What is the Equilibrium Shape of a String Suspended Between Two Points?

[CAF123]

Homework Statement

An inextensible string of length $\ell$, whose mass per unit length is $\rho$ is hung between two points at the same height a distance $d$ apart ($d < \ell)$. Using the Euler-Lagrange equations and the first integrals, find the shape of the string when it is in equilibrium.

Relate the constants of integration to $\ell$ and $d$.

Homework Equations

Need to find a function$F$ such that $I[y] = \int_{x_1}^{x_2} F(y,y',x)\,dx$ and $I[y]$ is minimized. For $F \neq F(x),$ $$y' \frac{\partial F}{\partial y'} - F = \text{const}$$

The Attempt at a Solution

Since l>d, the string will be elongated or distorted between the two ends. Suppose the shape is described by $y(x)$. In equilibrium, the net force on each segment of the string will vanish. Equivalently, the potential energy of each segment of the string is at a minimum and this is easier to deal with.

Take the zero of potential to be at the highest point of the string between the two supports. Then a small segment d$\ell$ of the string will be of length $\sqrt{1+y'^2}$. So $dm = \rho d\ell = \rho \sqrt{1+y'^2}$. Since zero of potential was assigned at the highest point, all points on the string are below this, so the potential energy of a segment is then $U = - \rho \sqrt{1+y'^2}gy$, with the y-axis vertical. Is this correct?

If I then use Euler-Lagrange, since $U \neq U(x)$, I use the formula in the relevant equations to get that the equilibrium length satisfies $y/\sqrt{1+y'^2} = \text{const}$, which when integrated gives $y$ ~ $\cosh x$, which seems reasonable given the form of this graph.

 

[BvU]

Looks good to me. Cosh is also called Catenary

 

[CAF123]

My integral is $$\int_{y_o}^{y} \frac{\text{d}y'}{\sqrt{\left(\frac{y'}{C}\right)^2 - 1}} = \int_{x_o}^x \text{d}x',$$ where $C$ is the constant on the RHS of the Euler-Lagrange first integral.

I set $y_o = x_o = 0$ arbritarily in my set up. Carrying the integration through, I obtain $C(\theta(y) - \theta(y_o)) = x - x_o$ using the substitution $y = C \cosh \theta$. My problem is there does not exist a $\theta$ satisfying $y_o = 0 = C \cosh \theta$, so I cannot change the integration limits.

 

[CAF123]

In terms of relating the constants of integration to l and d (the second part of the question), I can see one boundary condition that is y(x = 0, d) = 0. (it was given both ends were at the same height at both ends, and in my set up I chose these points to coincide with the x axis). I only had one integration constant, so this should be enough but I did not get any relation involving l.

 

[CAF123]

Is there something wrong with setting $x_o = y_o = 0$? Physically I think no, but if I am to use the substitution I did then maybe I need to change this.

 

[BvU]

My integral is $$\int_{y_o}^{y} \frac{\text{d}y'}{\sqrt{\left(\frac{y'}{C}\right)^2 - 1}} = \int_{x_o}^x \text{d}x',$$ where $C$ is the constant on the RHS of the Euler-Lagrange first integral.
I set $y_o = x_o = 0$ arbritarily in my set up. Carrying the integration through, I obtain $C(\theta(y) - \theta(y_o)) = x - x_o$ using the substitution $y = C \cosh \theta$. My problem is there does not exist a $\theta$ satisfying $y_o = 0 = C \cosh \theta$, so I cannot change the integration limits.

A few questions to get you unstuck again:
Whatever happened to $\ell$ itself ? Can't just disappear !
Why the integral from $y_0$ to $y$ (meaning from 0 to 0 ?) and not from $y'_0$ to $y'_{x_2}$ ?
And (from post 4): "y(x = 0, d) = 0" is really two boundary conditions, rght ?
Then: you introduce $\theta$ with $y = C \cosh \theta$; what is $\theta$ ?
Previously you had $y$ ~ $\cosh x$; would $y$ ~ $\cosh (x-d/2)$ be OK too ?

