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Derivatives and equilibrium position of a spring

  1. Nov 19, 2013 #1
    I determined the equilibrium point of a spring by setting the potential energy function U(r) equal to zero and solving for r. But I just looked at the guided solution, and they took the derivative of U(r) first, then solved for r.

    Is my approach correct? Can we solve for the equilibrium position of a spring without taking any derivatives?
  2. jcsd
  3. Nov 19, 2013 #2


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    hi lonewolf219! :smile:
    you must somehow have chosen your constant in such a way that that happened to work

    potential energy is only unique up to a constant

    eg with gravitational potential energy we often set it equal to 0 "at infinity", or at the level of the lab floor :wink:
  4. Nov 19, 2013 #3
    Thanks Tiny Tim! But I thought that a spring has zero potential energy and maximum kinetic energy at the equilibrium position ? Am I wrong?
  5. Nov 19, 2013 #4


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    Solving for the zero of the potential energy function is bogus, because you can always add an arbitrary constant to the potential energy without changing any physics. So if you got the right answer, you got lucky in your choice of zero point (which is pretty easy to do in a lot of simple systems). On the other hand, adding an arbitrary constant won't affect the derivative, so if you're going to use energy methods the derivative approach is always correct (which may be why they're teaching it to you).

    If you want a derivative-free solution, you can set the net force on the end of the spring to zero, solve for the tension in the spring required to meet that condition. That can be very hard to do in the general case of time-varying forces acting on the spring, such as if there's a weight bouncing around on the end. For that problem, you'll likely find that minimizing the potential energy by looking for the zeroes of the first derivative is easiest.
  6. Nov 19, 2013 #5


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    bogus! that's the word i was looking for!

    thanks, Nugatory :smile:
    it only has has zero potential energy at the equilibrium position if you define it that way …

    and if you know enough about the equilibrium position in the first place, to define it that way, then why did you ever need to solve anything?

    bogus! :rolleyes:
  7. Nov 19, 2013 #6
    Okay, thanks guys. I noticed it worked for another problem but as Nugatory mentioned, it must be luck with simple systems
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