 Solution

The kinetic energy is $T=m{\sum }_i{\dot{z}}_i^2/2$. Differentiating the quadratic potential energy gives the equations for small oscillations,Put $z=v\cos (\omega t+\phi )$ and $\lambda =m\ell {\omega }^2/\tau$. The eigenvalue problem is $Kv=\lambda v$. Its characteristic polynomial isThus a convenient set of normal modes isEach shape can be multiplied by an arbitrary sinusoidal time amplitude. In the lowest mode all beads move in phase. In the middle mode the central bead is fixed and the outer beads move oppositely. In the highest mode the two outer beads move together against the centre. These vectors are mutually orthogonal and span the three displacement coordinates.

These are the three-mass case of the normal modes of a uniformly loaded finite string, with discrete sine shapes $v_i^{(j)}=\sin (ij\pi /4)$. The sketches use straight string segments between masses and fixed zero displacements at the walls. The small-slope and constant-tension assumptions justify the quadratic energy and linear equations.