Question

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 122 N. During what time interval will a transverse wave travel the entire length of the two wires

Answer 1

Answer:

The time taken is  $t = 0.356 \ s$

Explanation:

From the question we are told that

  The length of steel the wire is  $l_1 = 31.0 \ m$

   The  length of the  copper wire is  $l_2 = 17.0 \ m$

    The  diameter of the wire is  $d = 1.00 \ m = 1.0 *10^{-3} \ m$

     The  tension is  $T = 122 \ N$

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              $t = t_s + t_c$

Where  $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

         $t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here  $\rho_s$ is the density of steel with a value  $\rho_s = 8920 \ kg/m^3$

   So

      $t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

      $t_s = 0.235 \ s$

 And

        $t_c$ is the time taken to transverse the copper wire which is mathematically represented as

      $t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here  $\rho_c$ is the density of steel with a value  $\rho_s = 7860 \ kg/m^3$

 So

      $t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

      $t_c =0.121$

So  

   $t = t_c + t_s$

    $t = 0.121 + 0.235$

    $t = 0.356 \ s$