Two springs and a mass between length L [closed]

I am trying to solve the problem that is depicted below;

I am writing the equations with state variables as $x_2$ and $v_2$within the matrix form as;

\begin{align*}\dot{x_2} &= v_2\\ \Sigma F_2 &= m_2a_2 \\ \dot{v_2} &= -\frac{k_1}{m_2}x_1-\frac{k_3}{m_2}x_3\end{align*}

from this point how can I construct my transition matrix for;

$$\begin{bmatrix} \dot{x_2}\\ \dot{v_2}\end{bmatrix} = A\cdot\begin{bmatrix} x_2\\ v_2\end{bmatrix}$$

edit: Only the block between the springs has mass. $x$ is denoted as displacement, so $x_1$ denotes the displacement of spring 1 by $x$.

closed as off-topic by Colin McFaul, Kyle Kanos, Brandon Enright, Dan, user10851 Jun 11 '14 at 20:11

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• how many masses are there? There is one in the title, maybe one in the drawing, and two in the equations. What are $x_1$, $x_2$, and $x_3$? – DavePhD May 9 '14 at 11:07
• @DavePhD please check my edit in the question. – Yirmidokuz May 9 '14 at 11:10
• I corrected the equation. – Yirmidokuz May 9 '14 at 11:15
• you should be able to rewrite the 3rd equation in terms of one displacement variable. – DavePhD May 9 '14 at 11:57
• how can I say?, you don't define what the three displacements are. You only need one displacement variable, x. Let x be the displacement from equilibrium. x=0 is equilibrium. Then write an equation for acceleration in terms of x, m, and the two ks. – DavePhD May 9 '14 at 12:16

I think the key point you are missing here is that the extension of the springs ($x_1$ & $x_3$) is determined by position of the block.

They must all be the same, i.e. $x_1=x_3=x_2$, otherwise the walls are moving!

Use this in the third equation and the result should become obvious.

• nivag, but shouldn't I include the length constraint L in to equation. I am getting confused at this point too. I am also curious about, if my sign convention is right in the equation 3. – Yirmidokuz May 9 '14 at 11:55
• let the equilibrium point be x=0. When the mass is displaced, both springs will exert force on the mass in the same direction. – DavePhD May 9 '14 at 12:05
• @Yirmidokuz Basically what DavePhD said. Your sign convention is fine as both the forces are in the same direction. The length is irrelevant as long as it doesn't change. As force on the spring is proportional to displacement there is no difference if there is no extension of the springs at equilibrium or lots. The equilibrium position will be a constant fraction of the length (1/2 for $k_1=k_3$). – nivag May 9 '14 at 12:44

\begin{align*}\dot{x} &= v \\ \dot{v} &=-\frac{k_1+k_3}{m}x\end{align*}
so the matrix $A$ can be written as;
$\begin{bmatrix} \dot{x} \\ \dot{v} \end{bmatrix} = \begin{bmatrix} 0&1\\ -\frac{(k_1 + k_3)}{m}&0\end{bmatrix} \cdot \begin{bmatrix}x\\v\end{bmatrix}$