# Spring between two masses: $ΔX$ is halved?

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.

• 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. Dec 3, 2021 at 19:24
• 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 Dec 3, 2021 at 19:27
• 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? Dec 3, 2021 at 19:45

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 $$4/80 = 2/40$$