From post 1: $dm = \rho d\ell = \rho \sqrt{1+y'^2}$, but you mean $\rho \sqrt{1+y'^2}\, dx$, right ? You can't have a differential on the left and not on the right hand side !
Similarly: $U = - \rho \sqrt{1+y'^2}gy$, is a value on the left and function on the right. You mean $U = \int_0^d F dx \$ with $F = - \rho \sqrt{1+y'^2}gy\$, right ?

 

[CAF123]

From post 1: $dm = \rho d\ell = \rho \sqrt{1+y'^2}$, but you mean $\rho \sqrt{1+y'^2}\, dx$, right ?

Indeed, so a small segment of string of length dl contributes $dU = -\rho \sqrt{1+y'^2} gy \,dx$ to the total potential energy of the system. To obtain the total potential energy of the system, integrate over all x. I.e $U = \int_0^d -\rho \sqrt{1+y'^2} gy \,dx$. Now, since the integrand is not a function of x, I can still apply the relevant Euler-Lagrange first integral.

A few questions to get you unstuck again:
Whatever happened to $\ell$ itself ? Can't just disappear !

Yes, I think I need to incorporate it into the boundary conditions.

Why the integral from $y_0$ to $y$ (meaning from 0 to 0 ?) and not from $y'_0$ to $y'_{x_2}$ ?

$y_o=0$ and $y$ stood for some arbritary y in the range from 0 to the y value corresponding to the lowest point of the string.

And (from post 4): "y(x = 0, d) = 0" is really two boundary conditions, rght ?

Yes.

Then: you introduce $\theta$ with $y = C \cosh \theta$; what is $\theta$ ?

I did not really think of a physical realization of $\theta$, I was using the substitution in order to solve the integral.

 

[BvU]

Ah, I'm sorry, you are doing the right thing. I got confused by your using $y'$ for $dy/dx$ on one occasion and for the name of an integration variable on another.

Indeed the "equilibrium length" (i.e. the functional that gives an extremum for $U$) satisfies the differential equation $${\rho g \, y \over \sqrt{1+y'^2} } = c, $$ which solves to $${dx\over dy} = { \bar c \over\sqrt{y^2-\bar c^2}}$$with $\bar c ={c\over \rho g}$.

Then you set $y = \bar c \,\cosh \theta$, hence $dy = \bar c \, \sinh \theta d\theta$ to get an integral over $\bar \theta$ :
$$x = \bar c \int_{\theta_0}^\theta {\sinh \bar\theta\over \sinh \bar\theta}\, d\bar\theta = \bar c \, \theta + b $$

But now I've painted myself in the same corner as you. There must be something wrong with our y: The minimum surface of revolution has the same differential equation and the same cosh answer, but there is of course a condition y > 0 (see e.g. page 6 of http://math.hunter.cuny.edu/mbenders/cofv.pdf) that we are lacking. We can shift the x such that e.g. b = 0, but I don't know about y.

And I also need $\ell$ back in, probably via the $\int ds$...

 

[BvU]

Found a solution that indeed makes use of the constraint, but I don't know if it is going to make you happy.

Credit to Frank Porter at CalTech for this link that combines Euler-Lagrange with Lagrange multipliers method for optimization with the $\int d\bar \ell = \ell$ constraint.

$\bar c$ and b as integration constants now get company from a third unknown (the Lagrange multiplier $\lambda$), so that y₀=0, y₂=0 and $\ell$ can be accommodated comfortably.

 

[CAF123]

I think I can use the boundary conditions to determine $\bar{c}$ and $b = -\bar{c}\theta_0$, but the problem is I cannot do this in practice because there is no theta satisfying coshθ=0.

Also, I am yet to find a boundary condition involving l. (as per the question requirements).

Edit: we typed past each other - thanks, but the solution uses Lagrange multipliers and I didn't cover this in my course. I am speaking to my professor on Thursday and I will ask him then unless someone else chips in. Then I will let you know.