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Equilibrium shape of string

  1. Apr 11, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    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##.

    2. Relevant 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), ## $$y' \frac{\partial F}{\partial y'} - F = \text{const}$$

    3. 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.
     

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    Last edited: Apr 11, 2014
  2. jcsd
  3. Apr 11, 2014 #2

    BvU

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    Looks good to me. Cosh is also called Catenary
     
  4. Apr 11, 2014 #3

    CAF123

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    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.
     
  5. Apr 12, 2014 #4

    CAF123

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    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.
     
  6. Apr 12, 2014 #5

    CAF123

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    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.
     
  7. Apr 14, 2014 #6

    BvU

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    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 ?
     
  8. Apr 14, 2014 #7

    CAF123

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    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.

    Yes, I think I need to incorporate it into the boundary conditions.
    ##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.
    Yes.
    I did not really think of a physical realization of ##\theta##, I was using the substitution in order to solve the integral.
     
  9. Apr 14, 2014 #8

    BvU

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    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 ##...
     
  10. Apr 15, 2014 #9

    BvU

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    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 accomodated comfortably.
     
  11. Apr 15, 2014 #10

    CAF123

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    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.
     
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