= Normal modes of a symmetric three-mass string
{title2=$\omega_a^2=2k/m_o,\quad\omega_\pm^2=k[m_o+m_c\pm\sqrt{m_o^2+m_c^2}]/(m_om_c)$}
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 $V=\tfrac12k[x^2+(x-z)^2+(y-z)^2+y^2]$ and the mass matrix is $M=\operatorname{diag}(m_o,m_o,m_c)$.
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 $m_om_c\omega^4-2k(m_o+m_c)\omega^2+2k^2$, 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.
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