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.


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?

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    $\begingroup$ Can you give us some idea of your background? Are you a mathematician - if you're considering solving equations of motion you're presumably happly with differential equations. Does this article on the 2D harmonic oscillator make any sense? $\endgroup$ – John Rennie Jun 14 '14 at 10:17
  • $\begingroup$ @JohnRennie I am studying mathematics, now dealing with ODE. I always have problems with example of physics. So to be honest, I do not understand a word of the link. $\endgroup$ – math12 Jun 14 '14 at 10:31
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    $\begingroup$ The harmonic oscillator is where the force is proportional to the displacement. A physical implementation of the one-dimensional h.o. is the spring pendulum. The normal pendulum isn't a harmonic oscillator (but can be approximated as one if the amplitude is sufficiently low). Solving a spherical pendulum is much harder than solving a harmonic oscillator, so you'd not normally substitute a spherical pendulum for a harmonic oscillator (rather, the other way round). $\endgroup$ – celtschk Jun 14 '14 at 12:42
  • $\begingroup$ Ok, so I now have the movement equations for a spherical pendulum. But I do not know how to find the movement equations for a harmonic oscillator... $\endgroup$ – math12 Jun 14 '14 at 12:48
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    $\begingroup$ See the edit of my answer. $\endgroup$ – celtschk Jun 14 '14 at 13:04

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.

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  • $\begingroup$ Are you using a spring pendulum now? Before I do this (thank you!) two questions remain: How can I express the coordinates $x$ and $y$? Why is the potential energy $ax^2+bxy+cy^2$? $\endgroup$ – math12 Jun 14 '14 at 13:13
  • $\begingroup$ Is a 2-dim. harmonic oscillator a spring pendulum that can move down and up and aditionally like a spherical pendulum? $\endgroup$ – math12 Jun 14 '14 at 13:20
  • $\begingroup$ $x$ and $y$ are just two directions in which the oscillator can be displaced. They don't even need to be orthogonal. The potential energy $ax^2+bxy+cy^2$ is just the most general possible quadratic form without linear or constant term. It has that form because that's what makes the system a harmonic oscillator (you could plug another potential in here, but then you'd no longer have a harmonic oscillator — well, unless your coordinates are something else than displacements, then it may be a harmonic oscillator in different coordinates). $\endgroup$ – celtschk Jun 14 '14 at 13:22
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    $\begingroup$ @math12: "Two-dimensional harmonic oscillator" means, by definition, "two-dimensional system with attractive quadratic potential". "Quadratic potential" means "potential that is a polynomial of degree 2 in the position", and "attractive" means "when you go away from the equilibrium position, the potential gets greater". $\endgroup$ – celtschk Jun 14 '14 at 13:41
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    $\begingroup$ It's the Hamiltonian of a harmonic oscillator; the Hamiltonian is used in the Hamilton formalism (another way to derive the equations of motion). Basically, $H=T+V$, except that $T$ is described with the momentum. Otherwise, that's a special case of the potential I've given (with $a=\tfrac12w_x^2$, $b=0$, $c=\tfrac12w_y^2$), achieved by using the main axes of the potential as coordinates. BTW, I just notice that I've got a sign error in the equations of motions; I'll correct it in a moment. $\endgroup$ – celtschk Jun 14 '14 at 13:46

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