 Solution

Taking the Hessian of the potential with respect to mass-weighted displacements givesThe equations of motion are $\ddot{\mathbf{q}}=-D_0\mathbf{q}$, so eigenvalues are squared angular frequencies. Normalized eigenvectors and frequencies areBoth vectors are in the q basis, and $\mu =m_a{m}_b/(m_a+m_b)$. 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 $\omega /(2\pi c)$.