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 $l=l_1+l_2$.

Image from book

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?

  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Feb 19, 2023 at 5:47
  • $\begingroup$ @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. $\endgroup$
    – Frunobulax
    Commented Feb 19, 2023 at 10:12
  • $\begingroup$ 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). $\endgroup$ Commented Feb 19, 2023 at 14:36

1 Answer 1


You have an extra force acting on $m$. Think about this, who is really pulling that mass?

  • $\begingroup$ 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. $\endgroup$
    – Frunobulax
    Commented Feb 19, 2023 at 10:09
  • 1
    $\begingroup$ @Frunobulax I mean, that's physics, right? Nature gives us the right answers and we make sure our problem setups give us what we want. Also, you're not ignoring the term, you're setting it to 0 to see it does what it should in that limit. $\endgroup$ Commented Feb 19, 2023 at 16:01
  • $\begingroup$ Well, it seems I'll have to adjust my mindset if I want to understand more physics. And, BTW, I meant to say "remove the first spring from the second equation" in my comment above. $\endgroup$
    – Frunobulax
    Commented Feb 19, 2023 at 17:30

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