a <simple pendulum> oscillation at fixed, statically stretched length $r_0$, with <angular frequency> $\omega_\theta=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the <elastic pendulum>.
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