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Wave Motion Physic

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L0 and its mass is m, much less than M. The "spring constant" for the cord is k. The block is released and stops at the lowest point. (Use L_0 for L0, M, g, and k as necessary.)
    (a) Determine the tension in the cord when the block is at this lowest point.


    (b) What is the length of the cord in this "stretched" position?


    (c) Find the speed of a transverse wave in the cord, if the block is held in this lowest position.

    2. Relevant equations
    v=sqrt(T/u)
    T=gu(L+x)

    3. The attempt at a solution
    I was really not sure how to start with this one. I tried to manipulate v=sqrt(T/u) and my other relevant equation. I really did not know what to do with this one . Please help.
     
  2. jcsd
  3. Dec 9, 2009 #2
    The tension in the cord will be equal to the weight of the mass M, as the cord is supporting this mass. (This ignores the mass of the cord itself.)
    Once you have this tension, you can calculate the extension of the cord using the spring constant.
    Next, the total length of the cord from its original length plus extension.
    You then have the formula for the speed of the wave in the string in terms of quantities you now know.
     
  4. Dec 9, 2009 #3

    Andrew Mason

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    Which part are you having difficulty with?

    For part a) do a free-body diagram with the forces summing to 0 (no acceleration). What tension must the cord have in order to provide the force needed to balance the weight of the block?

    b) is just a matter of determining the amount of stretch from the tension found in a) using Hooke's law

    c) is a matter of determining the speed of the wave from the tension and mass per unit length of the stretched cord using [itex]v = \sqrt{T/u}[/itex] where T is the tension and u is the mass per unit length.

    AM
     
  5. Dec 9, 2009 #4
    I guess that part that I an struggling with is part A. I tried to draw a force diagram earlier and then I just tried again. I know the tension has to be more than Mg be cause the rubber band is stretched. I just do not get how I can show this in a force diagram.
     
  6. Dec 9, 2009 #5

    Andrew Mason

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    The vertical components of the tension on both the left and right sides have to sum to the weight of the block. You can ignore the mass of the cord in this part. Stretch is not a factor. The cord will just stretch until the required tension is reached.

    AM
     
  7. Dec 9, 2009 #6
    So are there two equations?
    T=Ma and T-Mg=0 and then I set the two equal to each other and get T=Mg+Ma but the acceleration is equal to g so T=2Mg.
    Sorry I am having a tough time visualizing this correctly.
     
  8. Dec 9, 2009 #7

    Andrew Mason

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    I was confusing you because I didn't reread the problem when I gave you my last answer and had in mind that the cord was slung between two supension points with the block hanging. So the question is actually much easier than I suggested. Just ignore my references to the left and right tensions. There is only one tension here.

    [tex]T + mg = ma = 0[/tex] so [tex]T = -kx = -mg[/tex]

    where x is the displacement from L0 (ie the position of the end of the cord when no weight is bearing on the cord).

    AM
     
  9. Dec 9, 2009 #8
    The answer key for this problem has the tension equaling 2Mg. I thought that the tension would just be Mg as well. I guess this is what I don't understand. From the above information I think I can get parts B and C.
     
  10. Dec 9, 2009 #9

    ideasrule

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    I'm pretty sure the answer is Mg, not 2Mg, per the calculation of Andrew Mason
     
  11. Dec 9, 2009 #10

    Andrew Mason

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    Ok. I misread the question, again.

    When it says the block is released, the block drops with acceleration g and then undergoes simple harmonic motion. At the top, the acceleration is g down. So, by symmetry, the acceleration will be g up at the bottom, which means that T - mg = mg at the lowest point, so T = 2mg.

    From an energy perspective, the potential energy at the top relative to the bottom (when the block has fallen distance x) will be mgx. At the low point, this gravitational potential energy has been converted into spring potential energy, .5kx^2. So:

    [tex]mgx = \frac{1}{2}kx^2[/tex]

    [tex]kx = T = 2mg[/tex]

    AM
     
  12. Dec 9, 2009 #11
    Perfect. I got b and c from this thanks a lot. That cleared things up a lot, especially the potential energy part. Thanks a lot.
     
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