J.J. Thomson's Experiment, Help!


by SuicideSteve
Tags: j.j. thomson, mass/charge
SuicideSteve
SuicideSteve is offline
#1
Apr10-12, 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(d2x/dt2) + He(dy/dt) , m(d2y/dt2) - 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/H2e}(1 - cos([He/m]t))
y = {Em/H2e}([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.
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SuicideSteve
SuicideSteve is offline
#2
Apr10-12, 08:16 PM
P: 24
I'm sorry, I believe I have posted this in the wrong section, as I am asking for help.
HallsofIvy
HallsofIvy is offline
#3
Apr11-12, 07:43 AM
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Thanks
PF Gold
P: 38,900
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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