-1

Take a rubber band of length $l$ and mark a point $(\cdot)$ on it at $x^{th}$ distance assuming the elastic rubber band to be along the $x$ axis beginning from $x=0$.

My main query is: when the band is stretched, if it gets stretched horizontally and proportionally such that every point on the band elongates equally, then it would mean that when the end point is elongated to $p^{th}$ position, then the $(\cdot)$ moves to the $\big(\frac{xp}{l}\big)^{th}$ coordinate (I am considering $\text {x-axis}$ and this goes on as it stretches proportionally.

But what about the fact when the rubber band doesn't expand uniformly? Then it's not proportional.

Mathejunior
  • 193
  • I think this is a question about mathematics of transformations, not physics, and might be better on Mathematics SE. Asking how a rubber band stretches is physics, but forming equations for arbitrary non-linear transformations is mathematics. – sammy gerbil Mar 04 '17 at 19:15
  • Yeah I want to know stuffs relating to elasticity. Like, how to relate this with the elasticity of the band? Mathematically, it's already assumed that it stretches uniformly.. Else the mathematical version is doable for me. – Mathejunior Mar 04 '17 at 19:17
  • What do you mean by "moves to the $({2x}/p)^{th}$ part of the elongated string? So if x = l, then the end point moves to $({2l}/p)^{th}$ part of the elongated string? Also are you talking about linear elasticity in both parts of the question? Or you want to assume completely general relations? –  Mar 04 '17 at 19:22
  • The last line of your question asks about non-uniform expansion, which contradicts what you are now saying about uniform extension. If the extension is uniform, all distances along the band will expand by the same factor. What exactly do you want to relate with elasticity? Are you asking about Hooke's Law, or about Young's Modulus or some other elastic modulus? Please edit your question to make it clear what you are asking. – sammy gerbil Mar 04 '17 at 19:22
  • Please don't add colors to equations. – Kyle Kanos Mar 04 '17 at 19:35
  • Well, at a minimum don't add color without a really good reason. For instance in http://physics.stackexchange.com/a/301027 the symbols in the text are colored to match the vectors in the diagram so the color serves a specific purpose (and labeling the diagram is not a great alternative because it is already very busy). Even at that Rob only colored the symbol in the part of the text where he was talking about the figure. A nice piece of communication. – dmckee --- ex-moderator kitten Mar 04 '17 at 20:23
  • The question has been edited now. Someone please tell. – Mathejunior Mar 05 '17 at 07:28

1 Answers1

0

You can start with a simple model for an elastic line of unstretched length $L$ which is described by a elastic potential of the form $$ U=\frac{1}{2} \int_0^L k(x) \left(\frac{d u(x)}{d x} \right)^2 dx $$ where $u(x)$ is the displacement of a point of the line from its unperturbed position $x$ induced by the stretching. This give the equation (obtained by minimization of the potential energy as Euler-Lagrange equations for the lagrangian ${\cal L}=-U$) $$ \frac{d}{d x} \left( k(x) \frac{du}{dx} \right)=0 $$ This can be integrated easily, obtaining $$ u(x) = A \int_0^x \frac{dy}{k(y)} + B $$ where $A$ and $B$ are integration constants. Imposing the boundary conditions $$ u(0)=B=0 $$ and $$ u(L)=A\int_0^L \frac{dy}{k(y)}+B $$ we find $$ u(x) = \frac{u(L)}{\int_0^L \frac{dy}{k(y)}} \int_0^x \frac{dy}{k(y)} $$ which gives the displacement of each point as a function of its position, for a given displacement of the end point. If $k$ is constant the displacement is proportional to $x$ $$ u(x) = \frac{x}{L} u(L) $$ As an example of a different situation consider $k=k_0(1+\beta x)$. In this case $$ u(x) = \frac{u(L)}{\int_0^L \frac{dy}{1+\beta y}} \int_0^x \frac{dy}{1+\beta y} $$ which gives $$ u(x) = u(L) \frac{\log \left(1+\beta x\right)}{\log \left(1+\beta L\right)} $$ You see that the displacement of each point is proportional to the displacement of the end point, only the proportionality constant is position dependent.

GCLL
  • 893