Solution

The particle is born at rest inside a sinusoidal electric field $\mathcal{E}(t) = E \sin(\omega t)$
The electric force F = qE \sin(\omega t) modifies its relativistic momentum

$p = \dfrac{mv}{\sqrt{1 - v^2/c^2}}$

according to Newton's second law

$\frac{dp}{dt} = qE \sin(\omega t)$

Integrating with the initial condition $p(0) = 0$ we obtain

$p(t) = \frac{qE}{\omega}\bigl[1 - \cos(\omega t)\bigr]$

whose maximum value is reached when $\cos(\omega t) = -1$

$p_{\max} = \frac{2qE}{\omega}$

From the definition of p, the velocity is solved for in terms of momentum:

$v = \frac{p/m}{\sqrt{1 + \dfrac{p^2}{m^2 c^2}}}$

Substituting$p_{\max}$ and simplifying algebraically, we arrive at the maximum velocity, which is:

$\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$