Q:

Does the stretching of a spring depend on its rest length?

I don't understand if the stretching of a spring depends on its rest length besides its force constant $k$. I'll make an example to show my doubt.

Consider a vertical spring and a mass $m$ attached to it, such that
  the equilibrium elongation is $\delta$ from the rest length of the
  spring. If the spring is cut in half and and a new mass $2m$ is
  attached to the free extremum is the equilibrium elongation of the
  spring $2\delta$?

My attempt is :
$\delta=\frac{mg}{k}$ and since $k$ is independent from the total length of the spring, the half long spring still has $k$ as force constant, therefore the equilibrium elongation from the rest length should be $\frac{2mg}{k}=2\delta$.
But it turns out to be $\delta$ and not $2\delta$.
How can that be? What am I missing here?

A:

Your mistake is:

since $k$ is independent from the total length of the spring, the half long spring still has $k$ as force constant.

We can define a constant $K$ to be the force per relative extension of the spring:
$$ F = K \frac{x}{L} $$
where $x$ is the extension and $L$ is the original length. This constant $K$ is an intrinsic property of the string and doesn't depend on the length of the spring. It will be determined by things like the thickness of the wire, the spacing between the coils, the Young's modulus of the metal and so on.
However the force constant that we normally use, $k$, is defined as:
$$ k = \frac{K}{L} \tag{1} $$
so that the force is simply:
$$ F = kx $$
This force constant $k$ does depend on the length of the spring, as equation (1) says, so if you halve the length of the spring you double the value of $k$.

