How Much Does a Guitar String Stretch Under Tension?

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Homework Help Overview

The problem involves determining the amount a steel guitar string stretches when subjected to a specific tension. The string has a length of 76 cm and a diameter of 1.0 mm, with a tension of 2200 N applied to it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between force, Young's Modulus, and the stretch of the string, with some expressing uncertainty about how to calculate the necessary parameters. Questions arise regarding the definition and application of Young's Modulus, as well as the calculation of cross-sectional area.

Discussion Status

Participants have made progress in understanding the concepts involved, with some successfully identifying the value of Young's Modulus for steel and discussing how to calculate the cross-sectional area of the string. There is ongoing dialogue about unit consistency and the implications of calculations.

Contextual Notes

Some participants express concern about missing information and the need for clarity on definitions and formulas, while others provide links to resources for further understanding of Young's Modulus.

Jaklynn429
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Homework Statement



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?


Homework Equations



F=kx
K=YA/L

The Attempt at a Solution


I don't know how to approach this problem because I feel like I don't have everything i need! I know that F=2200N. but How do i get K to find delta x?
 
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What is Y? What does it depend on?
 
Y is Young's Modulus, which in my notes is just referred to as a measure of the substances inherent stiffness, and i don't know where to get this...
 
I understand that, but how is this going to help me in my problem? I'm sorry if I sound stupid...
 
Jaklynn429 said:
I understand that, but how is this going to help me in my problem? I'm sorry if I sound stupid...

Look at the formula for Young's modulus

Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L}
 
Would the stress be 2200 then?
 
Jaklynn429 said:
Would the stress be 2200 then?

No. But the Force 2200 N divided by the cross sectional area of a 1 mm diameter string would be though.
 
  • #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?
 
  • #11
Jaklynn429 said:
Okay. So now i have y=200. yay! So now i need area A which is [im assuming?] l*w? So .76*.001=A?

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

.76m is your L
 
  • #12
Jaklynn429 said:
Okay. So now i have y=200. yay! So now i need area A which is [im assuming?] l*w? So .76*.001=A?

By the way Y is 200 * 109N/m2
 
  • #13
I got my final answer to be .0106. Is this in meters, then? Do I need to convert to centimeters?
 
  • #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!
 
  • #15
Jaklynn429 said:
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!

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

Cheers
 
  • #16
* 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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