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]
