Is a periodic force capable of transporting a particle to large distances? I have a particle at rest. At $t = 0$ a periodic force like $F_0 \sin\omega t$ starts acting on my particle. Can such a force transport my particle to infinity when $t  \to \infty$? Please answer this question without solving the mechanical problem, just by intuition.
 A: Here is what my intuition says:
$m\ddot{x} = F_0\sin\omega t, x(0)=0, \dot{x}(0)=0$  
$\dot{x}(t)=\frac{F_0}{m\omega}-\frac{F_0}{m\omega}\cos\omega t$
$x(t)=\frac{F_0}{m\omega^2}\left(\omega t - \sin\omega t\right)$
Hmmmm... Yes it can!
A: Kostya has provided an excellent mathematical proof that it can. I will try to give a more intuitive explanation. Suppose the particle started with zero initial speed. Since the force is given as $F_0\sin(\omega t)$, the time for which the particle accelerates and decelerates would be the same (think about the graph of sin). Therefore for some time, the particle will gain speed. Soon the particle will decelerate and lose speed and instantaneously come to rest, before being accelerated again. Note that the in the whole cycle of acceleration and deceleration the particle doesn't change its direction. Thus the particle can be transported to any distance you want provided you wait long enough.
A: Can and must. The momentum at any given time is
$$
p(t)=\int_0^t F(t')dt'={F_0\over\omega}(1-\cos\omega t),
$$
which oscillates back and forth between zero and $F_0/\omega$ and averages to a positive value over any whole number of periods. 
A: A particle is transported by the velocity, $ds = vdt$, rather the acceleration, $ds = 1/2adt^2$ and hence force, because of the dt rather than $dt^2$ dependence.
