Question

Two wires are parallel, and one is directly above the other. Each has a length of 44.3 m and a mass per unit length of 0.0292 kg/m. However, the tension in wire A is 33.0 × 102 N, while the tension in wire B is 3.76 × 102 N. Transverse wave pulses are generated simultaneously, one at the left end of wire A and one at the right end of wire B. The pulses travel toward each other. How much time does it take until the pulses pass each other?

Answer 1

Answer:

0.09852 seconds

Explanation:

$\mu$ = Linear density = 0.0292 kg/m

$T_A$ = Tesnsion in string A = $33\times 10^2\ N$

$T_B$ = Tesnsion in string B = $3.76\times 10^2\ N$

t = Time taken till the pulses pass each other

Velocity of wave in a string is given by

$v_A=\sqrt{\dfrac{T_A}{\mu}}\\\Rightarrow v_A=\sqrt{\dfrac{33\times 10^2}{0.0292}}$

$v_B=\sqrt{\dfrac{T_2}{\mu}}\\\Rightarrow v_B=\sqrt{\dfrac{3.76\times 10^2}{0.0292}}$

$Distance=Speed\times Time$

$L=(v_A+v_B)t\\\Rightarrow t=\dfrac{L}{v_A+v_B}\\\Rightarrow t=\dfrac{44.3}{\sqrt{\dfrac{33\times 10^2}{0.0292}}+\sqrt{\dfrac{3.76\times 10^2}{0.0292}}}\\\Rightarrow t=0.09852\ s$

The time taken is 0.09852 seconds

Answer 2

Answer:

$t=0.0985\ s$

Explanation:

Given:

• length of each wire, $l_a=l_b=44.3\ m$

• linear mass density of each wire, $\mu_a=\mu_b=0.0292\ kg.m^{-1}$

• tension in wire A, $T_a=3300\ N$

• tension in wire B, $T_a=376\ N$

Now, the velocity of the wave pulse in the stretched spring is given as:

FOR A:

$v_a=\sqrt{\frac{T_a}{\mu_a} }$

$v_a=\sqrt{\frac{3300}{0.0292} }$

$v_a=336.175\ m.s^{-1}$

FOR B:

$v_b=\sqrt{\frac{T_b}{\mu_b} }$

$v_b=\sqrt{\frac{376}{0.0292} }$

$v_b=113.476\ m.s^{-1}$

Let the distance at which the wave of A meets the wave of B be x from left then the distance from B is (44.3-x) meters.

Now the time taken is constant:

$t=\frac{x}{v_a} =\frac{x-44.3}{v_b}$

$\frac{x}{336.175} =\frac{x-44.3}{113.476}$

$x=33.12\ m$ is the distance travelled by pulse in wire A in when the pulse in the wire B meets

Now the time taken to travel this distance by pulse in wire A:

$t=\frac{x}{v_a}$

$t=\frac{33.12}{336.175}$

$t=0.0985\ s$ is the time taken by the two waves to pass each other.