Center of mass of heavy spring 
If a heavy spring of uniform density and mass $m$ is hung vertically so that it is stretched by gravity under its own weight, where is the center of mass? 

I have had a few people tell me it is 1/2 the distance from the center of mass at equilibrium distance but this seems wrong to me. Thanks for any help.
 A: We can think of the spring as a series of masses and springs: 

Let's say we have $N$ masses, each of mass $m$. the total mass of the spring is $M = Nm$. each small spring will have a constant $k$. the springs are connected in a row, so that the total spring constant is: $$K = \frac {1}{\frac {1}{k} + \frac {1}{k} + ...} = \frac {k}{N}$$ Let's write the forces for each mass in equilibrium:
$$m_1:\ k\Delta x_1 - mg - k\Delta x_2 = 0$$
$$m_2:\ k\Delta x_2 - mg - k\Delta x_3 = 0$$
$$ \vdots$$
$$m_N:\ k\Delta x_N - mg = 0$$ 
Notice that each mass only feels it's own weight, and the springs attached to it. the last mass is only connected to one spring. 
We can add all $N$ equations to get:
$$ k\Delta x_1 - mNg = 0$$
$$ \Delta x_1 = \frac{Mg}{k}$$
To get $\Delta x_2$ we can plug this in to the second equation. in this manner we can get the general $\Delta x_i$: $$\Delta x_i = \frac {(M-(i-1)m)g}{k} = \frac {(N-i+1)mg}{k}$$ 
from here we can get anything we need, like the total elongation of the spring:
\begin{align}\Delta x &= \sum_{i=1}^{N} \Delta x_i \\&= \sum_{i=1}^{N} \frac {(N-i+1)mg}{k} \\&= \frac {(N+1)mg}{k} \sum_{i=1}^{N}  -\frac {mg}{k} \sum_{i=1}^{N} i \\&= \frac {N(N+1)mg}{k}  -\frac {mgN(N+1)}{2k} \\&= \frac {mgN(N+1)}{2k}\end{align} 
in the limit of many strings and small masses $N \gg 1$:
$$\Delta x = \frac {mgN^2}{2k} = \frac {Mg}{2K}$$ 
and this is the total elongation of the spring. if we want the center of mass it takes a bit more work but the concept is the same:
\begin{align} M_\text{cm} &=\frac{1}{M} \sum_{i=1}^N m\left(\frac {Li}{N} + \sum_{j=1}^i\Delta x_j\right)\end{align}
where $L$ is the length of the string without any force acting on it, so that $\frac {Li}{N}$ is the location of the $i$th mass when no forces act on the string. $\sum_{j=1}^i\Delta x_j$ is the total elongation of the spring up to the mass $i$.
\begin{align} M_\text{cm} &=\frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \sum_{i=1}^N\sum_{j=1}^i\Delta x_j\right) \\&=
 \frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \sum_{i=1}^N\sum_{j=1}^i\frac {(N-j+1)mg}{k}\right) \\&=
 \frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \frac {(N+1)mg}{k}\sum_{i=1}^N\sum_{j=1}^i - \frac {mg}{k}\sum_{i=1}^N\sum_{j=1}^ij\right) \\&= 
\frac {1}{M}m  \left(\frac {LN(N+1)}{2N} + \frac {(N+1)mg}{k}\sum_{i=1}^Ni - \frac {mg}{k}\sum_{i=1}^N \frac{i(i+1)}{2}\right) \\&= 
\frac {1}{M}m  \left(\frac {L(N+1)}{2} + \frac {N(N+1)^2mg}{2k} - \frac {mg}{2k} \left(\sum_{i=1}^Ni^2+\sum_{i=1}^Ni\right)\right) \\&= 
\frac {1}{M} m  \left(\frac {L(N+1)}{2} + \frac {N(N+1)^2mg}{2k} - \frac {mg}{2k} \left(\frac{N(N+1)(2N+1)}{6}+\frac{N(N+1)}{2}\right)\right) \\&= \frac {1}{M}m  \left(\frac {LN}{2} + \frac {N^3mg}{2k} - \frac {mg}{2k} \left(\frac{N^3}{3}+\frac{N^2}{2}\right)\right)\;\; \because N\gg 1 \\&=
   \frac {1}{M}\left(\frac {LM}{2} + \frac {N^3m^2g}{2k} -  \frac {m^2g}{2k}\frac{N^3}{3}-\frac {m^2g}{2k}\frac{N^2}{2}\right) \\&=
 \frac {1}{M}\left(\frac {LM}{2} + \frac {M^2g}{2K} -  \frac {M^2g}{6K}-\frac {Mg}{4NK}\right) \\&=
\end{align}
for $N\gg 1$ the last term is small:
$$\boxed{ M_\text{cm} =
 \frac {L}{2} + \frac {Mg}{3K}} $$
 if the total mass is small, we get that the center of mass of the hung spring is just the same as the center of mass of the spring at rest:$ \frac {L}{2}  $, that makes sense.  
I'm pretty sure my logic is correct, let me know if it's not, or if I have any calculation mistakes.  
A: Consider the unstretched length to be $\ell$ and the total stretching to be $\delta$. The center of mass is going to be located at $c = \frac{\ell+\delta}{2}$ if the cross section remains the same through out. But I suspect you want to consider the necking that happens when you stretch something so the final deflection won't be $$\delta = \frac{\rho g \ell^2}{2 E}$$ but something different. Above $E$ is the modulus of elasticity, $\rho$ is the mass density and $g$ is gravity.
Using Hook's Law the new section area under load $T$ is
$$ A' = \frac{(A E - \nu T)^2}{A E^2} $$
So the linear density of the rope is no longer $\lambda = \rho A$ but $$\lambda' = \rho \frac{(A E - \nu T)^2}{A E^2}$$
So the standard extension $u(x)$ due to gravity equation $\frac{{\rm d^2}u}{{\rm d}x^2} = \frac{g \lambda}{EA} =\frac{g \rho}{E} $ becomes
$$\frac{{\rm d^2}u}{{\rm d}x^2} = \frac{\rho g}{E} \left(1- \nu \frac{{\rm d}u}{{\rm d}x}\right)^2$$
This is solved by separation of variables (using $\epsilon =\frac{{\rm d}u}{{\rm d}x}$) as
$$ \int \frac{{\rm d}\epsilon}{(1-\nu \epsilon)^2} = \int \frac{\rho g}{E} {\rm d}x + K$$
$$ \frac{1}{\nu (\nu \epsilon-1)} = \frac{\rho g}{E}x + K$$
The constant of integration $K$ is found by setting the tension $T =(E A) \frac{{\rm d}}{{\rm d}x} u = (E A) \epsilon$ to zero when $x=\ell$. This makes $K=\frac{1}{\nu} - \frac{\rho g \ell}{E}$ and the above can be solved for the tension
$$ T(x) = E A \theta = \frac{E A \rho g (\ell-x)}{E+ \nu \rho g (\ell-x)} $$
Now the total deflection is found by $\delta = \int_0^\ell \frac{T}{E A}\,{\rm d}x$ which ends up being
$$ \delta = \frac{\nu \rho g \ell - E \ln \left(1+ \frac{\nu \rho g \ell}{E}\right)}{\rho g \nu^2} $$
To check this we take the limit of $\nu \rightarrow 0$ and it makes $\delta = \frac{\rho g \ell^2}{2 E}$ which is the standard extension with considering the necking of a stretched rope.
The expression above is a little difficult to deal with, and we can simplify it by considering situation where $E \gg \nu \rho g \ell$. 
$$\boxed{ \delta \approx \frac{3 \rho g \ell^2}{2 ( 3 E + 2 \nu g \rho \ell)} }$$
The center of mass is found by
$$ c \approx \frac{\ell+\delta}{2} $$
A: I had posted an answer earlier but I'm not convinced if it is right, luckily the Link I provided itself had an answer, which is summarized below

