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Homework Help: Potential energy for the harmonic oscillator

  1. Apr 22, 2010 #1
    Hey all,

    In the classical harmonic oscillator the force is given by Hooke's Law,
    F = -kx which gives us the potential energy function V(x)=(1/2)kx^2. Now I understand that the first derivative at the point of equilibrium must be zero since the slope at the point of equilibrium is zero. But what is the meaning of the second derivative at that point? Is it how fast the slope is changing? My professor says that it has to be positive or equal to zero but that for simple harmonic motion it is always positive. I don't see how you can have the first derivative of a function evaluated at some point be zero but the second derivative at the same point be nonzero.
     
  2. jcsd
  3. Apr 22, 2010 #2

    Doc Al

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    Staff: Mentor

    The slope of the potential energy function tells you the force, which you know must equal zero at equilibrium.
    Sure. Since slope relates to force, it's the rate of change of the force as you move past x = 0.
    That's true.
    Why do you think that? Plot the potential energy function and you get a parabola. The slope of the tangent to that curve happens to be zero at x = 0 (the first derivative is zero) but the slope certainly changes as you go from x < 0 to x > 0. The rate of change of the slope--the second derivative--is non-zero.
     
  4. Apr 22, 2010 #3

    Doc Al

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    Staff: Mentor

    Read the first couple of paragraphs of this page: http://www-users.math.umd.edu/~tjp/220%2002.2%20lecture%20notes.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Apr 22, 2010 #4
    I forgot that the potential energy can be written as the gradient of the force, this helps in understanding what is going on physically. Thanks!

    Ahhh... I think I understand now. Before I was thinking that if I took the derivative at the equilibrium point that it would be zero and that the derivative of zero is zero. But I have to compute the second derivative of the function first and then evaluate it at the point of equilibrium correct?

    Actually now that I think about it, if I take the second derivative of (1/2)kx^2 then you get kx, and plugging in a value of x taken to be the equilibrium point say x=b, you don't get zero, just some constant: kb. This would mean that the slope is zero since it's just a straight line which makes sense. But isn't it supposed to be zero? I'm confusing myself now, I'm forgetting my basic calculus.

    Thanks for your quick reply by the way.
     
  6. Apr 22, 2010 #5

    Doc Al

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    Staff: Mentor

    Careful. kx is the first derivative of (1/2)kx^2. And that equals zero at x = 0, the equilibrium point. For x = b ≠ 0, you'll get kb. Which should make sense, since as you move away from equilibrium, the restoring force increases. (The second derivative is just k, a constant.)
     
  7. Apr 22, 2010 #6
    OHH Ok now I really do get it. I was imagining a potential energy function that was symmetric about a point other than zero, for example the point x=b. But in that case the function would be different from (1/2)kx^2. My mistake.

    Thanks!

    Also thanks for the link it is a good review.

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