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

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

Step-by-step explanation:

From the question we are told that

The initial length is
`L = 1.00m`

The area is
`A = 0.500 mm^2 = (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 = (\sigma)/(e)`

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

`\sigma = (F)/(A)`

Substituting values

`\sigma = (1500)/(0.500*10^(-6))`

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

And e is the strain which is mathematically represented as

`e = (\Delta L)/(L )`

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

Substituting these into the equation above

`Y = (3.0*10^9)/((\Delta L)/(L) )`

Substituting values

`2.0 *10^(11) = (3.0*10^9)/((\Delta L)/(L) )`

`\Delta L = (3.0*10^9 * 1)/(2.0 *10^(11))`

`\Delta L = 0.015 \ m`