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

by SuicideSteve
Tags: j.j. thomson, mass/charge
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,346 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.