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Homework Help: Stretching in a guitar string! Please help?

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data

    A 76 cm long, 1.0 mm diameter steel guitar string must be tightened to a tension of 2200 N by turning the tuning screws. By how much is the string stretched?


    2. Relevant equations

    F=kx
    K=YA/L

    3. The attempt at a solution
    I dont know how to approach this problem because I feel like I dont have everything i need! I know that F=2200N. but How do i get K to find delta x?
     
  2. jcsd
  3. Oct 23, 2008 #2

    CompuChip

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    What is Y? What does it depend on?
     
  4. Oct 23, 2008 #3
    Y is Young's Modulus, which in my notes is just referred to as a measure of the substances inherent stiffness, and i dont know where to get this.....
     
  5. Oct 23, 2008 #4

    LowlyPion

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  6. Oct 23, 2008 #5
    I understand that, but how is this going to help me in my problem? I'm sorry if I sound stupid....
     
  7. Oct 23, 2008 #6

    LowlyPion

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    Look at the formula for Young's modulus

    [tex]Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L} [/tex]
     
  8. Oct 23, 2008 #7
    Would the stress be 2200 then?
     
  9. Oct 23, 2008 #8

    LowlyPion

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    No. But the Force 2200 N divided by the cross sectional area of a 1 mm diameter string would be though.
     
  10. Oct 23, 2008 #9

    HallsofIvy

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  11. Oct 23, 2008 #10
    Okay. So now i have y=200. yay! So now i need area A which is [im assuming?] l*w? So .76*.001=A?
     
  12. Oct 23, 2008 #11

    LowlyPion

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    A is cross section = π*R2 = π*(.5*10-3)2

    .76m is your L
     
  13. Oct 23, 2008 #12

    LowlyPion

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    By the way Y is 200 * 109N/m2
     
  14. Oct 23, 2008 #13
    I got my final answer to be .0106. Is this in meters, then? Do I need to convert to centimeters?
     
  15. Oct 23, 2008 #14
    I figured it out, converted to CM and got my final answer as 1.06 cm, and it was right! Thank you so much for all your help!
     
  16. Oct 23, 2008 #15

    LowlyPion

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    Hope you didn't get too stressed or that it was too much of a strain.

    Cheers
     
  17. Oct 24, 2008 #16

    CompuChip

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    * grins at Pions puns *

    Jaklynn, try to keep the units throughout your calculation, you can just consider them as ordinary variables. For example, if you try to do (0.76 m) * (0.001 m^2) you will get 0.00076 m^3. But m^3 is a unit of volume, not of area.
    Similarly, if you forget converting something, you will end up with something like: cm * m, instead of m * m = m^2. This makes it easier to spot your errors beforehand.
     
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