How Much Does a Guitar String Stretch Under Tension?

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SUMMARY

The discussion centers on calculating the stretch of a steel guitar string under tension, specifically a 76 cm long string with a diameter of 1.0 mm and a tension of 2200 N. Participants clarify the use of Young's Modulus (Y), which is 200 x 109 N/m2 for steel, and the formula for cross-sectional area (A) as π*(0.5 x 10-3)2. The final calculated stretch of the string is 1.06 cm after converting from meters. The discussion emphasizes the importance of unit consistency in calculations.

PREREQUISITES
  • Understanding of Young's Modulus and its application in material science.
  • Knowledge of basic physics formulas, specifically Hooke's Law (F = kx).
  • Familiarity with geometric calculations for cross-sectional area (A = πr2).
  • Ability to convert units accurately, particularly between meters and centimeters.
NEXT STEPS
  • Research the properties of different materials and their Young's Modulus values.
  • Learn about the applications of Hooke's Law in real-world scenarios.
  • Explore advanced topics in material deformation and stress-strain relationships.
  • Practice unit conversion techniques to avoid common calculation errors.
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Students studying physics, engineers working with materials, and anyone interested in the mechanics of tension and elasticity in materials.

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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