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A mass of 1 kg is hanging from a spring with a spring constant of 3 N/m. At a distance of 0.2 m below the equilibrium it has a velocity of 1 m/s in the upward direction. What is the amplitude of the oscillation?

  • college physics -

    F = -k y = -3 y
    when y = -.2, F = .6 N up
    w = sqrt (k/m) = 1.73 radians/s
    y = A sin w t
    v = w A cos w t
    a = -w^2 A sin w t = -w^2 y

    at time t
    v = w A cos w t = 1
    y = A sin w t = -.2
    a = F/m = .6/m = .6

    so
    1.73 A cos wt = 1
    A cos wt = .577
    and
    A sin wt = -.2
    A sin wt/A cos wt = -.2/.577 = - .346
    tan w t = -.346
    w t = -19.1 degrees = -.333 radians
    sin wt = -.327
    A = -.2/-.327 = .611
    check
    .611 cos -19.1 = .577 right
    v = .577 (w) = .577*1.73 = .998 right

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