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Mass-spring system (not exactly homework)

  1. Mar 11, 2014 #1
    Case of a spring with a mass,m, that has been stretch beyond the equilibrium in the positive x direction.

    mg + (-kx0) = 0
    k = mg/x0


    F = mg+ [-k(x0+x1)] = mg-kx0-kx1
    but since mg-kx0 = 0
    F = 0-kx1 = ma

    ma = -kx1
    a = -kx1/m

    How do I arrive at the corollary from a = - kx1/m to ω0 = SQRT[k/x]?
  2. jcsd
  3. Mar 11, 2014 #2


    User Avatar

    Staff: Mentor

    ##a = -kx_1## can be expressed as ##d^2x / dt^2 = -kx_1##

    Now it comes down to fitting a solution to this differential equation......
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