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Homework Help: Circular motion as a 2D harmonic oscillator

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    This is for my mechanics class. It seems like it should be easier than I'm making it.

    A single object of mass m is attached to the ends of two identical, very long springs of spring constant k. One spring is lined up on the x-axis; the other on the y-axis. Chpose your axes and positions of the springs so that the equilibrium position of the object is at x = y = 0. The springs are long enough that if the mass is at [tex] x\hat{i} + y\hat{j}[/tex], then the restoring force is [tex]-k(x\hat{i} + y\hat{j})[/tex]. Assume there is no damping in this problem, and feel free to make the substitution [tex]\omega = \sqrt{k/m}[/tex].

    Problem: Assume that the oscillations in the x and y directions have the same amplitude A and are in phase. Describe the object's path in terms of circular coordinates.


    There are more problems like this (one is out of phase by [tex]\pi/2[/tex], then one is out of phase and has a slightly different amplitude, and then a more general version of the problem), but I feel that if I can get this, I can get the rest.

    2. Relevant equations

    To make the transition between Cartesian and polar coordinates,

    [tex]
    x = r\cos\theta

    y = r\sin\theta

    x^2 + y^2 = r^2

    \theta = \tan^{-1}\frac{y}{x}
    [/tex]


    3. The attempt at a solution

    My first instinct was to solve these problems as two separate differential equations and then combine them. The two differential equations were

    [tex]
    \ddot{x} + \omega^2x = 0
    \ddot{y} + \omega^2y = 0
    [/tex]

    Solving these gives

    [tex]
    x(t) = A\cos(\omega t - \phi)
    y(t) = A\cos(\omega t - \phi)
    [/tex]

    In polar coordinates, it seems like that would be [tex] r = A\sqrt{2}\cos(\omega t - \phi)[/tex], but I can't figure out how to get the t out of there. I also tried making this into a complex problem, where [tex]z = x + iy = re^{i\theta}[/tex] (although this way, I cannot figure out how to bring in the phase shift). Taking the last equality makes the differential equation become

    [tex]\ddot{r} + 2i\dot{\theta}\dot{r} + r(\omega^2 - \ddot{\theta}) = 0[/tex]

    but if that's the correct approach, I have no idea how to solve it.

    Thanks for any help you can provide! :smile:
     
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 23, 2010 #2
    Hmm... If they are in phase then that looks right. You will get an oscillation back and forth along a single axis. The radius will oscillate. Check [tex]\theta(t)[/tex], you will see it not depend on time.
     
  4. Mar 23, 2010 #3
    Thanks for your help! Two questions, though. Why will it oscillate along a single axis? Do you mean the "radial" axis (i.e. staying at a constant radius?). Further, the tough part about this whole problem set is getting it in polar coordinates. That implies that I don't have explicit dependence on time, but rather on angle. I can't seem to get time out of the equation.

    Thanks again!
     
  5. Mar 23, 2010 #4
    Since the x and y are in phase, then they will have the same magnitude at all times, even though their magnitude is time dependent. So you can imagine a mass oscillating along the +x,+y axis through the origin and along the -x,-y axis, and finally back again. No circular motion, just a simple 1D oscillation with [tex]A\sqrt{2}[/tex] amplitude.
     
  6. Mar 23, 2010 #5
    I see that you're right. I think I may be going about describing it in polar coordinates the wrong way. I don't think I'm understanding that part correctly.
     
  7. Mar 23, 2010 #6
    Your polar coordinates are right. Although you never did solve for theta. You only solved for 'r'.
     
  8. Mar 23, 2010 #7
    I suppose solving for theta is where I'm confused (plus it's still time-dependent). As a function of time, my guess is that

    [tex]\theta - \tan^{-1}\frac{y(t)}{x(t)} = \tan^{-1}(1) = \pi/4[/tex]

    Where would that go in the equation? And is it okay to have explicit time dependence as well as implicit?

    Edit: Or would it simply be [tex]z = A\sqrt{2}\cos(\omega t - \phi)e^{i\pi/4}[/tex]? But there's still that time dependence...
     
  9. Mar 23, 2010 #8
    In polar coordinates you have two coordinates, 'r' and 'theta'. So you always need to solve for both. You just solved for theta and showed it is time independent. So basically your mass will oscillate along the Pi/4 angle, since the angle is fixed.

    When they are out of phase, then theta will become time dependent.
     
  10. Mar 23, 2010 #9
    Right. I guess that's why it's bothering me that I have the time component that I can't seem to get rid of.
     
  11. Mar 23, 2010 #10
    At least one of the coordinates need a time component. If they didn't, then the mass wouldn't be moving at all.
     
  12. Mar 23, 2010 #11
    Oh right. Silly me. Thanks so much for your help!
     
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