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Integration of an inverse sqrt composite function

  1. May 16, 2007 #1
    1. The problem statement, all variables and given/known data
    “Geologist A” at the bottom of a cave signals to his colleague “Geologist B” at the surface by pushing a 11.0 kg box of samples from side to side. This causes a transverse wave to propagate up the 77.0 m rope. The total mass of the rope is 14.0 kg. Take g = 9.8 m/s².

    How long does it take for the wave to travel from the bottom of the cave to the surface?[Hint: Find an analytic expression v(z) for the wave speed as a function of distance. Then use the fact that at any given point on the rope the time dt taken to travel a small distance dz is given by: dt=dz/v(z). Then integrate to obtain the total travel time. ]

    [​IMG]

    2. Relevant equations
    u = mR/z
    T(z) = u.z.g + mB.g
    v(z) = (T(z)/u)^(1/2)
    dt=dz/v(z)

    z = the length of the rope = L (used for integrating)

    3. The attempt at a solution
    v(z) = (T(z)/u)^(1/2)
    v(z) = ((mR/z).z.g + mB.g/(mR/z))^(1/2)
    v(z) = ((mR.g + mB.g)/(mR/z))^(1/2)
    v(z) = ([(mR.g)/(mR/z)] + [(mB.g)/(mR/z)])^(1/2)
    v(z) = ([(mR.g.z)/mR] + [(mB.g.z)/mR])^(1/2)
    v(z) = ([mR.g.z] + [(mB.g.z)/mR])^(1/2)

    dt=dz/v(z)
    dt=dz/([mR.g.z] + [(mB.g.z)/mR])^(1/2)

    Integration:

    ....f L
    t= | (1/([mR.g.z] + [(mB.g.z)/mR])^(1/2)).dz
    ....j 0

    ....f L
    t= | 2.([mR.g.z^2/2] + [(mB.g.z^2)/mR.2])^(1/2).dz
    ....j 0

    I think I integrated it properly but when substituted the values
    mB = mass of box
    mR = mass of rope
    g = 9.8 ms^2
    z = 77.0 m

    I didn't get the correct answer of t = 2.52s
     
    Last edited: May 16, 2007
  2. jcsd
  3. May 16, 2007 #2

    Dick

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    'u' in your solution is supposed to be the mass density of the rope. It's not mR/z. The rope doesn't have a variable density, it's mR/(total length of rope), a constant. Nice problem presentation, by the way.
     
    Last edited: May 16, 2007
  4. May 16, 2007 #3
    Hi Dick, z is the length of the rope (77.0m as in the problem) it was just the letter they used in the formula sheet so I carried it forth.

    How was my integration? I don't think I know how to integrate nested functions (I assume it's something like the reverse of the chain-rule?).
     
  5. May 16, 2007 #4

    Dick

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    z in your problem is the variable indicating length along the rope. I mean that u=14 kg/(77 m). It doesn't have the variable of integration in it. Until that gets fixed there isn't any point in discussing the integral.
     
  6. May 16, 2007 #5

    Dick

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    In your notation u=mR/L not mR/z.
     
  7. May 16, 2007 #6
    Oh, ok. Thanks for pointing that out :)

    So:-

    u = mR/L = mR/77
    T(z) = u.z.g + mB.g
    v(z) = (T(z)/u)^(1/2)
    dt=dz/v(z)

    z = the length of the rope = L (used for integrating)

    3. The attempt at a solution
    v(z) = (T(z)/u)^(1/2)
    v(z) = ((u.z.g + mB.g)/u)^(1/2)
    v(z) = ([(u.z.g)/u] + [(mB.g)/u])^(1/2)
    v(z) = ([g.z] + [(mB.g)/u])^(1/2)

    dt=dz/v(z)
    dt=dz/([g.z] + [(mB.g)/u])^(1/2)

    Integration:

    ....f L
    t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
    ....j 0

    ....f L
    t= | 2.([g.z] + [(mB.g)/u])^(1/2).(g.z^2)/2
    ....j 0

    Where:
    mB = mass of box
    mR = mass of rope
    g = 9.8 ms^2
    z = 77.0 m

    Is this correct?
     
    Last edited: May 16, 2007
  8. May 16, 2007 #7

    Dick

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    Your first integral looks just fine. I don't know how you got from there to the second one. The usual way to do an integration like this is to do a change of variable. Let v=gz+mBg/u.
     
  9. May 16, 2007 #8
    Ok, I was trying to use the chain rule lol ><

    So:-

    Integration:

    ....f L
    t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
    ....j 0

    ....f L
    t= | 2.([(g.z^2)/2] + [(mB.g.z)/u] + C)^(1/2)
    ....j 0

    Where:
    mB = mass of box
    mR = mass of rope
    g = 9.8 ms^2
    z = 77.0 m

    How am I doing? :smile:
     
  10. May 16, 2007 #9

    Dick

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    Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?
     
  11. May 16, 2007 #10
    Unfortunately not very familiar at all. I haven't really dealt with composite integrals. But I'll try:

    a = g.z
    b = (mB.g)/u
    1/(a+bz)^(1/2)

    v=(a+bz), so dv=b*dz.
    This turns the integral into 1/v^(1/2)*dv*(1/b).
    Integrating v^(-1/2).


    So:-

    Integration:

    ....f L
    t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz
    ....j 0

    ....f L
    t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u]dz.(1/[(mB.g)/u])
    ....j 0

    ....f L
    t= | 2.(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u].(1/[(mB.g)/u])
    ....j 0

    Where:
    mB = mass of box
    mR = mass of rope
    g = 9.8 ms^2
    z = 77.0 m
     
    Last edited: May 16, 2007
  12. May 16, 2007 #11

    Dick

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    Take a break and clear your head. While your at it look back at integration by substitution in a calc text. I can roughly see what you are trying to do - but you still seem to be trying to do some kind of a chain rule. And the (1/b) factor in the example becomes (1/g) in the problem, right? Do you see where it's coming from? And after the integration is done and the dz is gone you should also drop the integral sign - it looks pretty confusing otherwise.
     
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