= Solution
A <normal mode> is a small motion in which all coordinates oscillate at one common <angular frequency> with fixed relative <amplitudes>. For conservative linearized motion, these are <eigenvectors> of the mass-weighted <stiffness matrix>.
All three <springs> are initially stretched from $l$ to $2l$, giving equal <tension> $T_0=kl$ and zero resultant <force> on each <mass>. This prestress is essential for the transverse modes found below.
For the final <mixed-axis trajectory of a prestressed spring pair>, the initial displacements excite only the symmetric mode in each direction. Therefore, to first order in $\epsilon/l$,
$$
\boxed{x_1(t)=2l+\epsilon\cos(\sqrt{k/m}\,t),\qquad
y_1(t)=\epsilon\cos(\sqrt{k/(2m)}\,t).}
$$
The second <mass> has the same displacements about $4l$. Over the first transverse <period> $T_y=2\pi\sqrt{2m/k}$, write $s=t/T_y$:
$$
\frac{x_1-2l}{\epsilon}=\cos(2\pi\sqrt2\,s),\qquad
\frac{y_1}{\epsilon}=\cos(2\pi s),\qquad 0\le s\le1.
$$
The trajectory is not closed after $T_y$, because the <frequency> ratio is $\sqrt2$. It ends at approximately $(-0.858,1)$ in these scaled coordinates rather than $(1,1)$.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-gp2012-spring-trajectory.png]
{title=Prestressed spring trajectory}
{description=Original first-transverse-period trajectory of the first mass. Markers distinguish initial and final points and intermediate quarter-period positions; both normal-coordinate amplitudes equal the release displacement.}
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