Question

A steel wire of length 26.0 m and a copper wire of length 22.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 170 N. During what time interval will a transverse wave travel the entire length of the two wires?

Answer 1

0.298s

Step-by-step explanation:

Speed of wave in steel wire is

`V_z = \sqrt{(T)/(m/L) }`

`V_z = \sqrt{(T)/( [(p(Al))/(l)] ) }`

`V_z = \sqrt{(T)/(p\pi ((d)/(2))^2 ) }`

`V_z = \sqrt{(170)/((7860)\pi ((1* 10^-^3 )/(2))^2 ) } \\\\= 166m/s`

Speed of wave in copper wire is

`V_z = \sqrt{(170)/((8920)\pi ((1* 10^-^3 )/(2))^2 ) } \\\\= 156m/s`

The time required is

`t = t_1 + t_2\\t =(l_z)/(V_z) + (l_z)/(V_z)`

t =
`(26)/(166) + (22)/(155.77) \\\\= 0.298s`

Answer 2

0.3189 s

Step-by-step explanation:

The formula for calculating the speed of the wave in steel wire is:

`v_s = \sqrt{(T)/(m/l)}`

`v_s = \sqrt{(T)/((\rho(Al))/(l) ) }`

`v_s = \sqrt{(T)/((\rho \pi (d))/(2)^2 ) }`

`v_s = \sqrt{(170N)/(((7860 kg/m^3 \pi (1.00*10^(-3)))/(2)^2 ) }`

`v_s =147 m/s`

The time required is:

`t = t_1 + t_2`

=
`(I_s)/(v_s) + (I_c)/(v_c)`

=
`(26.0m)/(147 m/s) + (22.0m)/(154.9m/s)`

= 0.3189 s