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Laplace Transforms with an Undamped Spring with Unit Impulse

  1. Nov 5, 2013 #1
    Ok, so I start out with the basics and find k.

    F=ma=kx
    (1.5 kg)(9.8 m/s^2) = k (4.9 m)
    k = 3 N/m

    I also know from the problem that c=0 (no damping) and x(0) = 2m and x'(0) = 0.

    ƩF = ma
    ma = -cx' - kx, a = x''
    mx'' + cx' + kx = 0 and since c=0
    mx'' + kx = 0

    Nothing out of the ordinary or unusual...but where the heck/how the heck do I include the unit impulse? From reading the textbook I have the impression that

    mx'' + kx = ∂(t - 2∏)

    and laplace [∂(t - 2∏)] = e^(-2∏s)

    If my assumption is correct, I can do the problem.

    1.5x'' + 3x = ∂(t - 2∏)
    laplace (1.5x'') + laplace (3x) = laplace (∂(t - 2∏))
    X(s^2) -s x(0) - x'(0) + 3X = e^(-2∏s)
    X (s^2 + 3) - 2s - 0 = e^(-2∏s)
    X (s^2 + 3) = 2s + e^(-2∏s)
    X = [2s + e^(-2∏s)]/(s^2 + 3)
    inverse laplace (X) = inverse laplace [(2s + e^(-2∏s)) / (s^2 + 3)]
    x(t) = bleh.

    Wolfram alpha (which instructor says is a legit method to solve this) says that my laplace is invalid, so I think somewhere I don't understand the unit impulse thing.

    Pls help. kk thnx.
     
  2. jcsd
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