# Oscillation of elastic rope hanging from the ceiling

I have to explore the problem of oscillations of an elastic rope hanging from the ceiling, which can move both in vertical and horizontal direction. I'm planing to solve this using a finite differences method of some kind (the stability and how fast the method is is not important).

To get a grip on what I'm trying to do I'm first going to solve it in 1D (just vertical displacements).

The PDE we're solving is $\mu \frac{\partial^2 x}{\partial t^2} = \frac{\partial F(x)}{\partial x} + \mu g$, where $\mu$ is mass per unit length and F is the elastic force.

So I divide the rope into n small pieces, and $h = \frac{L}{n}$. Let's imagine them as n points connected with springs with coefficients k. We then need to express the equation in terms of finite differences where $x_{i,j}$ is the position of i-th element at time $j\Delta t$. So on each piece acts a force $F = k*((x_{i-1,j}-x_{i,j}) + (x_{i+1,j}-x_{i,j})-h) = k(x_{i+1,j}-2x_{i,j}+x_{i-1,j}-2h)$.

We need the derivative of the force: $\frac{\partial F(x)}{\partial x} = \frac{F(x+h)-F(x)}{h} = \frac{x_{i+2,j}-2x_{i+1,j}+x_{i,j}-2h-x_{i+1,j}+2x_{i,j}-x_{i-1,j}+2h}{h} = \frac{x_{i+2,j}-3x_{i+1,j}+3x_{i,j}-x_{i-1,j}}{h}$

The time derivative is $\frac{\partial^2 x}{\partial t^2} = \frac{x_{i,j+1}-2x_{i,j}+x_{i,j-1}}{(\Delta t)^2}$

So $\frac{x_{i,j+1}-2x_{i,j}+x_{i,j-1}}{(\Delta t)^2} = \frac{k}{\mu} \frac{x_{i+2,j}-3x_{i+1,j}+3x_{i,j}-x_{i-1,j}}{h} +g$

From here we can express $x_{i,j+1}$ and solve this recursively with double loop. But...

I'm not sure if this approach is okay. The recursion in time seems fine, you get the $x_{i,-1}$ from the initial condition for velocity and $x_{i,-1}$ from initial condition for position. But I don't know how to deal with the 4-part recursion that comes from the derivative of the force. Also when I tried to program this (I used $x_{-2} = x_{-1} = x_0 = 0$) I got some sort of oscillation (resembling a Bessel's function) even for no initial displacements (I just set initial $x_i$s a distance h apart). That's probably because h cancels out in $\frac{F(x+h)-F(x)}{h}$? Did I do something wrong in deriving this problem? Is there a better way to do it (in a way that I can translate it into a 2D problem later)?

I hope I didn't make any mistakes writing down the indexes. Any tips and ideas will be appreciated.

I have problems with your overall formulation. For an elastic rope, the tension in the rope is $$F=EA\frac{du}{dx}$$where u is the downward displacement relative to the upper end, A is the cross sectional area of the rope, and E is the Young's modulus. Then the force balance equation would read: $$\mu\frac{\partial ^2 u}{\partial t^2}=EA\frac{\partial ^2 u}{\partial x^2}+\mu g$$ The boundary conditions would be u = 0 at x = 0 and du/dx = 0 at x = L. The undisturbed tension would be $F_0(x)=\mu g (L-x)$, and the undisturbed displacement would be $u_0(x)=\frac{\mu g}{2EA}\left[L^2-(L-x)^2\right]$

Addendum 2D If we consider the material element between s and s + ds along the rope, the extension ratio of this material element is $$\lambda=\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}$$and the strain is $$\epsilon=\lambda-1=\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}-1$$In terms of displacements, this becomes:$$\epsilon=\sqrt{\left(1+\frac{\partial u}{\partial s}\right)^2+\left(\frac{\partial v}{\partial s}\right)^2}-1$$where u is the displacement of material point s in the x direction and v is the displacement in the y direction. The unit vector in the direction of the rope at material location s is given by: $$\mathbf{i_s}=\frac{\mathbf{i_x}dx+\mathbf{i_y}dy}{\sqrt{(dx)^2+(dy)^2}}=\frac{\mathbf{i_x}\frac{\partial x}{\partial s}+\mathbf{i_y}\frac{\partial y}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}=\frac{\mathbf{i_x}(1+\frac{\partial u}{\partial s})+\mathbf{i_y}\frac{\partial v}{\partial s}}{\sqrt{\left(1+\frac{\partial u}{\partial s}\right)^2+\left(\frac{\partial v}{\partial s}\right)^2}}$$ The tension force in the rope is then: $$\mathbf{F}=\frac{\mathbf{i_x}(1+\frac{\partial u}{\partial s})+\mathbf{i_y}\frac{\partial v}{\partial s}}{\sqrt{\left(1+\frac{\partial u}{\partial s}\right)^2+\left(\frac{\partial v}{\partial s}\right)^2}}EA\epsilon$$The acceleration vector at material point s is: $$\mathbf{a}=\frac{\partial ^2 u}{\partial t^2}\mathbf{i_x}+\frac{\partial ^2 v}{\partial t^2}\mathbf{i_y}$$

