= Normal modes of three vertically suspended masses with a branched spring connection
Let three equal masses $m$ move vertically, with one spring of stiffness $k$ attaching mass 1 to a fixed support and two further springs of the same stiffness attaching masses 2 and 3 separately to mass 1. For absolute downward displacements $q$ from equilibrium, the <small-oscillation mass and stiffness matrices> are
$$
M=mI,\qquad K=k\begin{pmatrix}3&-1&-1\\-1&1&0\\-1&0&1\end{pmatrix}.
$$
The <normal modes> have
$$
\begin{array}{c|c}
m\omega^2/k&\text{relative physical displacements}\\\hline
1&(0,1,-1)^T\\
2-\sqrt3&(1,(1+\sqrt3)/2,(1+\sqrt3)/2)^T\\
2+\sqrt3&(1,(1-\sqrt3)/2,(1-\sqrt3)/2)^T
\end{array}.
$$
Exchange symmetry of the two lower masses gives the antisymmetric mode with mass 1 stationary. In the symmetric sector, put $q=(A,B,B)^T$; the reduced equations give $\lambda^2-4\lambda+1=0$, producing the remaining two frequencies. The lower mode moves all masses in phase; in the higher mode the lower pair moves against the upper mass.
Using individual rope-length changes $y$ instead gives $q=(y_1,y_1+y_2,y_1+y_3)^T$. In those generalized coordinates $K=kI$ and $M=m\left(\begin{smallmatrix}3&1&1\\1&1&0\\1&0&1\end{smallmatrix}\right)$. Mode amplitudes must therefore be identified as physical displacements or rope-length changes before interpreting them. The model assumes taut elastic elements and small vertical motion; gravity shifts the equilibrium extensions.
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