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Trouble with this integral and only an hour to solve it

  1. Aug 9, 2004 #1
    Well first here is the question:

    A deep-sea diver is suspended beneath the surface of Loch Ness by a cable of length h that is attached to a boat on the surface . The diver and his suit have a total mass of m and a volume of V. The cable has a diameter of d and a linear mass density of mu. The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat.

    Calculate the tension in the cable a distance above the diver. The buoyant force on the cable must be included in your calculation. Take the free fall acceleration to be g.

    I solve this one, I got:

    [tex] F(x) = (\mu*x+m)*g-\rho_{water}*g*((d/2)^2*pi*x+V) [/tex]

    For the next one I can't solve it, the integral is nuts for me, I must be overlooking something

    The speed of transverse waves on the cable is given by v = sqrt(F/mu). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface. Take the free fall acceleration to be g.

    From what I understand I have to plug in my F(x) from part b and integrate, but I'm not sure how to integrate it since its just so huge. Any ideas?

    Thanks!

    Brian
     
  2. jcsd
  3. Aug 10, 2004 #2

    ehild

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    Homework Helper
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    Write [tex]F/\mu[/tex] in the form [tex]F/\mu=ax+b[/tex] .

    Remember that

    [tex]v=x'=\frac{dx}{dt}[/tex] .

    So you have got the differential equation

    [tex]\frac{dx}{dt}=\sqrt{ax+b}[/tex]

    with the starting condition x=0 at t=0.
    You can integrate it by separating the variables:

    [tex]\int_0^h\frac{dx}{\sqrt{ax+b}}=\int_0^T{dt}=T[/tex]

    where T is the time needed for the signal to reach the surface and h is the lenght of the cable. The integrand at the left side is

    [tex](ax+b)^{-1/2}[/tex] .

    I think you can do it...

    ehild
     
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