Answer:
The time taken is 
Explanation:
From the question we are told that
The length of steel the wire is 
The length of the copper wire is 
The diameter of the wire is 
The tension is 
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
Where 
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} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%20l_1%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here 
is the density of steel with a value 
So
![t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%2031%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B8920%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)
And
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} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%20l_2%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho_c%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here 
is the density of steel with a value 
So
![t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%2017%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B7860%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)
So
<span>the transverse wve that requires a medium to move is : sound waves
in order to travel, sound waves need to jumped on between one particle to another, no matter it's liquid, solid , and gases. Which explains why you wont hear a thing in outer space, simply because the sound waves got no particle to jumped into</span>
The frequency of sound waves goes up, so the pitch goes up as well. More frequent=higher pitched
Answer:
Explanation:
The solved solution is on the attach document
