= Solution
Take both displacement coordinates positive to the right. The <kinetic energy> and <elastic potential energy> are
$$
T=\frac12m_1\dot x_1^2+\frac12m_2\dot x_2^2,\qquad V=\frac k2[x_1^2+(x_2-x_1)^2+x_2^2].
$$
Thus the equations of motion are $m_1\ddot x_1=-2kx_1+kx_2$ and $m_2\ddot x_2=kx_1-2kx_2$. With $x=v\cos(\omega t+\varphi)$, the <generalized eigenvalue problem for small oscillations> is
$$
\left[k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}-\omega^2\begin{pmatrix}m_1&0\\0&m_2\end{pmatrix}\right]v=0.
$$
Its determinant gives $m_1m_2\omega^4-2k(m_1+m_2)\omega^2+3k^2=0$. Put $D=\sqrt{m_1^2-m_1m_2+m_2^2}$. The <two-mass three-spring normal modes> have
$$
\boxed{\omega_\mp^2=\frac{k(m_1+m_2\mp D)}{m_1m_2},\qquad v_\mp=(1,r_\mp)^T,\qquad r_\mp=\frac{m_2-m_1\pm D}{m_2}.}
$$
Both squared frequencies are positive because $(m_1+m_2)^2-D^2=3m_1m_2>0$. Also $D>|m_2-m_1|$, so $r_->0$ and $r_+<0$: the lower mode moves the masses in phase, and the higher mode moves them oppositely. A general mode is an arbitrary sinusoidal amplitude multiplying its fixed shape.
For equal masses these reduce to $v_-=(1,1)$, $\omega_-^2=k/m$ and $v_+=(1,-1)$, $\omega_+^2=3k/m$. For unequal masses the vectors need not be Euclidean-orthogonal; they are orthogonal in the mass inner product, $m_1+m_2r_-r_+=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2-2013-ib-two-mass-modes.png]
{title=In-phase and out-of-phase mode directions for an illustrative unequal-mass system; arrow lengths show relative displacements}
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