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Chapter 10: Problem 7

A 0.50 -m-long guitar string, of cross-sectional area $1.0 \times 10^{-6} \mathrm{m}^{2},\( has Young's modulus \)Y=2.0 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$ By how much must you stretch the string to obtain a tension of \(20 \mathrm{N} ?\)

Short Answer

Expert verified
Answer: The guitar string should be stretched by 5.0 micrometers to achieve the desired tension.

Step by step solution

01

Identify the relevant formula

We will use the definition of Young's modulus, which is given by: $$Y = \frac{F}{A} \frac{\Delta L}{L}$$ where \(Y\) is Young's modulus, \(F\) is the force (tension) applied, \(A\) is the cross-sectional area, \(\Delta L\) is the change in length, and \(L\) is the original length. We'll need to solve for \(\Delta L\).
02

Rearrange the formula for the desired variable

We want to find the change in length \(\Delta L\). First, we'll isolate the term \(\frac{\Delta L}{L}\): $$\frac{\Delta L}{L} = \frac{F}{Y \cdot A}$$ Now, we can solve for \(\Delta L\) by multiplying both sides by \(L\): $$\Delta L = L \cdot \frac{F}{Y \cdot A}$$
03

Insert given values into the formula

Now, we'll plug in the given values for the length \(L = 0.50\,\text{m}\), the force \(F = 20\,\text{N}\), Young's modulus \(Y = 2.0 \times 10^{9}\,\frac{\text{N}}{\text{m}^{2}}\), and the cross-sectional area \(A = 1.0 \times 10^{-6}\,\text{m}^{2}\): $$\Delta L = 0.50\,\text{m} \cdot \frac{20\,\text{N}}{(2.0 \times 10^{9}\,\frac{\text{N}}{\text{m}^{2}}) \cdot (1.0 \times 10^{-6}\,\text{m}^{2})}$$
04

Calculate the change in length

Perform the calculation: $$\Delta L = 0.50\,\text{m} \cdot \frac{20\,\text{N}}{(2.0 \times 10^{9}\,\text{N/m}^{2}) \cdot (1.0 \times 10^{-6}\,\text{m}^2)} = 5.0 \times 10^{-6} \,\text{m}$$
05

Interpret the result

The change in length required to obtain a tension of 20 N is \(5.0 \times 10^{-6}\, \text{m}\). This means the guitar string should be stretched by 5.0 micrometers to achieve the desired tension.

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