# Distance between coils on a massive spring

I'm trying to calculate the separation between coils on a massive spring with no weights attached under different conditions.

I will present my solution to the problem in the case of a uniform gravitational field, which gives answers which I know by experiment to be false. I would be great full if some one will point out a mistake in my reasoning.

Let $\rho$ be the linear density of the spring in it's unextended state, and $k$ its spring constant. Let $f(x)$ be the vertical component of position in the extended state of the point which was originally had vertical component of position $=x$. I'm not sure how to better phrase this, see diagram below: Let us consider find the resultant force acting on the piece (in the blue box) of the spring between $x$ and $x+h$ equating it to zero as the position we are considering is equilibrium. 3 Forces are acting:

• (Up) The tension from the part of the spring above it: $k*(f(x)-x)$ by Hooke's law

• (Down) Gravity: $\rho hg$

• (Down) Tension from the part of the spring below it $k*(f(x+h)-x-h)$. By analogy of the first force acting on the bottom part of the spring, and then using newton's 3rd law to conclude that the force must be equal and opposite.

Now the equation we get is:

$$k*(f(x)-x)=\rho hg+k*(f(x+h)-x-h)$$

Rearranging:

$$0=\frac{\rho g}{k}+\frac{f(x+h)-f(x)}{h}-\frac{h+x-x}{h}$$

And taking the limit as $h \rightarrow 0:$ $$0=\frac{\rho g}{k}+f'(x)-1$$

Rearranging and integrating:

$$f(x)=x*(1-\frac{\rho g}{k})+c$$

$c=0$ as $f(0)=0$ (the top of the spring is attached).

This means that the spring extends linearly, and the separation between successive coils are the equal. Yet intuitively the parts of the spring which are higher support more weight and therefore should be extended more! In the image below we can clearly see that the calculation is wrong and the intuition is correct. But where is the error in the calculation? This modifies your equations - there is an additional term $\rho g x$ that must appear.
• The expression appears on the left hand side of your equation... the force pulling on the top of your blue box is not simply $k(f(x)-x)$, but $k(f(x)-x)-\rho g x$. – Floris Oct 12 '15 at 22:30
• would the force pulling the bottom of the blue box also be changed to $k(f(x+h)-x-h)-\rho g (x+h)$? In the case $f(x)$ also seams to come out linear... I've also realised another mistake I made: the spring constant $k$ is dependant on the length on the spring – Michal Oct 12 '15 at 23:06