= Solution
Taking the Hessian of the potential with respect to mass-weighted displacements gives
$$
\boxed{D_0=k_1\begin{pmatrix}1/m_a&-1/\sqrt{m_am_b}\\-1/\sqrt{m_am_b}&1/m_b\end{pmatrix}.}
$$
The equations of motion are $\ddot{\boldsymbol q}=-D_0\boldsymbol q$, so <eigenvalues> are squared <angular frequencies>. Normalized <eigenvectors> and <frequencies> are
$$
\boxed{u_T=\frac{(\sqrt{m_a},\sqrt{m_b})}{\sqrt{m_a+m_b}},\quad\omega_T=0;\qquad
u_S=\frac{(\sqrt{m_b},-\sqrt{m_a})}{\sqrt{m_a+m_b}},\quad
\omega_S=\sqrt{k_1\left(\frac1{m_a}+\frac1{m_b}\right)}=\sqrt{\frac{k_1}{\mu}}.}
$$
Both vectors are in the q basis, and $\mu=m_am_b/(m_a+m_b)$. The translation corresponds to equal physical displacements $\delta y_a=\delta y_b$. The stretch has $m_a\delta y_a+m_b\delta y_b=0$ and opposite physical displacements with magnitude ratio $|\delta y_a/\delta y_b|=m_b/m_a$. Ordinary <frequencies> are $\nu=\omega/(2\pi)$; <spectroscopic wavenumbers> are $\omega/(2\pi c)$.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a-normal-modes.png]
{title=Original normal mode displacements for free and weakly surface-bound diatomic models}
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