= Solution
Resolve the initial displacement into the mode shapes from part (i):
$$
(0,a,0)^T=\frac{a}{2\sqrt2}(1,\sqrt2,1)^T-\frac{a}{2\sqrt2}(1,-\sqrt2,1)^T.
$$
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. Therefore
$$
\boxed{\begin{aligned}
z_1(t)=z_3(t)&=\frac{a}{2\sqrt2}[\cos(\omega_1t)-\cos(\omega_3t)],\\
z_2(t)&=\frac a2[\cos(\omega_1t)+\cos(\omega_3t)].
\end{aligned}}
$$
Here $\omega_1=\sqrt{\tau(2-\sqrt2)/(m\ell)}$ and $\omega_3=\sqrt{\tau(2+\sqrt2)/(m\ell)}$. 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.
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