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= (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 = (F L_0)/(EA)= ((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.