I am currently working on a problem involving the spring-mass system below and its total length as a function of $k_1/k_2$. I have written out the equations for each of the masses, but I am not sure how to combine them into a matrix equation, or even if I did the equation correctly.

enter image description here

Starting from the left ($m_1$): $$m_1\ddot x_1 +k_1(x_1-x_2)+k_2(x_1-x_3)=0 $$

Next mass ($m_2$): $$m_2\ddot x_2+k_1(x_2-x_3)+k_2(x_2-x_4)+k_1(x_2-x_1)=0$$

$m_3$: $$m_3\ddot x_3+k_1(x_3-x_2)+k_1(x_3-x_4)+k_2(x_3-x_1)=0$$

$m_4$: $$m_4\ddot x_4+k_1(x_4-x_3)+k_2(x_4-x_2)=0$$

What is the matrix equation for the equilibrium position of the blocks?

  • $\begingroup$ @Citut-Of course, not all the $x_n$'s can't be zero at the same time. And what is exactly meant by "the equilibrium position"? The position of what? Is the equilibrium situation the state in which all masses are at rest? I think (for example because the same applies to the other equations) the first equation doesn't express the fact that the mass is coupled to all other masses. You show four equations with eight unknown variables (the k's and m's), which is unsolvable. $\endgroup$ – descheleschilder Feb 17 '17 at 14:45
  • $\begingroup$ @descheleschilder Yes, that is what I meant by equilibrium position. As for the unknowns, the point of the exercise is to find the total length of the system as a function of $k_1$/$k_2$, in Matlab. I think I know how to do that part, but its getting the equilibrium matrix that is confusing me. $\endgroup$ – Citut Feb 17 '17 at 14:59
  • $\begingroup$ out of curiosity, where it the figure from? $\endgroup$ – ZeroTheHero Feb 17 '17 at 14:59
  • $\begingroup$ I'm not sure, I think it is out of a computational physics book. $\endgroup$ – Citut Feb 17 '17 at 15:02
  • $\begingroup$ -1. Not clear what you are asking. If "the equilibrium position" is when all the masses are at rest, then accleration is zero for all masses. There is no motion, and there are no equations of motion (except $\dot x=0$). You need to include the natural lengths of each spring. If the masses are horizontal then their values are irrelevant. The masses are only relevant if there is acceleration (which can include gravity). Please provide an accurate statement of the problem you are trying to solve. $\endgroup$ – sammy gerbil Feb 17 '17 at 18:16

It's easy to write the equations of motions in matrix form. The first one, for instance, you rewrite as $$ \ddot x_1= -\frac{k_1+k_2}{m_1} x_1 +\frac{k_1}{m_1}x_2 +\frac{k_2}{m_1}x_3\, . $$ Having done so for the other three degrees of freedom, you can then write $$ \frac{d^2}{dt^2}\left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4\end{array}\right) = M\left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4\end{array}\right) $$ where the first row of $M$ would be $$ M_{11}=-\frac{k_1+k_2}{m_1}, \quad M_{12}=\frac{k_1}{m_1}\, ,\quad M_{13}= \frac{k_2}{m_1}\, ,\quad M_{14}=0 $$ The equations of motions can be solved using normal mode analysis and initial conditions.

The issue is that this doesn't give you good information about the equilibrium positions. For this you would want to write the potential energy $U$ for the collection of springs, and then find the points where $\partial U/\partial x_k=0$ for $k=1,2,3,4$. (Unless we have a different definition of equilibrium position.)

  • $\begingroup$ Thanks! That's really helpful! I really need to find out what this question means by equilibrium position, because I don't think that our professor intended for us to need to write the potential energy and solve from there for equilibrium position. $\endgroup$ – Citut Feb 17 '17 at 15:56

It seems from your comments that this is a static problem not a dynamic one. The springs are at rest and there is no acceleration. Assuming the masses A, B, C, D are arranged horizontal then their values are irrelevant; they would only enter a static problem if they were arranged vertically.

Suppose the forces in the springs are $T_1$ to $T_5$. In the diagram I have drawn them all as compressive forces to be consistent, but this is not essential. If an individual value is later found to be -ve this indicates that the spring is in tension rather than compression.

enter image description here

Since the system is in static equilibrium then the net force on each mass must be zero. Analyzing the forces on each mass A to D in turn, and taking right as +ve, this gives the equations

These equations are not linearly independent : adding the 1st three gives the fourth. So we have 3 independent equations but 5 unknowns. In addition to forces being balanced the lengths AC and BD must be consistent with the lengths of the stretched or compressed springs :
where for each spring $L$ is natural length and $x=T/k$ is contraction.


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