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Potential energy in a two spring system

  1. Nov 3, 2009 #1
    A simple conceptual physics question on springs:

    Given a spring of length L0, with spring constant k, say this spring has been displaced to length L, thus its potential (stored) energy is given by:
    E = k(L-L0)^2/2

    What if this spring is broken down into two springs (half the length of the original), attached in series. The spring constant of the spring is a material property, correct? So, the spring constants should be the same no matter what the length of the spring is, i assume.
    So spring 1 has equilibrium length L0/2 as does spring 2, and they should both still have spring constants of k

    Displace the two spring set so that each spring is now at length L/2
    The total potential energy of this system is:
    E = E1 + E2 = k(L/2 - L0/2)^2/2 + k(L/2 - L0/2)^2/2
    E = k((L-L0)/2)^2 = k(L-L0)^2/4
    This doesn't match the total energy of the 1 spring case.

    Why not? What am I not understanding?

    Shouldn't the potential energy be the same for the 1 spring and 2 spring systems? If they are displaced the same overall and have the same overall equilibrium length and if the springs are made of the same material and thus have the same spring constant?

    Any explanations would be quite helpful!
    thank you!
     
  2. jcsd
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