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Homework Help: How to solve this equation

  1. May 11, 2005 #1
    Hey guys this is my question


    A long thin rod is clamped vertically at its lower end and a mass M is attached to its upper end. The coordinates (x, y) of any point on it satisfy the equation

    [tex] EI\frac{d^2x}{dy^2} = Mg(a-x)[/tex]

    where E, I and a are constants. Given that x = 0 when y = 0 and x = a when y = L show that

    [tex] x = a [1- \frac{sin\omega(L-y)}{sin\omega L}][/tex]

    where [tex]\omega^2 = \frac{Mg}{EI}[/tex]. (Hint: Let z = a - x).


    Now I have got this here:

    [tex] z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}[/tex]


    [tex] EI\left(-\frac{d^2z}{dy^2}\right) = Mgz[/tex]

    [tex] \frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0[/tex]

    I'm thinking that it's correct so far, How would I solve that?. ie. Continue on from here

    Could someone help me out please

    Last edited: May 11, 2005
  2. jcsd
  3. May 11, 2005 #2


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    Assuming what you have done is correct, the equation you have generated for z is a well known equation. If y were time, your equation would be saying acceleration is proportional to minus the displacement. This is the equation of a harmonic oscillator. Assume z as a function of y is harmonic (sine or cosine of ky) and take two derivatives. It will all fall into place.
  4. May 11, 2005 #3
    So I get this ?

    z = a - x = a*sinw(L-y) / sinwL

    dz/dy = -w*acosw(L-y) / sinwL

    d^2z/dy^2 = -w^2*asinw(L-y) / sinwL

    = -w^2 * 2

    If this is what you were reffering to, what happens now ?
  5. May 11, 2005 #4
    [tex] \frac{d^2z}{dy^2} + Cz = 0[/tex] [tex]C=\frac{Mg}{EI}[/tex]
    [tex] \frac{d^2z}{dy^2} =-Cz[/tex]
    [tex] z(y)=?[/tex]
  6. May 11, 2005 #5
    I'm not following :(((
  7. May 11, 2005 #6


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    It's hard to give any help without knowing what background you have, what kind of course, this is, etc. In particular, it seems very peculiar that you would be doing a problem like this without know how to solve a linear second order differential equation with constant coefficients.
  8. May 11, 2005 #7
    We have studied the things you just mentioned. linear, 2nd order, homogenous, inhomogenous, auxillary equations etc.

    It's just we never did anything like that, we were always dive, simpler questions. The lecturer just isn't on the same level as we are :(

    Edit: If someone did the working out, I'm sure I could figure it out and next time be able to do it myself :)
  9. May 11, 2005 #8
    OlderDan was trying to point out that the resulting equation was similar to that of a harmonic oscillator. da_willem then showed you the final differential equation you needed to solve to find z(y). The final result should be very similar to the equations of SHM.

    I haven't taken Diff Eq so I can't be more specific, sorry.
  10. May 11, 2005 #9


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    Where did the *2 come from? You mean *z. So

    d^2z/dy^2 + w^2*z = 0

    Compare this to your DE and identify omega. Put your expression for z into the defining equation for z and solve it for x.
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