 Normal modes of a symmetric three-mass string

Three particles are attached to a light taut string between fixed endpoints, with four equal equilibrium segment lengths $a$. The outer two masses are $m_o$ and the central mass is $m_c$. For small transverse displacement and tension $T$, set $k=T/a$. In coordinate order $(x,y,z)$ for the two outer particles and the central one, the potential is $V=\frac{1}{2}k[x^2+(x-z{)}^2+(y-z{)}^2+y^2]$ and the mass matrix is $M=\mathrm{diag}(m_o,m_o,m_c)$.

The antisymmetric normal mode is $(x,y,z)\propto (1,-1,0)$ with ${\omega }_a^2=2k/m_o$. In a symmetric mode $x=y=q$, the equations give $z/q=2-m_o{\omega }^2/k$ and $(2k-m_c{\omega }^2)z=2kq$. Their determinant is $m_o{m}_c{\omega }^4-2k(m_o+m_c){\omega }^2+2k^2$, whose roots give the displayed symmetric frequencies. Both are positive for positive masses. The physical order of points along the string is $x,z,y$; keeping it separate from the coordinate-vector order prevents a misleading mode sketch.