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Definite Integration of F=ma 
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#1
Jan3013, 07:07 PM

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 t1)[/tex]  I don't get where this came from, I understand the indefinite integration, but not where the "ωt1" came from. And the last step: [tex]\int\frac{dx}{dt}=\int^{t}_{0}[v_{0}\frac{a_{0}}{\omega}cost(\omega t1)]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? 


#2
Jan3013, 07:41 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,121

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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