= Frequency interlacing for two spring-coupled equal masses
{title2=$\omega_-<\omega_2<\omega_1<\omega_+$}
For equal masses $m>0$, positive end <spring constants> $k_1>k_2$ and coupling <spring constant> $\kappa>0$, the stiffness <matrix> is $K=((k_1+\kappa,-\kappa),(-\kappa,k_2+\kappa))$. Its <eigenvalues> divided by $m$ are the squared <normal mode> <angular frequencies> $\omega_\pm^2=(k_1+k_2+2\kappa\pm D)/(2m)$, where $D=\sqrt{(k_1-k_2)^2+4\kappa^2}$. Holding the opposite mass fixed gives $\omega_j^2=(k_j+\kappa)/m$. Since $D>k_1-k_2$, subtraction proves the title ordering. Holding a mass fixed retains the coupling stiffness; removing the coupling spring produces a different system.
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