A spring with non-negligible mass I see everywhere in the analysis of a spring-mass system of Simple Harmonic Motion, that each infinitesimal element on the spring of length $L$ is $\frac{vx}{L}$ where $v$ is the velocity of the block of mass on the end of the spring. It indeed makes intuitive sense that the closer the part of the spring is to the mass, the faster it moves, but how is this relation specifically derived? Is it mere experimental fact?
 A: Modeling heavy springs isn't easy. It becomes more and more complex the fewer assumptions you make. In the relation $\frac{x v}{\ell}$ there is a hidden assumption that would lead to an incorrect result (maybe).
First, here are some common (correct) assumptions that will help you get started:

*

*A helical spring behaves similarly to an elastic rod with longitudinal waves where the axial force $T$ relates to axial deflection $u(x,t)$ as $$T = k\, \ell \frac{\partial u}{\partial x} \tag{1}$$ where $k$ is the stiffness coefficient, and $\ell$ is the total length. This assumption eliminates any complexity with coils and torsional rigidity etc.


*The deflection is harmonic in time away from the equilibrium condition. This means that time must enter in the equation in the $\sin(\omega t)$ form, and if the spring is at rest, then $u(x,t)=0$.


*All parts of the spring move in-phase. This eliminates the need to model stress waves going back and forth along the rod. When the end of the rod is at maximum deflection, then each particle within the rod is also in it's maximum deflection. This assumption allows us to use separation of variables to solve the equation since we assume the solution to be of the following form $$ u(x,t) = p(t) s(x) \tag{2}$$ where $p(t)$ is a harmonic function of time only, and $s(x)$ is a shape function of position only.


*Each natural frequency shape consists of a single harmonic only. This means that the general deflection shape under forced vibration is a superposition of each mode shapes
$$ u(x,t) = \sum_n p_n(t) s_n(x) \tag{3}$$ where now $p_n(t)$ and $s_n(x)$ are separate functions for each harmonic.
In the original question the shape function was $s(x) = \frac{x}{\ell}$ and $p(t) = \delta (t)$ where $\delta$ is the deflection on the end as it varies with time. Take the time derivative with time and due to assumption 3 and that the end moves with $\dot{\delta} = v$, you have $$\frac{\partial u}{\partial t} = v\, s(x) = v \frac{x}{\ell} \tag{4}$$
So the shape function $s(x) = x/\ell$ describes the distribution of velocities along the rod.
The problem is that this shape function does not solve the equation of motion for the spring. This equation comes from the balance of forces on each particle on the rod. Let us look at the solutions, and we can figure out if $s = x/\ell$ fits in anywhere.
$$ k \ell\, \frac{\partial^2 u}{\partial x^2} = \frac{m}{\ell}\,  \frac{\partial^2 u}{\partial t^2} \tag{5a} $$
The above can be simplified using the wave speed of the material $c^2 = \frac{k}{m} \ell^2$ into
$$  \frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2}\,  \frac{\partial^2 u}{\partial t^2} \tag{5b} $$
Use the op shape function you gave $u(x,t) =C_1 \sin\left( \omega t\right) \left( \frac{x}{\ell} \right)$ and used in (5b) to get
$$ 0 = \frac{1}{c^2} \left( - C_1 \frac{\omega^2 x}{\ell} \sin\left( \omega t\right) \right) $$ which is obviously not correct.
THe proper shape function here should be $$s(x) = \sin \left( \frac{\Phi x}{\ell} \right) \tag{6}$$
where $\Phi$ is some constant to be determined by the end conditions of the rod (the mass attached).
A: To begin with, any elastic object that is immovably fixed on one end and stressed in tension will stretch in response. this is true for each infinitesimal cross-sectional element along the object's length, which will elongate by some amount dl. If you integrate these little dl's along the length of the object, you obtain the total elongation. since one end of the object is fixed, all the elongation is observed at its unfixed end. Because your question is sort of confusing to me, I'm not sure this answer addresses it; if not, let me know and I'll try again.
A: 
So first of all $L$ is the length of the spring at the time of measuring the speed, not the natural length.
As you can see in the first drawing we have a little spring with length $L$ and I chose an infinitesimal spring mass element that is $x_\text{spring}$ away from the fixed end.
In the second drawing we have the same spring but elongated with $dx_\text{body}$ so the new length is $L+dx_\text{body}$. We want to find out where the same spring particle is in the second drawing.
We know that the proportion $\frac{x_\text{spring}}{L}$ is kept at any time when the spring elongates, so the new position of the spring particle is $$\frac{x_\text{spring}}{L}(L+dx_\text{body})=x_\text{spring}+\frac{x_\text{spring}dx_\text{body}}{L}$$
$$\Rightarrow dx_\text{spring}=\frac{x_\text{spring}dx_\text{body}}{L}$$
Taking the time derivative we get the final answer $$v_\text{spring}=\frac{x_\text{spring}v_\text{body}}{L}$$
This is how I deduced it. Hope it clears the problem.
