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J.J. Thomson's Experiment, Help! 
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#1
Apr1012, 07:49 PM

P: 24

Okay, here is the problem I was given.
In Thomson's experimental determination of the ratio m/e of the mass to the charge of an electron, in which the electrons were subjected to an electric field of intensity E and a magnetic field of intensity H, the equations m(d^{2}x/dt^{2}) + He(dy/dt) , m(d^{2}y/dt^{2})  He(dx/dt) = 0 , were employed. If x=y=dx/dt=dy/dt=0 for t=0, show that the path is a cycloid whose parametric equations are: x = {Em/H^{2}e}(1  cos([He/m]t)) y = {Em/H^{2}e}([He/m]t  sin([He/m]t)) I have solved the differential equation by substituting 1 for the constants and come out with: x = 1  cost y = t  sint My problem is I can not figure out how to end up with the constants in the results. Any help is greatly appreciated, Thanks. Steve. 


#2
Apr1012, 08:16 PM

P: 24

I'm sorry, I believe I have posted this in the wrong section, as I am asking for help.



#3
Apr1112, 07:43 AM

Math
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Thanks
PF Gold
P: 39,693

Why did you "substitute 1 for the constants"? It doesn't really simplify anything.
Differentiating the first equation one more time, [tex]m(d^3x/dt^3) + He(d^2y/dt^2) = 0[/tex] and we can substitute He dx/dt for [itex]d^2y/dt^2[/itex] so we have [tex]m(d^3x/dt^3)+ H^2e^2dx/dt= 0[/tex] That has characteristic equation [itex]mr^3+ H^2e^2r= r(mr^2+ H^2e^2)= 0[/itex] which has roots 0, (He/m)i and (He/m)i so that the general solution is [tex]x(t)= C+ Dcos((He/m)t)+ E sin((He/m)t)[/tex] and then you can use [tex]\frac{d^2y}{dt^2}= He \frac{dx}{dt}[/tex] to find y. 


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