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Definite Integration of F=ma

by Astrum
Tags: definite, integration
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Astrum
#1
Jan30-13, 07:07 PM
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P: 274
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.

[tex]a=\frac{-eE}{m}[/tex] - reworking of F=ma

[tex]\frac{-eE}{m}sin(\omega t[/tex]

only interested in the x axis.

[tex]\int\frac{dv}{dt}=\int^{t}_{0}a_{0}sin(\omega t) dt[/tex]

This becomes: [tex]v(t)=v_{0}-\frac{a_{0}}{\omega}cos(\omega t-1)[/tex]
- 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:

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

= [tex]x_{0} + (v_{0}+\frac{a_{0}}{\omega})t-\frac{a_{0}}{\omega^{2}}sin(\omega t)[/tex]

Not sure where the final answer comes from. Could't you just integrate it twice, then tack on the definite integral?
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AlephZero
#2
Jan30-13, 07:41 PM
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I think some parentheses are in the wrong place.
$$\int_0^t a_0 \sin(\omega t)\, dt = \left[ -\frac{a_0}{\omega} \cos(\omega t)\right]_0^t$$
$$= - \frac{a_0}{\omega}(\cos(\omega t) - \cos 0)$$
$$= - \frac{a_0}{\omega}(\cos(\omega t) - 1)$$


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