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I've been tasked with describing the equations of motion of two bodies attached via three springs, as visualized below.

enter image description here

Let $x_1(t)$ and $x_2(t)$ denote the $x$-displacements of boxes $m_1$ and $m_2$ respectively, relative to the left wall. Then applying Newton's second law and Hooke's law should give us $$ m_1x_1'' = -k_1x_1 + k_2(x_2 - x_1)\qquad m_2x_2'' = -k_2(x_2 - x_1) + k_3(L - x_2) $$ where $L$ denotes the total length of all three springs (i.e. the length between the left and right wall, which is fixed!)

For some reason, however, my professor insists that there should be no constant $L$ in these equations of motion: we should have a homogeneous system $x' = Ax$ for some matrix $A$ (I know we could "augment" a system $x' = Ax + c$ to obtain a new homogeneous linear system $\tilde{x}' = \tilde{A}\tilde{x}$, but this is not what my question is concerning). My professor's suggestion was "review how elastic potential energy is calculated", and supposedly the constant $L$ should disappear.

I'm failing to see how this is so - it seems like the constant $L$ is always required in our calculation. Am I doing something wrong? Have I misinterpreted the question?

Qmechanic
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user3002473
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  • Is $L$ length of each spring? – Marko Gulin Jan 21 '22 at 17:12
  • @MarkoGulin $L$ is the total length of all three springs (or just the distance from one wall to the other wall), which I assume should be fixed. It's a constant which I believe needs to be introduced to describe the equations of motion (which my professor has suggested is untrue - no constant $L$ is necessary). – user3002473 Jan 21 '22 at 17:13
  • Why would you think that the force on m2 will depend on the length of the spring and not just on thea mount of compression/extension? Two spring with the same elastic constant will produce the same force for the same compression, no matter what is their length. – nasu Jan 21 '22 at 17:25
  • @nasu What do you mean, those are the same thing? All I'm doing is measuring the compression/extension of the third spring with respect to the wall on the right. The quantity $L - x_2$ is the compression/extension of the third spring. – user3002473 Jan 21 '22 at 17:27
  • I suggest $x_1$ and $x_2$ as the displacements from the respective equilibrium positions. – Farcher Jan 21 '22 at 17:50
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    Something is odd with the equations. If you say $x_1$ and $x_2$ are measured relative to the left wall, for $x_1 = x_2$ the spring 2 is completely compressed, and yet your equations imply there is no force contribution from the spring 2. Is it safe to assume that equilibrium exists for this system, i.e. there is a state in which all three springs to not exert any force? – Marko Gulin Jan 21 '22 at 17:55
  • @user3002473 No, x2 is the compression. x1 and x2 are measured from the equilibrium position. – nasu Jan 21 '22 at 17:56
  • Duplicate : Eigenvalue equation for kinetic and potential energy. – Frobenius Jan 21 '22 at 21:51
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    Does this answer your question? Eigenvalue equation for kinetic and potential energy – Frobenius Jan 21 '22 at 21:51
  • I don't think that's a duplicate. The proposed duplicate is about how eigenvalue equations arise in the first place, but this question deals with how we define coordinates for a coupled oscillation problem. – Michael Seifert Jan 24 '22 at 15:02

1 Answers1

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TL;DR The term $k_3 L$ from your question is not part of the differential equations, but rather $k_3 \Delta L$ where $\Delta L$ is difference between the total length $L$ and sum of lengths of all three springs in relaxed state. Problems like this usually assume $\Delta L = 0$ although it is not explicitly stated by the problem definition. Below I show why you should not neglect the $\Delta L$ term.

If $x$ denotes position measured from the left wall (as you have indicated in your question), then your (differential) equations for motion are not correct. Here is a simple check: for $x_1 = x_2$ the spring 2 is completely compressed, and yet your equations predict there is no force by the spring 2.

Let $x$ denote distance measured from the left wall. Then the left wall is at $x_L = 0$ and the right wall is at $x_R = L$. Nominal spring length (when there is no elongation) is denoted by $L_1$, $L_2$, and $L_3$. The spring forces are then defined as

$$F_1 = k_1 (L_1 - x_1) \qquad F_2 = k_2 (L_2 - (x_2-x_1)) \qquad F_3 = k_3 ((L-x_2)-L_3)$$

The equations of motion for both masses are

$$m_1 \ddot{x}_1 = F_1 - F_2 \qquad \text{and} \qquad m_2 \ddot{x}_2 = F_2 + F_3$$

which is expanded to

$$m_1 \ddot{x}_1 = k_1 (L_1 - x_1) - k_2 (L_2 - (x_2-x_1))$$

$$m_2 \ddot{x}_2 = k_2 (L_2 - (x_2-x_1)) + k_3 ((L-x_2)-L_3)$$

For reason that will be obvious later, let's now redefine the positions $x_1$ and $x_2$ as

$$\tilde{x}_1 = x_1 - L_1 \qquad \text{and} \qquad \tilde{x}_2 = x_2 - (L_1 + L_2)$$

