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PE function

  1. Feb 15, 2006 #1
    A particle of mass M is free to move in the horizontal plane(xy-planne here). It is subjected to force [itex]\vec F = -k\left(x\hat i + y\hat j\right)[/itex], where 'k' is a positive constant.
    There are two questions that have been asked here:
    1] Find the potential energy of the particle.

    [tex]\vec \nabla \times \vec F = 0[/tex]
    The given force is conservative and hence a potential energy function exists.
    Let it be U.
    [tex]F_x = -\frac{\partial U}{\partial x} = -kx[/tex]

    [tex]F_y = -\frac{\partial U}{\partial y} = -ky[/tex]

    [tex]U(x,y) = \frac{k}{2}\left(x^2 + y^2) + C[/tex]

    2]If the particle never passes through the origin, what is the nature of the orbit of the particle?

    I am not sure what the PE function tells about the trajectory of the particle. Explanation needed...
     
  2. jcsd
  3. Feb 15, 2006 #2

    George Jones

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    The potential energy can be used to find the Lagrangian, and then Lagrange's equation can be used to find the motion.

    Alternatively, [itex]m \ddot{x} = F_x = -kx[/itex] and [itex]m \ddot{y} = F_y = -ky[/tex] can be solved directly.

    These equations should look very familiar.

    Regards,
    George
     
  4. Feb 21, 2006 #3
    Thank you for replying.

    So this is a 2-dimensional harmonic oscillator. The general solution would be:
    [itex]x = A\cos(\omega_0 t - \alpha)[/itex] & [itex]y = B\cos(\omega_0 t - \beta)[/itex]

    So, the resultant path of these two SHMs would be an ellipse, right?
     
  5. Feb 21, 2006 #4

    George Jones

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    Yes.

    Regards,
    George
     
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