Frequency interlacing for two spring-coupled equal masses (${\omega }_{-}<{\omega }_2<{\omega }_1<{\omega }_{+}$)

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.