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Vertical string problem

  1. Mar 16, 2005 #1
    A rope of mass m and length L is suspended vertically. Show that a transverse wave pulse will travel the length of the tope in a time t = 2(L/g)^1/2.

    stuck... i tried:

    v = (T/u)^1/2 ; where T = m(X/L)g and u = m/L

    the X is the distance from the bottom of the string...

    so i got v= (xg)^1/2

    t= (L-X)/ (xg)^1/2

    which i can't simplify to t = 2(L/g)^1/2.

    Where did I go wrong here? logic is making sense to me but im no getting the answer. thanks.
     
  2. jcsd
  3. Mar 16, 2005 #2

    Doc Al

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    Staff: Mentor

    You found v as of function of x correctly, but then just multiplied by the distance to find the time. Since v is not constant, you must integrate:
    [tex]v = \frac{dx}{dt} = (xg)^{1/2}[/tex]
    [tex]t = \int_{0}^{L}\frac{dx}{(xg)^{1/2}}[/tex]
     
  4. Sep 16, 2010 #3
    can you please expain why T = m(X/L)g? Thanks!
     
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