= Solution
With equal masses m, the two mass-weighted Hessians are
$$
\boxed{D_{(1)}=\frac1m\begin{pmatrix}k_1+k_2&-k_1\\-k_1&k_1\end{pmatrix},\qquad
D_{(2)}=\frac1m\begin{pmatrix}k_1+k_3&-k_1\\-k_1&k_1+k_3\end{pmatrix}.}
$$
For a uniform displacement vector $t=(1,1)$,
$$
D_{(1)}t=\frac1m(k_2,0),\qquad D_{(2)}t=\frac{k_3}{m}(1,1).
$$
Therefore \b[uniform translation is not a <normal mode> in orientation (1)] when $k_2\ne0$: anchoring only one atom mixes translation with internal stretching. \b[It is an exact <normal mode> displacement in orientation (2)] because both atoms have identical restoring terms. Its <frequency> is nevertheless nonzero, $\omega_T=\sqrt{k_3/m}$; the antisymmetric stretch has $\omega_S=\sqrt{(2k_1+k_3)/m}$. The rigid surface breaks free translational invariance in both orientations. A vector resembling a uniform translation need not be a zero-frequency mode when the molecule is anchored.
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