= Solution
Use the <mode shapes> from part (b), and put $\omega=\sqrt{2k/m}$. The initial <velocity> decomposes as
$$
(0,0,\Omega)=\frac\Omega2(1,1,1)-\frac\Omega2(1,1,-1).
$$
There is no antisymmetric component because the initial state treats the equal <masses> identically. Initial displacements are zero, so the <zero mode> contributes a linear drift and the nonzero <normal mode> contributes a sine:
$$
\mathbf q(t)=\frac{\Omega t}{2}(1,1,1)
-\frac{\Omega}{2\omega}\sin(\omega t)(1,1,-1).
$$
In terms of actual angular positions,
$$
\boxed{\begin{aligned}
\theta_1(t)&=\frac\Omega2\left(t-\frac{\sin\omega t}{\omega}\right),\\
\theta_2(t)&=\frac{2\pi}{3}+\frac\Omega2\left(t-\frac{\sin\omega t}{\omega}\right),\\
\theta_3(t)&=\frac{4\pi}{3}+\frac\Omega2\left(t+\frac{\sin\omega t}{\omega}\right),\qquad
\omega=\sqrt{\frac{2k}{m}}.
\end{aligned}}
$$
The conserved <angular momentum> is initially $2m\Omega$ and the total <moment of inertia> is $4m$, so the common <angular velocity> must be $\Omega/2$, agreeing with the drift. The relative displacements remain of order $\Omega/\omega$, even when the common angle becomes large. Hence small $\Omega/\omega$ preserves the cyclic ordering and small spring extensions at arbitrarily late times. In the ideal arc-spring model the displayed energies are exactly quadratic in the relative angles, so this motion remains exact while that ordering persists.
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