Simple Harmonic Oscillator on a smooth surface

I feel I understand what happens, and how to solve the equation of motion x(t) for a mass attached to a spring and released from rest horizontally on a smooth surface. We typically end up with

[itex] x(t) = x_0 cos(ωt) [/itex]

as the solution, with [itex] x_0 [/itex]as the amplitude of the oscillation.

But I've been wondering what happens if the mass was released again from position [itex] x_0 [/itex]but with velocity [itex] v_0 [/itex] instead of being released from rest. Is x(t) the same or different to when the mass was released from rest?

Thanks in advance.