Question

An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a decrease in the wire's equilibrium length, increases the tension in the wire. Taking the cross-sectional area of the wire to be 5.75 10-6 m2, the density to be 2.70 103 kg/m3, and Young's modulus to be 7.00 1010 N/m2, what strain (ΔL/L) results in a transverse wave speed of 118 m/s

Answer 1

`(\Delta L)/(L) =5.37* 10^(-4)`

Step-by-step explanation:

Given:

• cross sectional area of the wire,
  `A=5.75* 10^(-6)\ m^2`

• density of aluminium wire,
  `\rho=2.7* 10^3\ kg.m^(-3)`

• young's modulus of the material,
  `E=7* 10^(10)\ N.m^(-2)`

• wave speed,
  `v=118\ m.s^(-1)`

We have mathematical expression for strain as:

`(\Delta L)/(L) =(\sigma)/(E)` ...............................(1)

and since,
`\sigma =(T)/(A)`

where, T = tension force in the wire

equation (1) becomes:

`(\Delta L)/(L) =(T)/(A.E)` ............................(2)

Also velocity ofwave in tensed wire:

`v=\sqrt{(T)/(\mu) }` ...................................(3)

where:
`\mu=` linear mass density of the wire

`\therefore \mu=\rho * A`

Now, equation (3) becomes

`v=\sqrt{(T)/(\rho * A) }`

`T=v^2.\rho * A` ............................(4)

Using eq. (2) & (4) for tension T

`v^2.\rho * A=A.E* (\Delta L)/(L)`

`(\Delta L)/(L) =(v^2.\rho)/(E)`

putting the respective values

`(\Delta L)/(L) =(118^2* 2.7* 10^3)/(7* 10^(10))`

`(\Delta L)/(L) =5.37* 10^(-4)`