= Solution
A <normal mode> is a motion in which all <displacement> coordinates oscillate at one <angular frequency> with fixed <amplitude> ratios. For small conservative oscillations, the symmetric stiffness and <mass> matrices give the <generalized eigenvalue problem for small oscillations>, $\mathsf K\mathbf u=\omega^2\mathsf M\mathbf u$. For positive <mass> and a stable equilibrium, its mass-orthogonal <eigenvectors> span the motion. A <superposition> of these modes, with phases and <amplitudes> fixed by initial <displacements> and <velocities>, gives the general solution; different modes need not share a phase.
Use upward end <displacements>. The <spring> compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The <potential energy> is
$$
V=\frac k2[(x_0-x_1)^2+(x_0-x_2)^2]+\frac{mg}{2}(x_1+x_2).
$$
Static <force balance> gives $2kx_0=mg$, cancelling the linear terms and leaving
$$
\boxed{V=\tfrac12k(x_1^2+x_2^2)+kx_0^2.}
$$
For a uniform beam, its central <moment of inertia> in the vertical plane is $ml^2/12$. To first order in the tilt, $\theta=(x_2-x_1)/l$. Its translational and rotational <kinetic energies> are
$$
T=\frac m8(\dot x_1+\dot x_2)^2+\frac m{24}(\dot x_2-\dot x_1)^2
=\frac m6(\dot x_1^2+\dot x_1\dot x_2+\dot x_2^2).
$$
Thus $\mathsf M=(m/6)\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $\mathsf K=k\mathsf1$. The <normal modes of a rigid beam on two end springs> have <eigenvectors>
$$
\boxed{\mathbf u_s=(1,1),\quad\omega_s=\sqrt{2k/m};\qquad
\mathbf u_a=(1,-1),\quad\omega_a=\sqrt{6k/m}.}
$$
The first translates without tilt; the second rocks about a stationary centre. Ordinary <frequencies> are $\omega_s/(2\pi)$ and $\omega_a/(2\pi)$.
Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static <spring> load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial <velocities> vanish because no <impulse> is imparted. Decomposing into the two <normal modes> gives
$$
x_1(t)=-\frac{Mg}{2k}(\cos\omega_st+\cos\omega_at),\qquad
x_2(t)=-\frac{Mg}{2k}(\cos\omega_st-\cos\omega_at).
$$
Only the symmetric mode contributes to the centre position:
$$
\boxed{z_{\rm CM}(t)-z_{\rm eq}=-\frac{Mg}{2k}\cos\left(\sqrt{\frac{2k}{m}}\,t\right).}
$$
The datum $z_{\rm eq}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll1$, ideal light vertical <springs> and negligible <damping>.
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