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CAF123
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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.
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