0
$\begingroup$

Let's say I have a spring with length of $80$m.
The spring is connected horizontally between two masses (on a table, without friction).
Lets say the string got narrower to $76$m.
Does it mean that $\Delta X$ of the two bodys is $2$m? or each one of the masses has $\Delta X=4$m?
It is a debate between a friend and me, he says its $4$m, I say its $2$m ( Think of an accordion, move it symmetrically to the inside, each one of the two sides moves equally half the $\Delta X$ ).

Any answers to our debate will be welcomed.

Improve this question
$\endgroup$
3
  • $\begingroup$ Better start thinking in terms of forces that act on the masses. The $\Delta x$ is irrelevant, the only relevant thing is force acting on each mass, which heavily depends on the system configuration. Draw exactly how masses are placed and what makes this $\Delta x$ and we might give you a better answer. $\endgroup$
    – Marko Gulin
    Dec 3, 2021 at 19:24
  • $\begingroup$ Well, I did draw it, but on my notebook.. I cant draw it here... I already answered my question in my notebook ( I can send a link, but I know stack site doesnt like links to picture, but to picture it here, although I cant do it currently... ) Anyway, back to question: I did to it in terms of forces, but I got ΔX halfed, and he did it not half. I cant really understand what you mean by forces will help me... I did use forces, but to find acceleration which is irrelevant to my question, since I asked only regarding ΔX $\endgroup$
    – Ben Shaines
    Dec 3, 2021 at 19:27
  • $\begingroup$ According to my friend, he says I have to watch as if the two masses are attached to a "wall". One of the masses is the wall and the other the is the second with the string that got narrower, which mean ΔX=4_m. and to look at both cases. but it doesnt seem logic to me.. why is it like that? $\endgroup$
    – Ben Shaines
    Dec 3, 2021 at 19:45

2 Answers 2

Reset to default
0
$\begingroup$

Consider a frictionless table, with the spring providing the only force on each mass. Considering the two masses as a system there is no external force on the system in the horizontal direction; therefore, the center of mass of the system does not change, and both masses each move the same distance in opposite directions, 2 cm for your example.

If the string is attached to a wall, the situation is different because the wall provides an external force.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Exactly, I was right then. The problem thing is my friend insists he is right ( regarding the 4cm ). How to explain it to him? He insists its true what he says, he says to look as if there is a wall ( and as you said, I said to him its something else ). $\endgroup$
    – Ben Shaines
    Dec 3, 2021 at 20:15
  • $\begingroup$ Show your friend a basics physics text that proves the center of mass of a system does not move if there are no external forces on the system; internal forces do not matter. The system is whatever you define it to be, here the system is the two masses and the spring (not an individual mass), and the spring force on each mass is a force internal to the system and does not affect the center of mass of the system. The wall is completely different; it does exert an external force. $\endgroup$
    – John Darby
    Dec 3, 2021 at 21:47
0
$\begingroup$

Since it is absolutely clear that the problem is symmetrical, you could introduce a fixed support at the centre point and it would make no difference. Each mass moves 2m so that is $\Delta x$. But you must realise then that the appropriate length when you are considering Hooke's law is not 80m but 40m. The proportional change in length is $4/80 = 2/40$

(I'm British so that's the way I spell)

Improve this answer
$\endgroup$

Your Answer

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

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