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

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

• cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

• density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

• young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

• wave speed, $v=118\ m.s^{-1}$

We have mathematical expression for strain as:

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

Also velocity ofwave in tensed wire:

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

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

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

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

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$