voko
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Very well. A couple of observations.
The gravitational potential energy in the equilibrium state has the term ## \frac 1 2 mga ##, which becomes zero in the limit. Note, however, that without taking the limit, this term is the gravitational potential energy of the slinky in its natural state, so the change in the gravitational potential energy and thus the change in the full potential energy are independent of the natural length.
Secondly, the full energy integral, which you never wrote, is $$ \int \limits_0^a \left( {AE \over 2}(y' - 1)^2 + \rho g y \right) dx $$ where I replaced ##l## with ##y## for readability. In the equilibrium state, this integral is minimal, which, using calculus of variations (Euler-Lagrange equation) leads to the differential equation $$ AEy'' = \rho g $$ Compare this with the equation for ##l(x)## that you obtained by balancing forces in infinitesimal elements.
The gravitational potential energy in the equilibrium state has the term ## \frac 1 2 mga ##, which becomes zero in the limit. Note, however, that without taking the limit, this term is the gravitational potential energy of the slinky in its natural state, so the change in the gravitational potential energy and thus the change in the full potential energy are independent of the natural length.
Secondly, the full energy integral, which you never wrote, is $$ \int \limits_0^a \left( {AE \over 2}(y' - 1)^2 + \rho g y \right) dx $$ where I replaced ##l## with ##y## for readability. In the equilibrium state, this integral is minimal, which, using calculus of variations (Euler-Lagrange equation) leads to the differential equation $$ AEy'' = \rho g $$ Compare this with the equation for ##l(x)## that you obtained by balancing forces in infinitesimal elements.