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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 $\dot{y}$?

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What exactly do you mean? Are you asking if there exists an equivalent set of equations that only involve $\dot x$ and $\dot y$? – joshphysics Mar 2 '13 at 17:37
Or if you just what to solve this system, make the following switch to complex coordinates $z=x+iy$. Its much easier this way, solving for $z$. – nijankowski Mar 2 '13 at 17:40
I like to make a simulation of a foucault pendulum so I need to know what is $\dot{x}$ and $\dot{y}$ are. so yeah that would be my question @joshphysics – user61835 Mar 2 '13 at 17:45
@user61835: then why aren't you just doing a finite difference method? All you need for that are the second derivatives as a function of the positions and the velocities. – Jerry Schirmer Mar 2 '13 at 18:04
@JerrySchirmer could you please explain how could i do that? – user61835 Mar 2 '13 at 18:11

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


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


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.

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Thanks for that, I think that would do. But what would be A and B? – user61835 Mar 2 '13 at 18:36
It says so in the text: $z(t=0)=2A$, thus $A$ is half the value of $z$ at $t=0$. – Rafael Reiter Mar 2 '13 at 18:46
As you can see from the solutions, it gives you the distance from the origin when $t=0$. When $t=0$, $x(t=0)=2A$ and $y(t=0)=0$, this gives you the point in the 2D plane of your simulation from where the pendulum starts. And don't worry about $B$ because the solutions are expressed only with the use of $A$. – nijankowski Mar 2 '13 at 18:50
I used these solutions to make a simulation of the pendulum in Mathematica some years ago. Take a look at these projects from Wolfram… – nijankowski Mar 2 '13 at 18:53

If I had to take a reverse-engineering guess, I would say that you want to solve the system by creating an autonomous system. It goes like this:

Define $u=\dot x$ and $v=\dot y$ (similar to a substitution). Then, your equations read:

$$\dot{u} = 2\omega \sin\lambda v - \frac{g}{L}x$$

$$\dot{v} = -2\omega \sin\lambda u - \frac{g}{L}y$$

$$\dot x=u$$

$$\dot y=v$$

Then, define $\mathbf x=(u,v,x,y)$. Your equation then reads

$$\frac{d\mathbf x}{dt}=\mathbf f(\mathbf x)$$

which you can solve using a standard numerical ODE (ordinary differential equation) solver, e.g. Runge-Kutta (tricky) or Euler (simpler, but much less accurate).

Here is a nice explanation for how to use Runge-Kutta for a system of equations (top of page 4).

This is what you need to make a simulation. As the answer by Nijankowski V. states, though, there is an analytical solution for the linear pendulum.

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If I follow this one, what happens with x and y? – user61835 Mar 2 '13 at 18:54
$x$ and $y$ are contained in $\mathbf x$. You do the calculation, and then the third and fourth entry of the vector are $x$ and $y$, and you can throw away the first and second entry. $x$ and $y$ are both functions of time, and they behave like a pendulum does: they oscillate. – Rafael Reiter Mar 2 '13 at 18:56
oh okay so I am just going to use Runge-Kutta with four variable, is that right? – user61835 Mar 2 '13 at 19:04
Yes, that's right. Please note the link I added on how to use Runge-Kutta with systems of equations. – Rafael Reiter Mar 2 '13 at 20:00

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