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What is the length of the string?

  1. Mar 18, 2016 #1
    1. The problem statement, all variables and given/known data

    A wave along a 15 gram string travels at 756 m/s when it is tightened to 189N?

    2. Relevant equations

    V = sqr root T/ m/l

    3. The attempt at a solution

    I got 3.30 x 10^-4
     
    Last edited by a moderator: Mar 18, 2016
  2. jcsd
  3. Mar 18, 2016 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    That looks kind of small for a length of string... What are the units?

    Can you show your work (including units) to solve the equation for the length? :smile:
     
  4. Mar 18, 2016 #3
    Your result seems the mass per unit length not length.
     
  5. Mar 18, 2016 #4
    I'm confuse, so what's the correct formula or way to solving for the length?
     
  6. Mar 18, 2016 #5

    berkeman

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    It looks like you wrote the correct equation, but the quantity in the denominator is the mass per unit length...

    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html
     
  7. Mar 18, 2016 #6
    Well, you already knew the formula to calculate the m/L (assuming string is uniform).So you can calculate mass per unit length(m/L) by just puttin v and T into equation. Then you can find L using mass(15 g).
     
  8. Mar 18, 2016 #7
    So mass is gram or I have to do it on kg
     
  9. Mar 18, 2016 #8
    You should first convert it to kg to be able to express length in meters. If you use gram you find length in terms of centimeters.
     
  10. Mar 18, 2016 #9
    So what's the correct formula?
     
  11. Mar 18, 2016 #10

    berkeman

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    Staff: Mentor

    You already wrote it in Post #1.

    Just re-arrange it to isolate the length on the lefthand side (LHS) of the equation, and substitute the numbers that you have been given in the problem statement. Be sure to convert to mks units as appropriate.

    Please show us your work, and your final solution so that we can check it. :smile:
     
  12. Mar 18, 2016 #11
    I got << Deleted by Mentor >> if no mistake. You should use kg or g depending on your preference to express length.
     
  13. Mar 18, 2016 #12
    I got 4.96x10^-6 am I correct?
     
  14. Mar 18, 2016 #13

    berkeman

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    Please don't do the student's work for them. That's one of the HH rules. He's almost got it figured out on his own now... :smile:
     
  15. Mar 18, 2016 #14

    berkeman

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    Almost certainly not. That length is in meters? Please show your detailed work, including units in your equations.
     
  16. Mar 18, 2016 #15
    L= .015m(189N/756m/s ^2)
     
  17. Mar 18, 2016 #16

    berkeman

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    What's that?
    Please use better parenthesis to show what is getting squared and what all the units are...
     
  18. Mar 18, 2016 #17

    berkeman

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    I think I see where your error is.

    You get that answer if you use the incorrect formula [tex]v = \sqrt{\frac{T}{\frac{L}{m}}}[/tex]

    Instead of the correct formula that you listed in Post #1 and is in my Hyperphysics link: [tex]v = \sqrt{\frac{T}{\frac{m}{L}}}[/tex]
     
  19. Mar 18, 2016 #18
    I got L = (mv^2)/T
     
  20. Mar 18, 2016 #19

    berkeman

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    Good. Now plug in the numbers you were given (and show units in square brackets [] or similar to make sure the units of the final answer are right), and solve! :smile:
     
  21. Mar 18, 2016 #20
    L= (.015m*756m/s ^2)/(189)
    L= 45.36 m
     
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