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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.

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IS there some nontrivial potential present? – Tim van Beek Feb 2 '11 at 11:41
No, no other potentials are present. – Vladimir Kalitvianski Feb 2 '11 at 11:51
Good question, cant think of an example – TROLLKILLER Feb 2 '11 at 12:16
Later on I will give you an example. It is a very common phenomenon. – Vladimir Kalitvianski Feb 2 '11 at 12:43
It can be done as long as the r.m.s power in the signal is not zero. That is the case for a sinusouidal force. How exactly is something I'm not sure about. Waiting for @Vladimir's explanation. – user346 Feb 2 '11 at 14:20
up vote 6 down vote accepted

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.

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So this is what we see/experience in the traffic jam: cars advance "step by step". – Vladimir Kalitvianski Feb 2 '11 at 14:35

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!

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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.

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Disagree with "must". For $p(0)=F_0/\omega$ it just stays around zero. – Kostya Feb 2 '11 at 16:48
We're given that the particle is initially at rest. – Ted Bunn Feb 2 '11 at 16:49

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.

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