Analytical mechanics: 2D isotropic harmonic oscillator

AI Thread Summary
The discussion centers on solving the equations of motion for a two-dimensional isotropic harmonic oscillator with given initial conditions. Participants suggest starting with constants of motion, including energy and angular momentum, and converting the problem into ordinary differential equations for x and y. The equations can be separated and solved individually, leading to solutions involving sine and cosine functions. One user encounters an imaginary solution for x, prompting a discussion about its relation to trigonometric functions. The conversation emphasizes the importance of applying initial conditions to solve for constants in the solutions.
swindhspectrum
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I'm stuck on this problem:
The initial conditions for a two-dimensional isotropic oscillator are as follows: t=0, x=A, y=4A, v=0i +3wAj (vector) where w is the angular frequency. Find x and y as functions of t.

Where do I even begin with this problem. I take it A = constant. Can anyone make this less intimadating?

thanks.
 
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By isotropic I assume you mean the force depends only on the distance from the origin, ie, F=kr, directed towards the origin.

You probably want to start by considering the constants of motion. You know U=1/2kr^2, and that the total energy E=U+KE is conserved. Also, since the system has rotational symmetry, angular momentum is conserved, ie, r X v (the cross product of the position and velocity vectors) is a constant, equal to its initial value for all time. With a little work, and using polar coordinates, you can use these equations to turn the problem into an ordinary differential equation involving r.

EDIT: Actually, I've gone through the work, and that DE is a pain to solve. It's not impossible, but there's probably an easier way. If I think of one I'll let you know.
 
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Right, so you might want to try the above approach, as it's similar to the methods you'll need to use on more difficult problems. But in this case, you can just separate the x and y equations of motion:

m\frac{d^2 \vec r}{dt^2} = -k \vec r

this gives:

\frac{d^2 x}{dt^2} = -k x

and

\frac{d^2 y}{dt^2} = -k y

These can be solved seperately. But note again that this method is not very general, and in most problems you'll need to do something like what I suggested in the last post.
 
thanks, ill try this out
 
i don't know if you (statusX) tried to solve these...
i am getting an imaginary answer for x (x=iwt). i haven't solved for y(t) yet. Isn't the imaginary answer somehow equivalent to a trig function? Does anyone know?
thanks
 
The solution to both equations will involve sines and cosines. Have you solved these types of equations before? I assume you meant to say you got something like x=e^iwt, which can be converted to sines and cosines using the http://mathworld.wolfram.com/EulerFormula.html" . You are solving a 2nd order ODE, so you should get two constants, which can be solved for by plugging in values for x and dx/dt at t=0.
 
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