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Homework Help: A wave speed of a hanging chain

  1. Jan 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Problem 5 from: http://www.swccd.edu/~jveal/phys274/images/hw01.pdf in case you dont understand my text.

    A chain of linear mass density u, and length L is hang-
    ing from a ceiling. There is a wave moving vertically
    along its length. a) Is the propagation speed constant?
    (Justify your answer.) b) Show that the amount of
    time it takes the wave to move along the full length is
    given by

    t=2[tex]\sqrt{\frac{L}{g}}[/tex]
    2. Relevant equations
    String waves speed: [tex]\frac{u}{T}[/tex][tex]\frac{\delta ^{2}y}{\delta t^{2}}[/tex]= [tex]\frac{\delta ^{2}y}{\delta x^{2}}[/tex]

    3. The attempt at a solution
    Ive spent 2 hours trying to use the forumula for a string waves speed but I really dont understand the concept of solving the partial differential equations.

    I know that the propagation speed is not constant because of gravity but i dont know how to apply that to the formula.

    btw used delta for partial derivatives.

    Thanks alot :)
     
    Last edited: Jan 29, 2010
  2. jcsd
  3. Jan 29, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: Make use of the results of problem 2.
     
  4. Jan 29, 2010 #3
    Yes, ive tried that approach but i think the awnser lies in the differential equation that i posted which is where the velocity of the wave in a string is derived from. Unless iam overcomplicating myself and iam not seeing something.

    Because i know that T=uLg and the time would be L/v but i still dont see where i would get a 2 from plugging that stuff in.
     
  5. Jan 30, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Show what you've tried.

    Careful. The tension--and thus the speed--varies along the chain. So neither of those two expressions are correct.

    Try this. Write the tension as a function of distance (x) from the bottom. Then set up and solve a simple differential equation, realizing that v = dx/dt.
     
  6. Jan 30, 2010 #5
    Ahhhh i finally got it. THANKS ALOT. My problem was that i didnt understand the concept well enough to understand that tension varies with the speed. Was easier than i thought.
     
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