Rest mass of phonon: is this concept definable? Phonons are obtaied by non-relativistic quantization of the lattice vibration. The dispersion relation is given by $\omega=c_s k$ where $c_s$ is the velocity of sound. What can we say about the mass of the phonon? I think it is not possible to compare this relation with the relativistic dispersion relation $E^2=p^2c^2+m^2c^4$ and conclude $m=0$. By mass, I do not mean the effective mass but the rest mass. Certainly, if the rest mass of phonon were zero it would have travelled with the velocity of light in vacuum. 
I think in the non-relativistic approximation of the  Einstein's energy momentum relation, the same $m$ appears in the non-relativistic kinetic energy $\frac{p^2}{2m}$. Therefore, we can still talk about rest mass in non-relativistic physics.
Moreover, phonon being a goldstone boson should have zero rest mass. 
Edit: How does one define the rest mass of the phonon?
 A: Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are not "gapped"). This does not mean that they travel with the speed of light; I guess one way to see that is that the lattice breaks Lorentz symmetry by giving us a preferred inertial frame. When formulating the theory of phonons we usually take the nonrelativistic limit $c \rightarrow \infty$ right from the start, so the speed of light never enters into any of the equations.
Phonons instead travel at the speed of sound $c_s$, which is the characteristic speed set by the lattice (if you compare the phonon dispersion to the relativistic dispersion relation you wrote down you see that $c_s$ replaces $c$, the speed of light).
Said differently, phonons are quasiparticles (=not true, elementary particles) that emerge in a theory with a lattice that breaks Lorentz symmetry, so your statement "if the rest mass of phonon was zero it would have traveled with the velocity of light in vacuum" does not apply to them.
A: Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations and phonons which follow one with something analogical to a mass term. 
Acoustic and optical phonons are roughly analogical to massless and massive particles. For acoustic phonons, a wave with very long wavelength becomes simply a translation of the whole lattice, so that the energy becomes zero. This is, indeed, like a Goldstone boson related with translational symmetry.   
For optical phonons, there is no such translational symmetry which forces the energy of waves with a long wavelength to go to zero. So they have non-zero energy even in the limit of infinite wavelength or momentum zero, similar to a mass term. Of course, this will not be exactly $E^2=p^2c_s^2 + m^2 c_s^4$, which is what one may obtain at best as an approximation near a minimum, but the most characteristic distinction remains: you need more than some minimal energy to create them.  
