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Spring Differential Equation w/ Damping

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data
    A spring with a spring constant of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion where t is time and y(t) is displacement with respect to time.

    2. Relevant equations

    [tex]
    10y'' + 4y' + 20y = 0 \\
    y(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \\
    e^{i\theta} = cos{\theta} + isin{\theta} \\
    y(0) = 1 \\
    y'(0) = 0 \\
    [/tex]

    3. The attempt at a solution
    The ODE's characteristic equation is:
    [tex]10\lambda ^2 + 4\lambda + 20 = 0 \\
    \text{or} \\
    5\lambda ^2 + 2\lambda + 10 = 0[/tex]
    Solving for lambda...
    [tex]\lambda = \frac{-1 \pm 7i}{5}[/tex]

    Therefore, y(t)—the solution to the ODE—is...
    [tex]y(t) = C_1e^{\frac{-1 + 7i}{5} t} + C_2e^{\frac{-1 - 7i}{5} t} \\
    \\
    = C_1e^{\frac{-1}{5}t}e^{\frac{7i}{5}t} + C_2e^{\frac{-1}{5}t}e^{\frac{-7i}{5}t} \\
    = e^{\frac{-1}{5}t} (C_1e^{\frac{7i}{5}t} + C_2e^{\frac{-7i}{5}t}) \\
    = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t}) \\
    = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t})[/tex]
    Since cos[x] is even and sin[x] is odd...
    [tex]
    = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{7}{5}t} - C_2isin{\frac{7}{5}t}) \\
    = e^{\frac{-1}{5}t} ( (C_1 + C_2)cos{\frac{7}{5}t} + (C_1 - C_2)isin{\frac{7}{5}t}) \\
    \text{Let:} \\
    A = (C_1 + C_2), B = i(C_1 - C_2) \\
    \text{So...} \\
    y(t) = e^{\frac{-1}{5}t} (Acos{\frac{7}{5}t} + Bsin{\frac{7}{5}t})
    [/tex]

    Now we use the initial conditions to solve for A and B.

    [tex]
    y(0) = e^{\frac{-1}{5}(0)} (A\cos{\frac{7}{5}(0)} + B\sin{\frac{7}{5}(0)}) = 1 \\
    = A\cos 0 + B\sin 0 = 1 \\
    A = 1 \\
    y'(t) = \frac{-1}{5}e^{\frac{-1}{5}t}(A\cos{\frac{7}{5}t} + B\sin{\frac{7}{5}t}) +
    e^{\frac{-1}{5}t} (\frac{-7}{5}A\sin{\frac{7}{5}t} + \frac{7}{5}B\cos{\frac{7}{5}t})
    [/tex]

    We end up w/: B = 1/7

    So...
    [tex]
    y(t) = e^{\frac{-1}{5}t} \left(\cos \frac{7}{5}t + \frac{1}{7}\sin \frac{7}{5}t \right)[/tex]

    (I'm not sure why the LaTeX stuff is showing up in boxes like that; I've tried a few things but it's always coming up).

    This answer's, apparently, wrong; I'm not sure why. Halp, please.
     
    Last edited by a moderator: Mar 31, 2013
  2. jcsd
  3. Mar 31, 2013 #2

    Zondrina

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    Homework Helper

    ##y(t) = c_1e^{\frac{-1 + 7i}{5} t} + c_2e^{\frac{-1 - 7i}{5} t} \\
    = c_1e^{-\frac{t}{5}}(cos(\frac{7t}{5}) + isin(\frac{7t}{5})) + c_2e^{-\frac{t}{5}}(cos(\frac{7t}{5}) - isin(\frac{7t}{5}))##

    Find y'.

    Now use those initial conditions.
     
  4. Mar 31, 2013 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Wolfram Alpha's answer agrees with yours. It could be you're entering the answer incorrectly if it's graded by a computer. I expect, however, it's because the coefficient of y'' in your differential equation is wrong. It's suppose to be the mass, not the weight of the object.
     
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