• How would that translate to a 2D case? Would it be correct to use 2 equations equations of this type, one for $u_y$ and the other for $u_x$ (minus the gravity ofc)? – nunca13 Sep 4 '18 at 13:49
• I assume you are neglecting the bending stiffness of the rope, correct? I think it's against the forum rules for me to provide the full 2D derivation. But, to stimulate your thinking, I would let s represent the undeformed distance coordinate measured along the rope, and I would let x(s,t), y(s,t), and z(s,t) parametrically represent the cartesian coordinates of the material point s at time t. Anyway, this is how I personally would set it up. – Chet Miller Sep 4 '18 at 14:40
• Yes, bending is neglected. I tried to work with the parametrisation you described before starting with the 1D model first, but I went about it the wrong way I guess. I did something like what I described in my question, except it was quite terrible - a recursion with 4 elements and it was not even linear. I because I used $(ds+du)^2 = (x_{i+1} -x_i)^2+ (y_{i+1} -y_i)^2$and from there $F = k*du$, then I made separate equations for x and y directions, that looked like this: $\mu x_{tt} = \frac{\partial}{\partial s}(F(s)\frac{dx}{ds})$ and similar for y direction. – nunca13 Sep 4 '18 at 15:07
• You can't use the straight spring equation for this, without taking into account the effect of spring length on spring constant (which is captured by the stress-strain relation). Since you tried some effort, I will continue a little with the 2D analysis. I will enter the addendum soon. – Chet Miller Sep 4 '18 at 16:18
• I'm now actively trying to wrap my head around what you told me, especially in the original answer. So if the force is $F = EA\frac{du}{ds}$ and pointing in the direction of s at location of ds, the component in x direction is $EA\frac{du}{ds}\frac{dx}{ds}$. Now I'm having trouble working out 1. What the equation will be (since we then get 2 equations for x and y projection of the force, on both there will be $\frac{\part^2 u}{\part s^2} and the derivative of x/y with respect to s. What will be on the left side? And 2. I can't seem to figure out the connection between dx/dy and ds – nunca13 Sep 4 '18 at 16:27 Oscillations of an elastic rope I try to solve this problem with a different Ansatz, because i think that the finite element method (FEM) might be too complicated. Ansatz: The length of the pendulum rope$L$depend on the "strain coefficient"$\etaEquations: \begin{align*} &\text{Position vector}\\ &\vec{r}=\left[ \begin {array}{c} L \left( \eta \right) \cos \left( \vartheta \right) \\ L \left( \eta \right) \sin \left( \vartheta \right) \end {array} \right]\\\\ & \text{Kinetic energy }\\ &T=\frac{1}{2}\,m\,\dot{\vec{r}}^2=\frac{1}{2}\,m\,L(\eta)^2\dot{\vartheta}^2\\\\ & \text{Potential energy }\\ &V=m\,g\,L(\eta)\,\cos(\varphi) \end{align*} How we calculateL(\eta)? \begin{align*} &L(\eta)=L_0+\frac{1}{\eta}\,\sqrt{\vec{r}^2}=L_0+\frac{1}{|\eta|}\,L(\eta)\quad \Rightarrow\\ &L(\eta)=L_0\,\frac{|\eta|}{1+|\eta|}\\ &\text{,so if the strain coefficient is very stiff\eta\mapsto\inftywe get } \quad L(\eta)=L_0 \quad \text{as it schuld be} \end{align*} Figure:L(\eta)$versus$\eta\$

The equation of motion : \begin{align*} &\frac{d^2 \vartheta}{dt^2}+\frac{g}{L_0}\frac{1+|\eta|}{|\eta|}\,\sin(\vartheta)=0 \end{align*}

• I have not encountered the strain coefficient before. I would like to understand some more about this solution rather than just blindly implementing it. Could you perhaps elaborate or to have a link or a book recommendation where I could find some theory behind it? – nunca13 Sep 4 '18 at 17:17
• Perhaps you can find some information en.wikipedia.org/wiki/Young%27s_modulus – Eli Sep 4 '18 at 17:46
• Please don't vandalize your post by saying "you took it out". If you don't want to display the answer, press the "Delete" button and delete it instead. – Kyle Kanos Sep 4 '18 at 20:10