Questions: A 0.490 -m-long guitar string, of cross-sectional area 1.00 x 10^-6 m^2, has Young's modulus Y=2.00 GPa. By how much must you stretch the string to obtain a tension of 16.1 N?

A 0.490 -m-long guitar string, of cross-sectional area 1.00 x 10^-6 m^2, has Young's modulus Y=2.00 GPa. By how much must you stretch the string to obtain a tension of 16.1 N?

Transcript text: A 0.490 -m-long guitar string, of cross-sectional area $1.00 \times 10^{-6} \mathrm{~m}^{2}$, has Young's modulus $Y=2.00 \mathrm{GPa}$. By how much must you stretch the string to obtain a tension of 16.1 N ?

Solution

Solution Steps

Step 1: Identify the Given Values

We are given the following values:

Length of the guitar string, $ L = 0.490 \, \text{m} $
Cross-sectional area, $ A = 1.00 \times 10^{-6} \, \text{m}^2 $
Young's modulus, $ Y = 2.00 \, \text{GPa} = 2.00 \times 10^9 \, \text{Pa} $
Tension, $ F = 16.1 \, \text{N} $

Step 2: Use Young's Modulus Formula

Young's modulus $ Y $ is defined as: \[ Y = \frac{F / A}{\Delta L / L} \] where $ \Delta L $ is the change in length (stretch) of the string.

Step 3: Rearrange the Formula to Solve for $\Delta L$

Rearrange the formula to solve for $\Delta L$: \[ \Delta L = \frac{F L}{A Y} \]

Step 4: Substitute the Given Values

Substitute the given values into the formula: \[ \Delta L = \frac{16.1 \, \text{N} \times 0.490 \, \text{m}}{1.00 \times 10^{-6} \, \text{m}^2 \times 2.00 \times 10^9 \, \text{Pa}} \]

Step 5: Calculate $\Delta L$

Perform the calculation: \[ \Delta L = \frac{16.1 \times 0.490}{1.00 \times 10^{-6} \times 2.00 \times 10^9} \] \[ \Delta L = \frac{7.889}{2.00 \times 10^3} \] \[ \Delta L = 0.0039445 \, \text{m} \]

Step 6: Convert $\Delta L$ to Millimeters

Convert the result from meters to millimeters: \[ \Delta L = 0.0039445 \, \text{m} \times 1000 \, \text{mm/m} = 3.9445 \, \text{mm} \]