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Simple Harmonic Motion

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Two-dimensional SHM: A particle undergoes simple harmonic motion in both the x and y directions
    simultaneously. Its x and y coordinates are given by
    x = asin(ωt)
    y = bcos(ωt)

    Show that the quantity x[itex]\dot{y}[/itex]-y[itex]\dot{x}[/itex] is also constant along the ellipse, where here the dot means the derivative with respect to time. Show that this quantity is related to the angular momentum of the system.

    2. Relevant equations
    L = mv x r

    3. The attempt at a solution
    Hi, so for the first part it is pretty simple and my answer is -abω, unless i made a dumb mistake which I don't think I did.

    It's the second part that is giving me issues. How do I show that it is related to angular momentum? I tried doing this

    L = [itex]\sqrt{L^{2}_{x}+L^{2}_{y}}[/itex]

    then L[itex]_{x}[/itex] = m[itex]\frac{∂x}{∂t}[/itex] x r

    where r = [itex]\sqrt{x^{2}+ y^{2}}[/itex]

    and then plugging everything in. I was hoping all the cos and sin were going to cancel out but it got really huge and messy. I didn't think it was supposed to be that hard so can anyone tell me if I am going in the right direction or if there is something I am missing?

    Thanks
     
  2. jcsd
  3. Sep 28, 2013 #2

    TSny

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    Gold Member

    Use the vector definition of angular momentum of a particle: ##\vec{L} = \vec{r} \times \vec{p}= m\; \vec{r} \times \vec{v}##, where ##\vec{r} = x\hat{i} + y\hat{j}##. (Note, the order of the cross product is important. Thus, ##\vec{L} = m\;\vec{v} \times \vec{r}## is not correct.)
     
  4. Sep 28, 2013 #3
    Right whoops.

    so then v = dx/dt i + dy/dt j

    and when we cross them we get m(xdy/dt - ydx/dt) = L correct?
     
  5. Sep 28, 2013 #4

    TSny

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    Yes. Good.
     
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