# 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?

• Without telling what $v$ is, this question cannot be answered. I don't know where to fit $v$ because I don't know what it is. – Yashas Mar 1 '17 at 19:04
• I'm trying to figure that out too. The few things I've read about this problem just toss $v$ into the expression. – user146639 Mar 1 '17 at 19:08
• $v$ is the velocity of the end of the spring. If you substitute $x$ as $L$, you get $u$ = $v$. From the definition of $u$ you have given, $v$ must therefore be the velocity of the end of the spring. – Yashas Mar 1 '17 at 19:10
• Hmm well if that is what $v$ is I still do not understand why it's linearly proportional or what do to if that was not the case. – user146639 Mar 1 '17 at 19:33
• Have a look a "effective mass of a spring" en.m.wikipedia.org/wiki/Effective_mass_(spring–mass_system) – Farcher Mar 1 '17 at 20:12

• I see, so uniform stretching means that all of the $dm$'s should be spaced out equally according to $x/L$. What happens if the spring was not uniformly stretched? – user146639 Mar 1 '17 at 19:47