= Solution
For $M=2m$, take the mode shapes $v_0=(1,1,1)$, $v_a=(1,0,-1)$ and $v_s=(1,-1,1)$. Their <angular frequencies> are $0$, $\omega=\sqrt{k/m}$ and $\sqrt2\omega$. The initial <displacement> decomposes as
$$
(0,d,-d)=\frac d4v_0+\frac d2v_a-\frac{3d}{4}v_s.
$$
All initial <velocities> vanish, so the zero-mode coefficient stays constant and the two oscillatory coefficients use cosines. Thus
$$
\boxed{\begin{aligned}
x_1(t)&=\frac d4+\frac d2\cos(\omega t)-\frac{3d}{4}\cos(\sqrt2\omega t),\\
x_2(t)&=\frac d4+\frac{3d}{4}\cos(\sqrt2\omega t),\\
x_3(t)&=\frac d4-\frac d2\cos(\omega t)-\frac{3d}{4}\cos(\sqrt2\omega t).
\end{aligned}}
$$
The conserved <centre of mass> is $(x_1+2x_2+x_3)/4=d/4$. Omitting the zero mode would incorrectly force this to zero. The two vibrational frequencies are in the ratio $\sqrt2$, so there is no common nonzero period.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-triatomic-initial-motion.png]
{title=Triatomic-chain motion from the specified initial displacement, including its constant centre of mass}
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