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Wave pulse along a wire

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    The figure shows two masses hanging from a steel wire. The mass of the wire is 60.0 g. A wave pulse travels along the wire from point 1 to point 2 in 24.0 ms.

    What is mass m?

    [​IMG]

    2. Relevant equations
    3. The attempt at a solution

    L=8.0m

    [tex]m_{string}=60g=0.06kg[/tex]

    [tex]
    \mu=m_{string}/L=0.0075 kg/m
    [/tex]

    [tex]v=4.0m/24ms=166.667m/s[/tex]

    [tex]v=\sqrt{T/\mu}=\sqrt{T/0.0075kg/m}[/tex]

    [tex]T=\sqrt{(2mgsin\theta)^2+(2mgcos\theta)^2)[/tex]

    Using [tex]\theta=40 degrees[/tex], I got m to be 10.6 kg. However, this answer is wrong. Can anyone tell me where I made the mistake? Thanks!
     
  2. jcsd
  3. Dec 5, 2008 #2

    LowlyPion

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    Aren't you only interested in the horizontal tension in the wire from 1 to 2?
     
  4. Dec 5, 2008 #3
    So you mean I should keep using 8.0m for L but use 2mg cos 40 for T?

    I tried doing that-->setting 2mg cos 40 for T and then solving for T.

    I got 13.9 kg, which is still not right...

    Am I supposed to use 4m for L too?
     
    Last edited: Dec 5, 2008
  5. Dec 5, 2008 #4

    LowlyPion

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    Why is it 2mgCos40 ?

    But as to the μ, you simply calculate the density as you did. The density can be considered not to change between the various segments.
     
  6. Dec 5, 2008 #5
    Well, I thought that it was 2 mg cos 40 since each of the weights exerts a force of mg cos 40 on the string...if not, I'm confused as to how I should find T
     
  7. Dec 5, 2008 #6

    LowlyPion

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    Consider the free body diagram of 1 of the masses in isolation.

    Isn't the force of the tension required to hold that mass in equilibrium equal to just mgCos40°?
     
  8. Dec 5, 2008 #7
    Ah I figured it out. T=mg/tan(40)...
     
  9. Dec 5, 2008 #8

    LowlyPion

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    As I suggested before I think you are only interested in the horizontal Tension between points 1 and 2.
     
  10. Dec 5, 2008 #9
    I figured out the correct answer using T=mg/tan(40) so I am pretty sure that is the correct value for the tension in that part :)
     
  11. May 11, 2011 #10
    996gt2 is right. The answer is 17.8.
     
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