After setting up Newton's Law for the Lorentz Force on a particle with arbitrary velocity in a B-field directed along the z-axis, we found the coupled differential equations (below). We got as far as the y equation of motion but just wrote down the x solution. The complete solution is below.

We started with the differential equations:
<math> (1)\ddot{y}=-\omega\dot{x} \;\;\; (2)\ddot{x}=\omega\dot{y} \;\;\; </math> where <math> \omega=\frac{qB}{m} </math>

Integrating with respect to time once gives:
<math> (3)\dot{y}=-\omega x +C_1 \;\;\; (4)\dot{x}=\omega y + C_2 </math>

Substitute eq. 4 into eq. 1: which yields
<math> \ddot{y}=-\omega^2 y - C_2 </math>

Solution is of the form (check by direct substitution):
<math> (5)y=-Acos(\omega t+\phi)+y_0 \;\;\; y_0 = -C_2/\omega </math>

Take the time derivative of eq. 5 and substitute into eq. 3. This shows you why the amplitude, A, and phase, <math>\phi</math>, must be the same in both solutions and why you get the sin function.
The time derivative of eq. 5 is:
<math> (6)\dot{y}=A \omega sin(\omega t+\phi)
</math>

Substitute into 3:
<math> A \omega sin(\omega t+\phi) = -\omega x +C_1 </math>

Now just solve for x:
<math> x= -Asin(\omega t+\phi)+x_0
x_0=C_1/\omega </math>

The wiki works! -Dan K