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How to show that the motion graph of a linear oscillator is an ellipse.

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The motion of a linear oscillator may be represented by means of a graph in which x is abscissa and dx/dt as ordinate. The histroy of the oscillator is then a curve
    a)show that for an undamped oscillator this curve is an ellipse
    b) show (at least qualitatively) that if a damping curve is introduced on gets a curve spiraling into origin.
    2. Relevant equations
    3. The attempt at a solution
    a) I got that
    x(t)=Asin(wt-α)
    v(t)=wAcos(wt-α)
    Another student told me to "elimate the t's" to get
    x2/A2 +X2/(Aw)2 =1
    and that is total energy is E=1/2KA2 and w2=k/m then
    x2/(2E/k) +X2/(2E/m) =1


    First of all, I dont under stand how eliminated his t's. I do get that he found the eq of an ellipse, but how do I go from an eq with X and w to one with x and dx/dt?

    b)I have no sweet clue
     
  2. jcsd
  3. Oct 7, 2012 #2
    I suspect that his answer should read

    [tex]
    \frac{x^2}{A^2} + \frac{v^2}{A^2w^2} = 1
    [/tex]

    To arrive at this rearrange the x and v equations so that only the sin and cosine functions are left on the right hand side. Then square both equations and add. \

    For part b you will want to add a damping constant to the equation of motion:

    [tex]
    m \frac{d^2x}{dt^2} = -kx - c \frac{dx}{dt}
    [/tex]

    You will need to find solutions to this equation. From there you can find v, and plot x vs v.
     
  4. Oct 7, 2012 #3

    Ok a) makes total sense now
    b) In is c the damping constant? (we're using b) we have found in class that the solution to his comes in the fourm Aej(pt+α) is this what you mean?
     
  5. Oct 7, 2012 #4
    Yep, that's what I'm talking about.
     
  6. Oct 7, 2012 #5
    Ok then but Im still not sure where v is going to come from...
     
  7. Oct 7, 2012 #6
    Once you solve for x(t) then the velocity is just the derivative.
     
  8. Oct 8, 2012 #7
    Oh right. I get it. Thanks alot :)
     
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