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Frequency of the 7th Overtone

  1. Oct 9, 2007 #1
    1. The problem statement, all variables and given/known data

    5.10m, 0.733kg wire is used to support two uniform 230N posts of equal length. Assume that the wire is essentially horizontal and that the speed of sound is 344m/s. A strong wind is blowing, causing the wire to vibrate in its 7th overtone.

    [​IMG]
    What is the frequency of the sound this wire produces?

    2. Relevant equations

    fn=nv/2L
    v=sqrt(F/linearmass)
    sumF=0

    3. The attempt at a solution

    -split the force vectors into their components
    -multiplied the force along the x axis from one of the posts by two to find the tension in the string
    -used this value to find velocity
    -used velocity to find the frequency of the 7th overtone
    -got answer wrong

    help?
     
  2. jcsd
  3. Oct 9, 2007 #2

    learningphysics

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    What tension did you get in the string?
     
  4. Oct 9, 2007 #3
    Balancing torques will be correct, I suppose.
     
  5. Oct 9, 2007 #4
    T=250.5n
     
  6. Oct 9, 2007 #5

    learningphysics

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    That's not what I get. How did you get that?
     
  7. Oct 9, 2007 #6
    T=250.5
    v=sqrt(250.5/(0.733/5.1))
    v=41.75
    f(6)=6(v/2L)
    f(6)=24.6

    thats my initial work
    however it is incorrect

    what was done wrong?
     
  8. Oct 9, 2007 #7
    230cos(57)=125.3

    125.3X2=250.5
     
  9. Oct 9, 2007 #8
    Tension is not balancing forces. Its their torques which are balancing.
     
  10. Oct 9, 2007 #9

    learningphysics

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    This isn't correct. The 230N isn't the tension/compression in the post... it is just the weight of the post. As Sourabh N said, use torque about one of the pivots (either one doesn't matter which one)... net torque about the pivot is 0.
     
  11. Oct 9, 2007 #10
    ok I understand to use torque
    but how can I find the torque without the length of the pole?
    maybe I am unaware of a method to calculate torque
     
  12. Oct 9, 2007 #11

    learningphysics

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    The length will cancel... write out the equation for torque about the pivot...
     
  13. Oct 9, 2007 #12
    torque= rXF

    I don't seem to understand how to use this formula without knowing the position of the forces.
    is the force in this equation acting on the string?
    where are the 230N of the pole acting?
     
  14. Oct 9, 2007 #13

    learningphysics

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    The 230N force is acting at the center of the post (midpoint)... The tension is acting at the end of the post...
     
  15. Oct 9, 2007 #14
    ok so the tension balances the torque.

    and I can find torque with the formula torque=rXF

    and F is 230N down

    the position of this force is at the midpoint of the post

    therefore, I can't get a numerical value for torque...

    r(pivot to force)=0.5(length of pole)???
     
  16. Oct 9, 2007 #15

    learningphysics

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    Torque is r X F. This can be interpreted as force times the perpendicular distance from the line of force to the pivot... ie the torque due to the weight is 230N*(L/2)cos57... if we take clockwise positive and counterclockwise negative and look at the left pivot... torque due to the weight is -230*(L/2)cos57

    What is the torque due to tension about the left pivot? The force is T. What is the perpendicular distance from the line of force to the pivot?

    The sum of these 2 torques is 0.
     
  17. Oct 9, 2007 #16
    the torque on the right pivot is 115(L)cos57
    the torque on the left pivot is -115(L)cos57

    so if I use torque=Fd

    and the torque is the above calculated values...and the force is the tension in the wire and d=(L)sin57

    then F(right pivot)= (115(L)cos57)/(sin57(L))
    F=74.68N

    so the total tension is 74.68X2=149.36N

    the use this value for the tension in the string

    is that correct?
     
  18. Oct 9, 2007 #17

    learningphysics

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    You don't need to multiply by 2. The tension is just 74.68N.
     
  19. Oct 9, 2007 #18
    T=74.68N

    v=sqrt(74.68/(0.733/5.1))
    v=22.79m/s

    f=6(v/2L)
    f=13.4Hz

    what is wrong with those calculations?

    the answer is still wrong
     
  20. Oct 9, 2007 #19

    learningphysics

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    I think you should be multiplying by 8 not 6.
     
  21. Oct 9, 2007 #20
    oops silly mistake

    thank you
     
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