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Take both displacement coordinates positive to the right. The kinetic energy and elastic potential energy areThus the equations of motion are $m_1{\ddot{x}}_1=-2k{x}_1+k{x}_2$ and $m_2{\ddot{x}}_2=k{x}_1-2k{x}_2$. With $x=v\cos (\omega t+\phi )$, the generalized eigenvalue problem for small oscillations isIts determinant gives $m_1{m}_2{\omega }^4-2k(m_1+m_2){\omega }^2+3k^2=0$. Put $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$. The two-mass three-spring normal modes haveBoth squared frequencies are positive because $(m_1+m_2{)}^2-D^2=3m_1{m}_2>0$. Also $D>|m_2-m_1|$, so $r_{-}>0$ and $r_{+}<0$: the lower mode moves the masses in phase, and the higher mode moves them oppositely. A general mode is an arbitrary sinusoidal amplitude multiplying its fixed shape.

For equal masses these reduce to $v_{-}=(1,1)$, ${\omega }_{-}^2=k/m$ and $v_{+}=(1,-1)$, ${\omega }_{+}^2=3k/m$. For unequal masses the vectors need not be Euclidean-orthogonal; they are orthogonal in the mass inner product, $m_1+m_2{r}_{-}{r}_{+}=0$.

To find the two-mass three-spring normal modes, choose both displacements positive to the right, measured from mechanical equilibrium. The middle spring's extension changes by $x_2-x_1$. Hooke's law and Newton's second law therefore giveIn a normal mode both displacements have the same angular frequency: take $x_j=A_j\cos (\omega t+\varphi )$. A nonzero amplitude vector exists when the determinant vanishes:Writing $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$, the two angular frequencies areThe corresponding eigenvectors can be specified bySince $D>|m_2-m_1|$ for positive masses, the lower normal mode ${\omega }_{-}$ has $A_2/A_1>0$: both bodies move in the same direction at any instant. The higher normal mode ${\omega }_{+}$ has $A_2/A_1<0$: they move in opposite directions. Their amplitudes are generally unequal. For the useful check $m_1=m_2=m$, these reduce to ${\omega }_{-}=\sqrt{k/m}$ with $A_2=A_1$, and ${\omega }_{+}=\sqrt{3k/m}$ with $A_2=-A_1$. Each angular frequency has units $\mathrm{rad}{\mathrm{s}}^{-1}$.