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ODE with base excitation caused by a half sine wave

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    The suspension system of a car traveling on a bumpy road has a stiffness of ##k = 5\times 10^6## N/m and the effective mass of the car on the suspension is ##m = 750## kg. The road bumps can be considered to be periodic half sine waves with period ##\tau##. Determine the displacement response of the car. Assume the damping of the system to be negligible. Hint: The Fourier series representation of the bumpy road, ##y(t)##, is given by
    $$
    y(t) = \frac{1}{\pi} + \frac{1}{2}\sin(2\pi t) - \frac{2}{\pi}
    \Big(\frac{\cos(4\pi t)}{3} + \frac{\cos(8\pi t)}{15} +
    \frac{\cos(12\pi t)}{35} + \cdots\Big)
    $$

    2. Relevant equations
    $$
    \frac{2}{\pi}
    \Big(\frac{\cos(4\pi t)}{3} + \frac{\cos(8\pi t)}{15} +
    \frac{\cos(12\pi t)}{35} + \cdots\Big) = \frac{2}{\pi}\sum_{j = 1}^{\infty}\frac{\cos(4j\pi t)}{4j^2 - 1}
    $$

    3. The attempt at a solution
    If I solve each piece separately and use the superposition principle, the first two terms work as solution but I can't verify that the final piece is so I am assuming something is wrong.
    \begin{align*}
    750\ddot{x} + 5\times 10^6x
    &= \frac{1}{\pi} + \frac{1}{2}\sin(2\pi t) - \frac{2}{\pi}
    \sum_{j = 1}^{\infty}\frac{\cos(4j\pi t)}{4j^2 - 1}\\
    750\ddot{x} + 5\times 10^6x &= \frac{1}{\pi}\\
    x_{p1}(t) &= \frac{1}{5\times 10^6\pi}\\
    750\ddot{x} + 5\times 10^6x &= \frac{1}{2}\sin(2\pi t)
    \end{align*}
    Here we will let ##x_{p2} = A\cos(2\pi t) + B\sin(2\pi t)## Then
    $$
    A\cos(2\pi t)\big[5\times 10^6 - 3000\pi^2\big] +
    B\sin(2\pi t)\big[5\times 10^6 - 3000\pi^2\big]
    = \frac{1}{2}\sin(2\pi t)
    $$
    Therefore, ##A = 0## and ##B = \frac{1}{10\times 10^6 - 6000\pi^2}##.
    \begin{align*}
    x_{p2}(t) &= \frac{1}{10\times 10^6 - 6000\pi^2}\sin(2\pi t)\\
    750\ddot{x} + 5\times 10^6x
    &= \frac{-2}{\pi}\sum_{j = 1}^{\infty}\frac{\cos(4j\pi t)}{4j^2 - 1}
    \end{align*}
    No matter how I have tried to solve this final piece, the solution doesn't seem to work out though. I end up getting ##A_j = \frac{-2}{\pi(k-16\pi^2j^2m)}## and ##B = 0##. Then
    $$
    \frac{-2}{\pi(k-16\pi^2j^2m)}\cos(4\pi jt) + 0\sin(4\pi j t) = \cos(4\pi j t - \phi_j)
    $$
    Therefore, ##\phi = \tan^{-1}(0/A) = 0## and ##R = \sqrt{A^2 + B^2} = \frac{2}{\pi(k-16\pi^2j^2m)}\cos(4\pi jt)##.
    $$
    x_{p3} = \frac{2}{\pi}\sum_{j = 1}^{\infty}\frac{\cos(4\pi j t)}{k-16\pi^2j^2m}
    $$
    and
    $$
    x_p(t) = \frac{1}{5\times 10^6\pi} + \frac{1}{10\times 10^6 - 6000\pi^2}\sin(2\pi t) + \frac{2}{\pi}\sum_{j = 1}^{\infty}\frac{\cos(4\pi j t)}{5*10^6-12000\pi^2j^2}
    $$
    but I am unable to verify
    $$
    \frac{2}{\pi}\sum_{j = 1}^{\infty}\frac{\cos(4\pi j t)}{5*10^6-12000\pi^2j^2}
    $$
    as a solution so I dont think it is correct.
     
  2. jcsd
  3. Oct 26, 2014 #2

    vela

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    I think you made a sign error, but otherwise, everything looks fine. How are you trying to verify the solution?
     
  4. Oct 26, 2014 #3
    Differentiating twice x_{p3}times m + k times x_{p3} = RHS sum
     
  5. Oct 26, 2014 #4

    vela

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    You don't need to do the next step of finding a phase angle. You can write down the solution immediately:
    $$ x_{p3} = \sum_{j=1}^\infty [A_j \cos(4\pi jt) + B_j \sin(4\pi jt)] = \sum_{j=1}^\infty-\frac{2}{\pi(k-16\pi^2j^2m)}\cos (4\pi j t).$$
    The sign error crept in here. You should have gotten ##\phi_j = \pi## because you ended up with R = -A. In fact, you need to be a little more careful. The way you defined R, you have R = |A| because the principal square root is always positive, so depending on the sign of A, R could be +A or -A.
     
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