ODE with base excitation caused by a half sine wave

1. Oct 26, 2014

Dustinsfl

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. Oct 26, 2014

vela

Staff Emeritus
I think you made a sign error, but otherwise, everything looks fine. How are you trying to verify the solution?

3. Oct 26, 2014

Dustinsfl

Differentiating twice x_{p3}times m + k times x_{p3} = RHS sum

4. Oct 26, 2014

vela

Staff Emeritus
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.