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Finding spring constant of hung mass

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a certain mass m is hung from a spring,

    Find spring constant.

    mg=kx or 1/2kx^2 = mgh?

    These two give very different results for k, but I can't tell which is right or wrong!
     
  2. jcsd
  3. Nov 4, 2012 #2
    That isn't enough information to find the spring constant. There must be more to the question.
     
  4. Nov 4, 2012 #3
    i mean, which is the correct approach.

    EDIT: assume the spring start stretching from rest.

    you can let m=1, the h=1
     
  5. Nov 4, 2012 #4
    The correct approach depends on what information you have.

    The basic equation is Hooke's Law F = -kx, which you use if you know the force and the extension of the spring.

    The other equation gives the Energy stored in a spring 1/2kx2 and sets it equal to the potential energy of a mass hanging at a height h.
    It's not been written correctly. To make sense, h would be the same as x, in which case it's the same formula as the first.
     
  6. Dec 10, 2012 #5
    The reason why the two answers are different is that in the Hooke's Law equation, F represents the AVERAGE force, not the peak force. Force is not constant as the the spring extends from its relaxed length to its new equilibrium point with the weight hanging on it.

    The problem assumes that the spring starts from the relaxed length with zero force. The force at the maximum extension is the peak force (mg). Since k is a constant, the relationship between force and distance is linear. So, the average force is 1/2 of the peak force (1/2 * m * g).

    The two equations are now exactly the same.
     
  7. Dec 10, 2012 #6

    PhanthomJay

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    Actually, that's not quite right, Hookes law represents not the average force, but the force at a given displacement x. The energy approach appears to fail you because it assumes the mass is dropped from rest, which causes the spring to extend to twice as much than the first case where the mass is slowley lowered, before rebounding in simple harmonic motion. That means what you call h in the energy approach is actually 2x, where x is the displacement when the mass is slowly lowered by an external force to its equilibrium position, and then released, with no harmonic motion (no oscillation).
    As an example, assume m = 1 and k = 1. You place the object on a hanging spring and slowly lower it with your hand underneath. When the spring force and weight equalize, that is, when kx = mg, you no longer need to support the mass, and it hangs there by itself at equilibrium, and x = mg/k = 10. Now if instead you just dropped the weight, using conservation of energy since there is no force applied by your hand, then mgx = 1/2kx^2 at the bottom of the drop when there is no speed to the mass. Solving, mg = 1/2kx, or 2mg = kx, thus, x = 2mg/k = 20, which is twice as much an extension than the first case. k of course is still k, k = 1, but the force in the spring is not mg, but 2 mg at that low point. .
     
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