Helping a confused particle!

From the definition of electric field, $\textbf{F}=q\textbf{E}$ and, $\textbf{F}=m\textbf{a}$ So, $\textbf{a}=\frac{\textbf{F}}{m}=\hat{\imath}\frac{qE_{0}}{m}\cos(\frac{\pi}{2}+\omega t)=\hat{\imath}a_{0}\cos(\frac{\pi}{2}+\omega t)$ Now, we can integrate it to get the velocity of the particle, $\textbf{v}=\int_{0}^{t}\textbf{a}dt'=\hat{\imath}a_{0}\int_{0}^{t}\cos(\omega t'+\frac{\pi}{2})dt'$ $\therefore\textbf{v}=\hat{\imath}\frac{a_{0}}{\omega}(\cos \omega t-1)$ We can further integrate it to get the displacement. But, we don't need to do that for solving this problem. Notice that $-2\leq(\cos \omega t-1)\leq0$ . Hence, $\textbf{v}$ always works in the direction of $-\hat{\imath}$ . So, the particle will drift towards that direction and its motion will follow a start-and-stop rhythm. [It's like pushing a toy car every time it comes to a stop.Though the analogy is not completely perfect!]

From the equation $E=E_{0}\cos (\frac{\pi }{2}+\omega t)$ , we can see that after $t$ starts growing from $0$ , the $\cos (\frac{\pi }{2}+\omega t)$ will decrease more and more until it reaches $-1$ at some moment $t_{1}$ , so the particle will be drifting towards the direction of $-\imath$ .

In fact, the particle will be accelerating with maximum acceleration at $t_{1}$ , because $\left | E \right |$ will be increasing over that period. After $t_{1}$ , the $\left | E \right |$ will be decreasing but still forcing particle to drift towards the direction of $-\imath$ until some moment $t_{2}$ when $\cos (\frac{\pi }{2}+\omega t)$ will equal $0$ .

Actually, at $t_{2}$ , the particle will have the greatest speed. After that, acceleration produced by $E$ will have opposite direction from velocity of a particle, so it will slow it down but not enough to make it oscillate around one point. Generally, it will continue to drift towards the direction of $-\imath$ and my prediction is that it will have kind of stop-go movement.