Let the spring of Young's modulus $E$, density $\rho$, with a section $S$
$E$ and $\rho$ are related by
$$E=\frac{k\rho L^2}{m}$$
$$\rho=\frac{m}{SL}$$
It is convenient to use as reference position, the spring as it would be if the gravity did not act: that
is, vertical and uniform. Let us consider two sections $S$ and $S’$ of the spring. Two points of the
spring $P$ on $S$ and $Q$ on $S‘$, at the reference positions $x$ and $x+\Delta x$, will be displaced, under the effect
of the gravity, by the quantities $y$ and $y+\Delta y $ respectively  
The mass element $\Delta x $ from $P$ to $Q$ is subjected to gravity and two tensions $T$ and $T+\Delta T$
$\Delta T+\rho S g \Delta x=0$
Since the extension of the element $\Delta y$ is given by the applied force $T+\Delta T$, we have
$$T+\Delta T=SE \frac{\Delta y}{\Delta x}$$
Processing to the limit as $\Delta x→0, dT/dx = − ρSg, T = S E \frac{dy}{dx}$, we have then: 
$$SE\frac{d^2 y}{dx^2}=\rho S g$$
with the boundary conditions: 
$y=0$ for $x=0$, $\frac{dy}{dx}=0$ for $x=L$
That is, the tension vanishes at the lower end, this gives
$y=\frac{\rho g L}{E}x-\frac{\rho g}{2E}x^2$
so,
$$y=\frac{mg}{kL}x-\frac{mg}{2kL^2}x^2$$
for $x=\frac{L}{2}$ i.e for the center
$y=\frac{3mg}{8k}$
So the center of mass will be at the point $$\frac{L}{2}+\frac{3mg}{8k}$$
