Physics

An object with mass 3.0 kg is executing simple harmonic motion, attached to a spring with spring constant k =210 N/m. When the object is 0.23 m from its equilibrium position, it is moving with a speed of 0.72 m/s.
(a) Calculate the amplitude of the motion.
m

(b) Calculate the maximum speed attained by the object.
m/s

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  1. F = m a = -kx

    if x = A cos w t
    v = dx/dt = -Aw sin w t
    a = d^2x/dt^2 = -A w^2 cos w t = -w^2 x

    so
    -m w^2 x = -k x
    w^2 = k/m
    w^2 = 210/3 = 70
    w = sqrt(70) = 8.37

    when |x| = .23 , |v| = .72
    .23 = A cos 8.37 t
    .72 = -A (8.37) sin 8.37 t
    so
    cos 8.37 t = .23/A
    sin 8.37 t = .086/A

    cos^2 + sin^2 = 1
    .0529 + .007399 = A^2
    A^2 = .060299
    A = .246 m
    Vmax = Aw = .246(8.37) = 2.06 m/s

    check my arithmetic !!!

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    posted by Damon

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