[Physics] Foucault pendulum

Question

The equations of motions for a Foucault pendulum are given by:

$$\ddot{x} = 2\omega \sin\lambda \dot{y} – \frac{g}{L}x,$$
$$\ddot{y} = -2\omega \sin\lambda \dot{x} – \frac{g}{L}y.$$

What are the equations describing $\dot{x}$ and $\dot{y}$?

Answer 1

I'll work a little backwards, but arrive at a form for $x(t)$ and $y(t)$, which you can use for your simulation.Having those differential equation and making the switch to complex coordinate $z=x+iy$ you get the following diff. equation

$\ddot{z}+2i\omega\dot{z}\sin{\lambda}+\omega_{p}^2z=0$

with $\omega_{p}^2=\frac{g}{L}$. For this kind of diff. equation you take a solution of the following type

$z(t)=Z_{0}(t)e^{-i\omega\sin{\lambda}t}$. Inserting this into the eq. above you arrive at

$\ddot{Z}(t)+(\alpha^2+\omega_{p}^2)Z(t)=0$. Where $\alpha=\omega\sin{\lambda}$ and $\alpha^2$ is tiny when we compare it to $\omega_{p}^2$ so we can neglect it. So, this leave you with $\ddot{Z}(t)+\omega_{p}^2Z(t)=0$. And the solution for this has the following general expression

$Z(t)=Ae^{i\omega_{p}t}+Be^{-i\omega_{p}t}$

and the complete solution is now

$z(t)=e^{-i\alpha t}(Ae^{i\omega_{p}t}+Be^{-i\omega_{p}t})$

Here you can see that are two special cases which correspond to harmonic oscillations of the pendulum, when $A=B$ and $A=-B$. In the first care you find

$z(t)=2Ae^{-i\alpha t}\cos{\omega_{p}t}$ and in the second case

$z(t)=2ie^{-i\alpha t}\sin{\omega_{p}t}$

The first solution corresponds to the initial condition $z(t=0)=2A$ and the second to $z(t=0)=0$. In these solutions, you can see that the exponential factor is due to the Coriolis force. To get rid of these exponential you apply Euler formula. After doing this, to find the "real" trajectories, you take the real and imaginary part of $z$. Hence, for the first solution you find

$Re(z)=x(t)=2A\cos(\alpha t)\cos(\omega_{p}t)$

$Im(z)=y(t)=-2A\sin(\alpha t)\cos(\omega_{p}t)$

You can do the same for the last solution. Having these, you can simply find $\dot{x}(t)$ and $\dot{y}(t)$. Hope it helps.