 Solution

With equal masses m, the two mass-weighted Hessians areFor a uniform displacement vector $t=(1,1)$,Therefore uniform translation is not a normal mode in orientation (1) when $k_2\ne 0$: anchoring only one atom mixes translation with internal stretching. 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.