# Oscillations of a free hanging chain

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1. Mar 21, 2016

### kregg34

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

2. Relevant equations

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)

3. The attempt at a solution

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?

Last edited: Mar 21, 2016
2. Mar 21, 2016

### haruspex

It doesn't seem right to have a dy/dx term in your very first equation. I would have thought it should be the standard string vibration equation, except for a factor x in the tension term.

3. Mar 21, 2016

### kregg34

On the question it actually gives you that equation, you have to derive it, and thats what it is

4. Mar 21, 2016

### TSny

For this to be the correct equation, I think you need to take x as increasing upward from the lower end of the string.

5. Mar 21, 2016

### kregg34

oh sorry, thats how I derived it but I wrote it wrong. x is measured from the free end of the chain upwards

6. Mar 21, 2016

### TSny

No. To get the equation for s(ξ), you will need to start with -((ω^2)/g)u(x) = u'(x) + xu''(x) and transform it to the equation for s(ξ) by using the definition ξ = √x. The chain rule for derivatives will be useful here.

7. Mar 21, 2016

### kregg34

Not sure I understand. Cause I thought the transformations of u(x) were basically saying u(x) = s(√x). So I took the derivative of both sides with respect to x and substituted u'(x) and u''(x) in

8. Mar 21, 2016

### TSny

In obtaining equation 1, it appears to me that you are simply replacing u'(x) by s'(ξ) and replacing u''(x) by s''(ξ).

Take a specific example. Suppose u(x) = sin(√x). So, with ξ = √x, you have s(ξ) = sin(ξ).
What is u'(x)? What is s'(ξ)? Does u'(x) = s'(ξ)?

Note that you have claimed u'(√x) = (u'(√x))/(2√x). If you cancel u'(√x) from both sides, you are left with 1 = 1/(2√x).

9. Mar 21, 2016

### kregg34

u'(x) = cos(√x)/(2√x)
s'(ξ) = cos(ξ) (dξ/dx)
and ξ = √x so
dξ/dx = d√x/dx = 1/(2√x) aren't they the same? or should I write
s'(ξ) = cos(ξ) (dξ/dξ) = cos(ξ) with respect to ξ instead of x?

10. Mar 21, 2016

### TSny

OK, I see where you're coming from. You are interpreting all primes as meaning derivative with respect to x. However, since s is considered a function of ξ, the notation s'(ξ) would normally denote the derivative of s with respect to ξ. So, in the example, s'(ξ) = cos(ξ).

What you need to do here is convert the differential equation involving u and x into a differential equation involving s and ξ. In the differential equation for s, derivatives should be with respect to ξ and the symbol x should not appear.

11. Mar 21, 2016

### kregg34

Figured it out, thanks man!!

Last edited: Mar 21, 2016
12. Mar 21, 2016

Good work!