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Homework Help: Spherical Oscillator

  1. Mar 24, 2008 #1


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    I'd really appreciate some speedy advice on this question. I need to know whether the expressions for energy and the angular momentum are correct, as I seem unable to produce the final expression desired from what I have derived.

    Here's the problem:

    1. The problem statement, all variables and given/known data

    The Lagrangian is

    [tex]L = \mu \nu^{2}/2 - kr^{2}/2[/tex]

    i. Calculate the angular momentum vector, L, proving that L is constant and that, unless L vanishes, the oscillation occurs in the x,y plane.

    ii. Using [tex]x = x_{0}cos(\omega t), y = y_{0}sin(\omega t)[/tex], show the trajectory is an ellipse. Express the energy, E, and then both the eccentricity and the parameter in terms of E and Lz.

    iii. Representing the trajectory in polar coordinates by [tex]x = rcos(\phi), y = rsin(\phi)[/tex], show that you get the simple relation

    cos (2\phi) = \frac{1 - \nu^{2} / (2\xi^{2})}{\sqrt{1 - \nu^{2}}} ,
    r = \sqrt{ \frac{2E}{k} }\xi ,
    \nu = \frac{ L_{z}\omega }{E}

    3. The attempt at a solution

    For i. I found that [tex]L = Lz = \mu \omega x_{0}y_{0}sin(\phi_{x}-\phi_{y})[/tex] (assuming L is not zero), by using three sine equations for the x,y and z components (with angle offsets) and taking [tex]L = m(r \times v)[/tex]; I chose my axis so that x and y could be rewritten as [tex]x = x_{0}cos(\omega t), y = y_{0}sin(\omega t)[/tex], and then find I can write [tex]Lz = \mu \omega x_{0}y_{0}[/tex].

    For ii., it seemed all that was required was to square both expressions and combine thus:

    [tex]\frac{x}{x_{0}}^{2} + \frac{y}{y_{0}}^{2} = 1[/tex]

    Writing the energy [tex]E = \mu \nu^{2}/2 + kr^{2}/2[/tex] and substituting the expressions for x and y, I found the expression to boil down to [tex]E = \frac{(\mu \omega^{2})}{2} (x_{0}^{2}+y_{0}^{2})[/tex] where [tex] \omega^{2} = k/m[/tex].

    I'm guessing I need to go ahead and solve this equation for x0 (assuming it's right) by using Lz to get rid of y0, and hope that something which looks like the eccentricity and the parameter will emerge (?).

    Part iii. has me really stuck. I substitute [tex]x = rcos(\phi), y = rsin(\phi)[/tex] into the ellipse equation, which looks like a promising start, use double angle results to turn the cos and the sine into [tex]cos (2\phi)[/tex], and eventually obtain (after some simply rearranging):

    [tex]cos (2\phi) = \frac{1-r^{2}/2(1/x_{0}^{2}+1/y_{0}^{2})}{r^{2}/2(1/x_{0}^{2}-1/y_{0}^{2})}[/tex]

    Multiplying numerator and denomiator by [tex]x_{0}y_{0}[/tex] and using my expressions derived earlier for Energy and Lz I arrive at:

    [tex]cos (2\phi) = \frac{L_{z}^{2}/\mu^{2}\omega^{2}-r^{2}/2(2E/\mu\omega^{2})}{r^{2}/2(y_{0}^{2}-x_{0}^{2})}[/tex]

    It looks promising, but once the first term on the numerator has been made one, it seems no manipulation can make this equation correspond to the one given. I can only presume my energy, angular moment or ellipse equations are wrong? But are they?

    Many thanks for your insights!
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 26, 2008 #2
    Hey, Can I just ask how you got the expression for [itex]E=\frac{(\mu\omega^2)}{2}(x_0^2+y^2_0)[/itex]
  4. Mar 26, 2008 #3
    I've been working on this problem as well, and I think that the expression for the energy should have a k term


    Because if you put the terms for xo and yo in for r I can't see why the k term would disappear. Also for your expression for the ellipse the denominators should be squared.

    I haven't gone any further with this yet, but the method seems like it should work from what you've explained.
  5. Mar 26, 2008 #4
    Can I just ask if we are all handing this in tomorrow for a certain classical mechanics module at a certain Scottish University!?!

    I tried using Maple to get the expression for E which comes out as:

    I can't seem to get very far with the next question and haven't even looked at the last one. Oh Classical Mechanics!

    has anyone had any luck getting the eccentricity out?
    Last edited: Mar 26, 2008
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