Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 7 / ii

Resolve the initial displacement into the mode shapes from part (i):There is no component along $(1,0,-1)$: the initial displacement is invariant under exchanging beads 1 and 3, whereas that mode is antisymmetric. Zero initial velocities select cosine time dependence. ThereforeHere ${\omega }_1=\sqrt{\tau (2-\sqrt{2})/(m\ell )}$ and ${\omega }_3=\sqrt{\tau (2+\sqrt{2})/(m\ell )}$. At zero time these expressions give $(0,a,0)$ and all three velocities vanish.

The motion is a superposition of the two symmetric normal modes, with opposite signed modal amplitudes. The outer beads always move together. Equivalently, $z_2=a\cos [({\omega }_1+{\omega }_3)t/2]\cos [({\omega }_3-{\omega }_1)t/2]$, displaying the modulation from the two frequencies. The result describes the linearized string while the initial amplitude and subsequent slopes remain small compared with the bead spacing.