# How does my reference point affect the regresion of data about Hooke's law?

A vertical spring is used (subject to a universal support) whose length is $20$ $cm$ when no force is acting on it. We put on a weight of $1$ $N$ and the new length is $22$ $cm$. When writing the data to use regression, the first point will be the force $1$ $N$ and the deformation $22-20=2$ $cm$. Then I increase the weight in $1$ $N$ more; as a result, $2$ $N$ is the net force deforming the spring. Given that weight, the new length is 24 cm; therefore the second point will be the force $2$ $N$ and $4$ $cm$. However: If I picked the length of $22$ $cm$ as the point of reference, and knowing the next measurement, what would be the first point at my regression?

I'm sort of confused. This is my answer: the first point would be $(2 N - 1 N)= 1 N$ and the deformation of X: $24cm-22cm= 2cm$.

Maybe it is a basic question, but I want to be sure. For a Hookean spring:

$F=k\Delta L$. Here $k$ is the spring constant and $\Delta L$ is the spring deformation:

$\Delta L=L-L_0$, with $L_0$ the length of the spring when no force acts on it and $L$ the length of the spring when a force $F$ acts on it.

In your case for $F=0$, $L=L_0=20cm$ and $\Delta L=0$.

For $F=1N$, $L=22cm$ and $\Delta L=2cm$.

For $F=2N$, $L=24cm$ and $\Delta L=4cm$.

So you always need to correlate $F$ with $\Delta L$.

You are correct, because for a vertical spring, adding a weight has the effect of extending the spring to a new equilibrium position with the same constant $k$ (see Why does the acceleration $g$ due to gravity not affect the period of a vertically mounted spring? ). So your first point will be the extra weight and the extra length: $(1N,2cm)$

• Thank you man, just one question. I learnt bodies experience deformation due to its own weigth. Should I take this in consideration as a new deformation? I guess since it 's very small, we could neglect it. Am I wrong?
– Omar
Sep 4 '15 at 22:10
• You are welcomed, and you are right.
– user83548
Sep 4 '15 at 22:19