= Solution
At the zone centre, the <dynamical matrix> commutes with the point-group representation on displacements. An <irreducible representation> of <dimension> two enforces a doublet; distinct one-dimensional representations need not have equal frequencies. The tetragonal <point group> $D_{4h}$ has two-dimensional $E_g$ and $E_u$ representations, so <optical phonons>[optical modes] transforming that way are doubly degenerate at the centre. Atomic positions and the displacement representation would be needed to assign every <optical phonon branch>[optical branch] to a particular representation; the point-group label alone is not a complete mode decomposition.
For propagation along the tetragonal fourfold axis, the two transverse acoustic polarizations transform together and have equal <sound speeds> and frequencies. For example $\omega_{{\rm TA},x}=\omega_{{\rm TA},y}\simeq v_T|q_z|$, whereas the longitudinal speed is independent. An optical doublet may split away from the centre along a direction whose <little group> has only one-dimensional representations. It need not remain degenerate at every nearby <wavevector>.
All <irreducible representations> of $D_{2h}$ are one-dimensional, so the orthorhombic <point group> enforces no analogous optical doublet or equality of transverse <sound speeds>. The three acoustic frequencies are nevertheless all zero at the exact centre in both <crystals>, because of translations; that common zero is not evidence that their slopes are equal. Accidental equalities remain possible. For a polar <crystal>, a direction-dependent macroscopic <electric field> also distinguishes longitudinal from transverse infrared-active limits; an analytic zone-centre <symmetry> classification alone does not remove LO–TO splitting.
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