= Solution
When the block is free, $x$ becomes a dynamical <generalized coordinate> and its <kinetic energy> $M\dot x^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant gives
$$
\boxed{\mathcal L_2=\frac12(M+m)\dot x^2+ml\dot x\dot\theta+\frac12ml^2\dot\theta^2-\frac12mgl\theta^2.}
$$
The coordinate $x$ is cyclic. Its conserved <canonical momentum> is the total horizontal <momentum>,
$$
P=(M+m)\dot x+ml\dot\theta.
$$
The two linear <Euler-Lagrange equations> are
$$
(M+m)\ddot x+ml\ddot\theta=0,\qquad l\ddot\theta+\ddot x+g\theta=0.
$$
Eliminating $\ddot x$ gives
$$
\boxed{\ddot\theta+\Omega^2\theta=0,\qquad \Omega^2=\frac{(M+m)g}{Ml}.}
$$
Thus the <pendulum with a freely sliding support> has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution is
$$
\boxed{\theta(t)=C\cos\Omega t+D\sin\Omega t,\qquad
x(t)=X_c+Vt-\frac{ml}{M+m}\theta(t).}
$$
Here $V=P/(M+m)$ and $X_c+Vt$ is the horizontal <centre of mass> coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to\infty$ restores $\Omega\to\sqrt{g/l}$.
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