Answer:
The time taken is [tex]t = 0.356 \ s[/tex]
Explanation:
From the question we are told that
The length of steel the wire is [tex]l_1 = 31.0 \ m[/tex]
The length of the copper wire is [tex]l_2 = 17.0 \ m[/tex]
The diameter of the wire is [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]
The tension is [tex]T = 122 \ N[/tex]
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
[tex]t = t_s + t_c[/tex]
Where [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as
[tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_s[/tex] is the density of steel with a value [tex]\rho_s = 8920 \ kg/m^3[/tex]
So
[tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_s = 0.235 \ s[/tex]
And
[tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as
[tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_c[/tex] is the density of steel with a value [tex]\rho_s = 7860 \ kg/m^3[/tex]
So
[tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_c =0.121[/tex]
So
[tex]t = t_c + t_s[/tex]
[tex]t = 0.121 + 0.235[/tex]
[tex]t = 0.356 \ s[/tex]
