A naive question on the "continuously" degenerate ground states of 1d phonons?

In general, the gapless Goldstone mode is related to the "continuously" degenerate ground states. The Mexican hat potential is an example (see the logo of this SE website), where the bottom circle is the continuous ground-state manifold. Another example is the ferromagnetic (FM) states of the FM Heisenberg model, where the continuous ground-state manifold is the 2d sphere corresponding to the global rotation of spin directions.

All the above examples have a classical picture: The ground-state manifold or "continuously" degenerate ground states related by the continuous symmetry transformations.

My question is: Is there also a classical picture of "ground-state manifold" of the 1d oscillating atoms (phonons)? As we know, under periodic boundary conditions and in the large size limit, the excitation spectrum of phonons is gapless which is an example of Goldstone theorem.

Edit:

I think I made a wrong statement in my comments below: "It seems that the phonon vacuum state does NOT break any symmetry".

In fact, the phonon vacuum state (one of the degenerate ground states) breaks the global continuous translation symmetry, no matter the system size is finite or infinite.

Thus, it seems that there always exist both "spontaneously breaking of the continuous translation symmetry" and "continuously degenerate ground states", even when the system size is finite. But the gapless excitation can only happen when the system size is infinite (since the momentum $k$ is discrete when the system size is finite).

Therefore, the Goldstone theorem is a theorem for the infinite system ?? And how to intuitively understand the relation between "infinite size" and "gaplessness" without looking in the momentum space??

Thanks in advance.