4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.