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kregg34
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Homework Statement
I am trying to find an equation for a free hanging chain of mass m and length L. The chain is hanging vertically downwards where x is measured vertically upwards from the free end of the chain and y is measured horizontally.
Homework Equations
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I derived this differential equation for the chains motion,
(1/g)(second derivative of y with respect to t) = (derivative of y with respect to x) + x(second derivative of y with respect to x)
Trial solution that was given is,
y = u(x)cos(ωt) where x ⇒ ξ and ξ = √x and u(x) ⇒ s(ξ)
The Attempt at a Solution
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By putting y into the equation above I get,
-((ω^2)/g)u(x) = u'(x) + xu''(x) where the ' means the derivative with respect to x
I think its implied that -((ω^2)/g)s(ξ) = s'(ξ) + xs''(ξ) ... equation 1
After changing the variable to s(ξ) I'm suppose to get a Bessel's equation of order zero I think.
so from the change of variables,
u(x) ⇒ s(ξ)
s'(ξ) ⇒ u'(√x) = (u'(√x))/(2√x) right?
and from quotient rule I got,
s''(ξ) ⇒ u''(√x) = (u''(√x) - u''(√x)(1/√x)) / (4x)
plugging these into equation 1 I get,
(x^2)s''(ξ) + x^(3/2)s'(ξ) + (((ω^2)x^2)/g)s(ξ) = 0
but the bessel equation looks like
(x^2)s''(ξ) + (x)s'(ξ) + constant(x^2)s(ξ) = 0 which has x instead of x^(3/2)?
Not sure what I did wrong?
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