# Two springs soldered together, equations of motion

I know there are already several questions about springs in series, but I think this one is different. It is from an exercise from a German book ("Physik mit Bleistift" by Hermann Schulz) and it goes like this: Two springs (with spring constant $$\kappa_i$$ and unstressed length $$l_i$$ each) are soldered together with the solder joint having mass $$m_0$$. (See image.) The task is to show in two ways that for $$m_0=0$$ the system behaves like one spring with $$\kappa=\kappa_1\kappa_2/(\kappa_1+\kappa_2)$$ and $$l=l_1+l_2$$.

The first way is to assume that the position $$y$$ of the solder joint will always be such that the potential energy is minimized. That part I could solve: The potential energy is $$\kappa_2/2 \cdot (x - y - l_2)^2 + \kappa_1/2\cdot (y - l_1)^2$$ which - as a function of $$y$$ - has a minimum at $$y_0 = (\kappa_1 l_1 - \kappa_2 l_2 + \kappa_2 x)/(\kappa_1 + \kappa_2)$$. If you replace $$y$$ by $$y_0$$ in the above term for the potential energy and simplify you get $$1/2\cdot\kappa_1\kappa_2/(\kappa_1+\kappa_2) \cdot (x-l_1-l_2)^2$$ which is what is expected. So far, so good.

Now, the second way to solve this is supposed to be the following: Assume that $$m_0$$ is not zero and set up the two equations of motion for $$m$$ and $$m_0$$, then let $$m_0\ddot{y}$$ be zero and eliminate $$y$$. The book has no worked-out solutions, only hints. The hint in this case says that the resulting equation will be $$m\ddot{x}=-\kappa(x-l_1-l_2)$$ and that $$m_0\to0$$ won't help as it would result in oscillations of $$m_0$$ with a frequency tending to infinity.

My first question is in how far it is legitimate to set up an equation of motion for a mass only to later postulate that there is no mass and no force. To an amateur like me this seems like a sleight of hand.

Now, as to the solution, I would think that two forces act on $$m_0$$ and we thus have $$m_0\ddot{y} = -\kappa_1(y-l_1)+\kappa_2(x-y-l_2)$$ and likewise $$m\ddot{x} = -\kappa_1(y-l_1)-\kappa_2(x-y-l_2)$$. If, as suggested, I set $$m_0\ddot{y}=0$$ in the first equation of motion and solve for $$y$$, then I get the same result as $$y_0$$ above. If I put this into the second equation of motion, I get almost the expected result, except that there's a factor of $$2$$ in there.

So I obviously have a conceptual misunderstanding here and at least one of the equations must be wrong. It is apparently not the case that both springs act on $$m_0$$ and $$m$$ in the way I described it. But what is really happening, i.e., what would happen if $$m_0$$ weren't zero?

• Hi Frunobulax. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. Commented Feb 19, 2023 at 5:47
• @Qmechanic The question was closed as homework and then reopened after editing. FWIW, this is not homework. I'm not a student. I was just reading a book and stumbled across a conceptual problem in its exercises. Commented Feb 19, 2023 at 10:12
• It does not matter whether or not the question is actually a homework exercise or not. Please read the policies that @Qmechanic has linked. IMHO the tag applies here and I've added it again. That being said, I think the question is fine and on-topic (hence I voted to reopen). Commented Feb 19, 2023 at 14:36

You have an extra force acting on $$m$$. Think about this, who is really pulling that mass?
• OK, thanks, I see now that I get the desired result if I remove the second spring from the second equation. I'm a bit frustrated because, as a mathematician, it often seems to me that physics problems require a certain approach that you can only verify to be correct if you know what the answer is. I also still don't see why we can on the one side work with $m_0\ddot{y}$ and then in the middle of the computation ignore this term. Commented Feb 19, 2023 at 10:09