physics

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 2.0 m/s. Determine the amplitude A of the motion.

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asked by Kate
  1. x = A sin 2 pi f t
    v = 2 pi f A cos 2 pi f t
    max speed is when cos = 1 or - 1
    vmax = 2 pi f A = 2

    F = m a = -kx

    a = -(2 pi f)^2 A sin 2 pi f t
    = - (2 pi f)^2 x
    so
    .2 (-2pi f)^2 x = -120 x
    so
    2 pi f = sqrt (k/m) but you knew that :)
    2 pi f = sqrt(120/.2) = 24.5
    but
    2 pi f A = 2
    24.5 A = 2
    A = .0816 or about 8 centimeters

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    posted by Damon

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