Question

Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011Y=2.0×1011 pascals. How far ( ΔLΔLDelta L) would such a string stretch under a tension of 1500 newtons?

Answer 1

Answer:

The extent to which it would stretch  is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

    The initial length is  $L = 1.00m$

     The area is  $A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2$

     The Young modulus of the steel is  $Y = 2.0*10^{11} Pa$

     The tension   is  $T =1500 N$

The Young modulus is mathematically represented as

       $Y = \frac{\sigma}{e}$

Where $\sigma$ is the stress which is mathematically represented as

           $\sigma = \frac{F}{A}$  

Substituting values

            $\sigma = \frac{1500}{0.500*10^{-6}}$  

           $\sigma = 3.0*10^9 N/m^2$  

And  e is the strain which is mathematically represented as

            $e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

             $Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values  

           $2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

          $\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

         $\Delta L = 0.015 \ m$