A Question regarding Klepnner and Kolenkow's mechanics: the electron under an electromagnetic wave I came across the example 1.11 in Kleppner and Kolenkow's mechanics' book. There in the example using integration the motion of an ionospheric electron is found under the influence of an oscillating electric field.
The problem statement is:
The ionosphere is a region of electrically neutral gas, composed of posi·
tively charged ions and negatively charged electrons, which surrounds
the earth at a height of approximately 200 km (120 mi). If a radio wave
passes through the ionosphere, its electric field accelerates the charged
particle. Because the electric field oscillates in time, the charged
particles tend to jiggle back and forth. The problem is to find the motion
of an electron of charge -e and mass m which is initially at rest, and
which is suddenly subjected to an electric field E = Eo sin wt (w is the
frequency of oscillation in radians per second).
And after all the process, the result says:
*x0  = v0 = 0, so we have x(t) = (a0/ω) t − (a0/ω2) (sinωt).
The result is interesting: the second term oscillates and corresponds
to the jiggling motion of the electron, which we predicted. The first
term, however, corresponds to motion with uniform velocity, so in addition
to the jiggling motion the electron starts to drift away. Can you see
why?*
Now I did not see why the electron will drift. But I tried understanding, and so looked at it from a different angle and thought what if instead of a sine varying field (as in the problem) we had cosine varying field (E = E cos(wt))
And the result I obtained after similar boundary conditions were:
So, employing this change in the oscillations, I do not get any term that describes any drift. Instead I get a constant position (independent of time) term, implying at t=o electron was there.
So how does one reconcile these two results?
 A: Your derivation is correct.  The solution to $\ddot{x} = a_0 \cos (\omega t)$, with $\dot{x}_0 = x_0 = 0$ does not lead to a drift term, while the solution to $\ddot{x} = a_0 \sin (\omega t)$ (and the same initial conditions) does.
The best explanation I can come up with for why this is is that in the original problem, you are demanding that the velocity vanish when the force is zero;  while in the modified case, you are demanding that the velocity vanish when the force is at a maximum.  For conventional simple harmonic motion, the velocity is zero when the acceleration is at a maximum (and vice versa), so in your modified case you just get something like simple harmonic motion (albeit "off-center".)  But in the original case, we have to "add in" an extra term to ensure that the velocity is zero at the same time that the force is zero.
A: The interpretation I'd give in the "sine" case is that the arrival of the electric field imparts some energy/momentum that is absorbed by the particle (note the wave is not a perfect sine wave but has an associated step function pulse).
Since the electric field changes discontinuously in your "cosine" example, so I don't think it's very physical. Roughly speaking, I think you could say that the momentum imparted by the arrival of a sinusoidally varying EM wave is canceled by the momentum imparted by discontinuously changing the EM field. If you added a constant negative offset $-E_0$ for $t>0$ to the cosine field so that the field were continuous, there would be a net "secular" (non-oscillating) motion. Alternatively, if you added a "linear leading edge" field so that the field were $E=0$ from $-\infty < t < -T$, then $E=E_0 \left(1 + t/T\right)$ for $-T < t < 0$, then $E = E_0 \cos \omega t$ for $t > 0$ (then the field is continuous and connects to your cosine force), you would also find that there was secular motion associated with the "turn on" period of the field. Presumably this secular motion vanishes in the limit $\omega T \rightarrow 0$ (which is the limit you solved).
