homework and exercises - Kinetic energy of a massive spring

Suppose we had a spring-mass system where the spring isn't assumed to be massless (has mass $M$) and is of length $L$. One end of the spring is held fixed and the other end I guess is left to freely oscillate. Here, I am told that the spring is assumed to uniform and stretches uniformly. If I want to find the kinetic energy of the spring, we have to set up an expression for it

$$dT_{\text{spring}} = \frac{1}{2}u^{2} dm$$

where $dT_{\text{spring}}$ is the kinetic energy of an infinitesimal part $dm$ somewhere along the spring and $u$ is its corresponding velocity. Since the spring is uniform, I can find its mass density

$$ \lambda = \frac{dm}{dx} \longrightarrow dm = \lambda dx = \frac{M}{L}dx$$

so that $$dT_{\text{spring}} = \frac{1}{2}u^{2}\frac{M}{L}dx $$

The one step I am not understanding is how $u= \frac{x}{L}v$, where $v$ is, I think, the velocity of some point that has been displaced by the stretching of the spring (please correct me here if I'm wrong). Why is the velocity of a piece $dm$ linearly proportional to $v$ and how can I derive that expression mathematically, i.e. if $u = \alpha v$, how do I find $\alpha$ and why? Something is not registering in my head and I feel like it has to do with the fact that the spring is assumed to be uniform. That then begs the question: what if it wasn't? What would I do in that case?

Answer

Imagine the spring is horizontal. Imagine that it is anchored on the left side

Describe the left side as x = 0. Describe the right side as x = L.

The assumption is that the spring stretches uniformly. So, if the left side is anchored, and the right side moves at 4 ft, then the middle must move 2 ft. The part of the spring 1/4 of the way from the left side move at 1 ft, and the part of the spring that 3/4 of the way from the left side must move at 3 ft.

Using this argument, you could describe the displacement (the amount of movement) of any individual point on the spring as d(x) = (x/L)D where D is the distance the right side has moved. That fits the assumption of uniform stretching.

Now you can take the time derivative of that and get an expression for the velocity everywhere along the spring, as a function of x, with the velocity at the right side as a parameter.