The electric field between plates is

$E=\frac{\sigma }{{\epsilon }_0}$ (Constant)

The angular frequency of SHM doesn't change by a constant force, only the equilibrium position will be affected.

$\omega =\sqrt{\frac{2k}{m}}$

The amplitude of SHM can be calculated by finding the distance of the equilibrium position from the rest position (starting position) of the particle.

$2kA=\frac{q\sigma }{{\epsilon }_0}$

$\Rightarrow A=\frac{q\sigma }{2k{\epsilon }_0}$

So the particle will cross its equilibrium position with a speed

$v=\frac{q\sigma }{{\epsilon }_0}\sqrt{\frac{k}{2m}}$