Definite Integration of F=ma

1. Jan 30, 2013

Astrum

This isn't a HM question, and I'm asking for an explanation.

This is "The effect of a Radio Wave on an Ionospheric Electron"

The integration is weird, I don't follow what is being done.

$$a=\frac{-eE}{m}$$ - reworking of F=ma

$$\frac{-eE}{m}sin(\omega t$$

only interested in the x axis.

$$\int\frac{dv}{dt}=\int^{t}_{0}a_{0}sin(\omega t) dt$$

This becomes: $$v(t)=v_{0}-\frac{a_{0}}{\omega}cos(\omega t-1)$$
- I don't get where this came from, I understand the indefinite integration, but not where the "ωt-1" came from.

And the last step:

$$\int\frac{dx}{dt}=\int^{t}_{0}[v_{0}-\frac{a_{0}}{\omega}cost(\omega t-1)]dt$$

= $$x_{0} + (v_{0}+\frac{a_{0}}{\omega})t-\frac{a_{0}}{\omega^{2}}sin(\omega t)$$

Not sure where the final answer comes from. Could't you just integrate it twice, then tack on the definite integral?

2. Jan 30, 2013

AlephZero

I think some parentheses are in the wrong place.
$$\int_0^t a_0 \sin(\omega t)\, dt = \left[ -\frac{a_0}{\omega} \cos(\omega t)\right]_0^t$$
$$= - \frac{a_0}{\omega}(\cos(\omega t) - \cos 0)$$
$$= - \frac{a_0}{\omega}(\cos(\omega t) - 1)$$