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Spring Problem, Differential Equations

  1. Oct 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping constant of 2 lb-s/ft. If the mass is set in motion from its equilibrium position with a downward velocity of 2 in/s, find its position u at any time t. Assume the acceleration of gravity g = 32 ft/s2.

    2. Relevant equations


    3. The attempt at a solution

    I solve for k and get 64, and solve for the mass and get 32/64, so my differential equation is 0.5y'' + 2y' + 64y = 0, I solve for r and get c1*e^(-2t)*cos(2t*sqrt(31)) + c2*e^(-2t)*sin(2t*sqrt(31))

    My initial position is 0, so y(0) = 0, and my initial velocity is -2, so y'(0) = -2

    So substituting, I get

    0 = c1*e^0*cos(0) + c2*sin(0)

    0 = c1

    Now for y',

    y' = -c1*e^(-2t)*sin(2t*sqrt(31))*2sqrt(31)) + -2c1*e^(-2t)*cos(2t*sqrt(31)) + c2*e^(-2t)*cos(2t*sqrt(31))*2sqrt(31) - 2*c2*e^(-2t)*sin(2t*sqrt(31))

    -2 = -c1*e(0)*0 - 2c1e^(0)*cos(0) + c2*e^(0)*cos(0)*2sqrt(31) - 2*c2*e^(0)*sin(0))

    -2 = -2c1 + c2*2sqrt(31)

    But c1 is 0, so -2 = c2*2sqrt(31), and so c2 = -1/sqrt(31)

    So my final equation is -1*e^(-2t)*sin(2t*sqrt(31))/sqrt(31)

    But when I pick that as an option, the computer marks it wrong. I see some options with a 12sqrt(31) on the bottom, but I don't think that's it.
     
  2. jcsd
  3. Oct 30, 2014 #2

    SteamKing

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    Homework Helper

    You have to be careful with units here. If the initial velocity is -2 in/s, then you can't just blindly plug y'(0) = 2 into your equation, because the damping constant was given as 2 lb-s/ft. The initial velocity should be -2/12 ft/s, to convert in/sec to ft/sec.
     
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