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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

Respuesta :

Answer:

[tex]\frac{\Delta L}{L} =5.37\times 10^{-4}[/tex]

Explanation:

Given:

  • cross sectional area of the wire, [tex]A=5.75\times 10^{-6}\ m^2[/tex]
  • density of aluminium wire, [tex]\rho=2.7\times 10^3\ kg.m^{-3}[/tex]
  • young's modulus of the material, [tex]E=7\times 10^{10}\ N.m^{-2}[/tex]
  • wave speed, [tex]v=118\ m.s^{-1}[/tex]

We have mathematical expression for strain as:

[tex]\frac{\Delta L}{L} =\frac{\sigma}{E}[/tex] ...............................(1)

and since, [tex]\sigma =\frac{T}{A}[/tex]

where, T = tension force in the wire

equation (1) becomes:

[tex]\frac{\Delta L}{L} =\frac{T}{A.E}[/tex] ............................(2)

Also velocity ofwave in tensed wire:

[tex]v=\sqrt{\frac{T}{\mu} }[/tex] ...................................(3)

where: [tex]\mu=[/tex] linear mass density of the wire

[tex]\therefore \mu=\rho \times A[/tex]

Now, equation (3) becomes

[tex]v=\sqrt{\frac{T}{\rho \times A} }[/tex]

[tex]T=v^2.\rho \times A[/tex] ............................(4)

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

[tex]v^2.\rho \times A=A.E\times \frac{\Delta L}{L}[/tex]

[tex]\frac{\Delta L}{L} =\frac{v^2.\rho}{E}[/tex]

putting the respective values

[tex]\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}[/tex]

[tex]\frac{\Delta L}{L} =5.37\times 10^{-4}[/tex]

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