 Solution

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 $\mathrm{rad}{\mathrm{s}}^{-1}$.