Note that these substitutions do not affect the corresponding time-derivatives (accelerations) since spring length at relaxed state are constant

$$\frac{d^2}{dt^2} \tilde{x}_1(t) = \frac{d^2}{dt^2} x_1(t) \qquad \text{and} \qquad \frac{d^2}{dt^2} \tilde{x}_2(t) = \frac{d^2}{dt^2} x_2(t)$$

With these substitutions, the equations of motion now become

$$\boxed{m_1 \ddot{x}_1 = -k_1 \tilde{x}_1 + k_2 (\tilde{x}_2 - \tilde{x}_1)} \qquad \text{and} \qquad \boxed{m_2 \ddot{x}_2 = -k_2 (\tilde{x}_2 - \tilde{x}_1) - k_3 \tilde{x}_2 + k_3 \Delta L}$$

where $\Delta L = L - (L_1 + L_2 + L_3)$. The problem definition does not say anything about the spring lengths, so the affine term $k_3 \Delta L$ must be part of the equation. Problems like this are usually simplified such that sum of individual lengths equals total length in which case $\Delta L = 0$.

Although this is not part of the original question, for completeness of the solution I discuss it here. The system is in equilibrium when forces exerted by all springs are balanced, in which case $\ddot{x} = 0$

$$ \left\{ \begin{array}{ll} (k_1 + k_2) \tilde{x}_1 - k_2 \tilde{x}_2 = 0 \\ k_2 \tilde{x}_1 - (k_2 + k_3) \tilde{x}_2 = -k_3 \Delta L \end{array} \right. $$

from which solution for equilibrium immediately follows:

$$\tilde{x}_1 = \frac{k_2 k_3}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L \quad \text{and} \quad \tilde{x}_2 = \frac{k_2 k_3 + k_3 k_1}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L$$

When all three springs can be relaxed at the same time then $\Delta L = 0$ and the equilibrium is $\tilde{x}_1 = 0$ and $\tilde{x}_2 = 0$. In this case $\tilde{x}$ is a position relative to the equilibrium point.

Marko Gulin
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  • Thanks! I see now what I was doing wrong - at one time I was calculating force relative to the left wall, and at another I was calculating relative to the equilibrium position, thus the confusion. This is much clearer! Now, I think we are to assume there is an equilibrium position, but how would we check if we weren't sure? – user3002473 Jan 21 '22 at 18:07
  • In the case there were no equilibrium position, is the only assumption we have to drop simply that $L_1 + L_2 + L_3 = L$, in which case the formula for $\tilde{x}_2''$ would have a constant? – user3002473 Jan 21 '22 at 18:09
  • @user3002473 You cannot check if there is an equilibrium position, this needs to be defined in the problem statement. If there is no equilibrium position, then the equality you mentioned does not hold and we would need to know all the lengths in order to solve the problem. – Marko Gulin Jan 21 '22 at 18:12
  • So in case there were no equilibrium position, does it still make sense to say that $\tilde{x}_1$ and $\tilde{x}_2$ are the measurements "relative to the equilibrium"? Is this saying it's relative to the equilibrium of the springs (as you say, the lengths of zero elongation), rather than the equilibrium of the system? – user3002473 Jan 21 '22 at 18:14
  • Even if there is no equilibrium, I would still write differential equations in terms of $\tilde{x}_1$ and $\tilde{x}_2$. The only difference is that in the expression for $F_3$ we would not be able to replace $L$ by the sum of lengths, and we would end up having an affine component in the differential equation. – Marko Gulin Jan 21 '22 at 18:18
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    Would you have started out with the equations in terms of $\tilde{x}_1$ and $\tilde{x}_2$? I could've guessed at the first equation (as I basically did in the original question), but I'd have to write it out in terms of absolute distances ($x_1, x_2$) in order to figure out what the constant term in the second equation was. I don't really understand how someone could've intuitively reached the equations in terms of $\tilde{x}_1, \tilde{x}_2$ from the get-go - we don't even know if an equilibrium position exists in general! – user3002473 Jan 21 '22 at 18:21
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    @user3002473 You are right, for me it was more intuitive to start from absolute positions, and then I realized I could make substitutions such that differential equations are written in a nicer form. It is easy to get confused when you start thinking in relative terms, there are a lot of things to keep in mind at the same time. – Marko Gulin Jan 21 '22 at 18:25
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    Thanks for all the help! I really appreciate your insights! – user3002473 Jan 21 '22 at 18:26
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    @user3002473 I have edited the answer to be more general. Now it does not assume all three springs can be relaxed at the same time. – Marko Gulin Jan 21 '22 at 20:39
  • Thanks for including that, I was messing around with it after our discussion, glad to see it here! – user3002473 Jan 22 '22 at 00:27