How can I find the motion equations of the 2-dim harmonic oscillator? First of all: I am no physicist, so I am rather helpless.
I need to find the moving equations of the 2-dim. harmonic oscillator.
If it is possible it should be rather elementary, because, as I said, I am no physicist. I've heard that ONE possibility is Hamilton mechanics, but to be honest, I hardly know what that is!
Maybe you can give me help? I would be very thankful.
Update:
For the general spherical pendulum I get
$$
\ddot{\theta}=\sin\theta\cos\theta\dot{\varphi}^2+\frac{g}{l}\sin\theta,~~\ddot{\varphi}=\frac{-2\cos\theta\dot{\theta}\dot{\varphi}}{\sin\theta}.
$$
Isn't the 2-dim. harmonic oscillator a special spherical pendulum so that I maybe can use this results?
 A: The trick with the two-dimensional harmonic oscillator is to recognize that there are two directions so that movement in one direction is independent of the movement in the other (if the harmonic oscillator is rotationally symmetric, any two orthogonal directions will do). If you plot the equipotential lines of the oscillator potential (that is, the potential energy if the mass is at that point), it consists of ellipses; the main axes of those ellipses give those two directions.
In each of the directions, the equation of motion is just the equation of motion of a one-dimensional harmonic oscillator. So you solve the two one-dimensional harmonic oscillators separately.
If you don't want to use such a shortcut, you can also calculate it directly using any of the usual methods, like Lagrange formalism or Hamilton formalism.
Here's how you would do it in Lagrange formalism:
Step 1: determine the kinetic and potential energy of the 2D harmonic oscillator.


*

*Kinetic energy: $T = \tfrac{1}{2}m(\dot x^2+\dot y^2)$
Here $x$ and $y$ are the coordinates, and the dot describes the time derivative, that is, $\dot x$ and $\dot y$ are the components of the velocity.

*Potential energy: $V = ax^2 + bxy + cy^2$
Here $a$, $b$ and $c$ are general constants (with the restriction that $a>0$, $c>0$ and $2ac-b^2>0$). This is the most general two-dimensional harmonic oscillator potential with the restriction that the minimum is at $x=y=0$ (and the value there is $0$, but a constant term in the potential doesn't change the equations of motion).
Step 2: From the kinetic and potential energy, you calculate the Lagrange function. That step is trivial: The Lagrange function is always $L=T-V$, that is in this case,
$$L = \tfrac12m(\dot x^2+\dot y^2) - ax^2-bxy - cy^2$$
Step 3: To derive the equation of motion, you just plug this lagrange equation into the Euler-Lagrange equatons (of the second type): For each coordinate $q$ (that is here, $x$ and $y$), the equation of motion reads
$$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q} = \frac{\partial L}{\partial q}$$
So for $x$, we get $\partial L/\partial\dot x = m\dot x$ and $\partial L/\partial x = 2ax + by$, and thus
$$m\ddot x = -(2ax+by)$$
and analogously
$$m\ddot y = -(bx + 2cy)$$
Those are the equations of motion.
