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A 200g mass is attached to a spring and allowed to execute simple harmonic motion with a period of 0.25seconds. If the total energy of the system is 2.0 joules, find the force constant of the spring and the amplitude of the motion.

  • physics -

    F = -k x
    x = A sin w t
    v = Aw cos wt
    a = -Aw^2 sin wt = -w^2 x
    but
    F = m a
    -k x = m(-w^2 x)
    so
    w^2 =k/m
    (2pi)^2/T^2 = k/m
    k = m (2pi)^2/T^2 = 0.200* (2pi)^2/0.25^2
    k = 126 Newtons/meter
    now back to get A
    when at max x, energy = (1/2) k x^2
    = (1/2) k A^2
    so
    (1/2) k A^2 = 2 Joules
    we know k, solve for A

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