Question

Consider a steel guitar string of initial length l=1.00 meter and cross-sectional area a=0.500 square millimeters. the young's modulus of the steel is y=2.0×1011 pascals. how far ( δl) would such a string stretch under a tension of 1500 newtons?

Answer 1

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where 
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase in length of the object.

For the guitar string in the problem, we have $L_0 = 1.00 m$, $A=0.500 mm^2 = 0.5 \cdot 10^{-6}m^2$, $E=2.00 \cdot 10^{11}Pa$, and the force applied is $F=1500 N$. Substituting the numbers into the formula, we find:
$\Delta L = \frac{F L_0}{EA}= \frac{(1500 N)(1.00 m)}{(2.0\cdot 10^{11}Pa)(0.5\cdot 10^{-6}m^2)}=0.015 m$

So, the string stretches by 0.015 m under a tension of 1500 N.