3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Yashas Mar 1 '17 at 19:04
  • $\begingroup$ I'm trying to figure that out too. The few things I've read about this problem just toss $v$ into the expression. $\endgroup$ – user146639 Mar 1 '17 at 19:08
  • $\begingroup$ $v$ 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. $\endgroup$ – Yashas Mar 1 '17 at 19:10
  • $\begingroup$ 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. $\endgroup$ – user146639 Mar 1 '17 at 19:33
  • $\begingroup$ Have a look a "effective mass of a spring" en.m.wikipedia.org/wiki/Effective_mass_(spring–mass_system) $\endgroup$ – Farcher Mar 1 '17 at 20:12
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – user146639 Mar 1 '17 at 19:47
  • 1
    $\begingroup$ If you cannot assume it is uniform, you must some other (non-linear) relationship for displacement as a function of position along the length... then you substitute that relationship for the simpler one given here. $\endgroup$ – Floris Mar 1 '17 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.