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Oscillations of a free hanging chain

  1. Mar 21, 2016 #1
    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. jcsd
  3. Mar 21, 2016 #2

    haruspex

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    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.
     
  4. Mar 21, 2016 #3
    On the question it actually gives you that equation, you have to derive it, and thats what it is
     
  5. Mar 21, 2016 #4

    TSny

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    For this to be the correct equation, I think you need to take x as increasing upward from the lower end of the string.
     
  6. Mar 21, 2016 #5
    oh sorry, thats how I derived it but I wrote it wrong. x is measured from the free end of the chain upwards
     
  7. Mar 21, 2016 #6

    TSny

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    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.
     
  8. Mar 21, 2016 #7
    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
     
  9. Mar 21, 2016 #8

    TSny

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    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).
     
  10. Mar 21, 2016 #9
    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?
     
  11. Mar 21, 2016 #10

    TSny

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    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.
     
  12. Mar 21, 2016 #11
    Figured it out, thanks man!!
     
    Last edited: Mar 21, 2016
  13. Mar 21, 2016 #12

    TSny

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    Good work!
     
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