## J.J. Thomson's Experiment, Help!

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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 I'm sorry, I believe I have posted this in the wrong section, as I am asking for help.
 Recognitions: Gold Member Science Advisor Staff Emeritus Why did you "substitute 1 for the constants"? It doesn't really simplify anything. Differentiating the first equation one more time, $$m(d^3x/dt^3) + He(d^2y/dt^2) = 0$$ and we can substitute He dx/dt for $d^2y/dt^2$ so we have $$m(d^3x/dt^3)+ H^2e^2dx/dt= 0$$ That has characteristic equation $mr^3+ H^2e^2r= r(mr^2+ H^2e^2)= 0$ which has roots 0, (He/m)i and -(He/m)i so that the general solution is $$x(t)= C+ Dcos((He/m)t)+ E sin((He/m)t)$$ and then you can use $$\frac{d^2y}{dt^2}= He \frac{dx}{dt}$$ to find y.

 Tags j.j. thomson, mass/charge