Two wires are parallel and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m

Question

3 years ago

11

Two wires are parallel and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m

. However, the tension in the wire A is 5.70 102 N, and the tension in the wire B is 2.50 102 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

Physics

2 answers:

Grace [21] · Answer 1 · 2022-07-05T16:25:18+03:00

Answer:

t = the time required for the waves to pass each other = 0.178s

Explanation:

For Wire A: Ta = 5.70×10² N, μa = μb = 0.02kg/m, La = 50m

For Wire B: Tb = 2.50×10² N, μa = μb = 0.02kg/m, Lb = 50m

Let the time required to elapse before both waves pass each other be t.

First let us calculate the speed of the wave in both wires.

V = :√(T/μ)

Wire A: Va = √(Ta/μa) = √(5.70×10²/0.02)

Va = 169m/s

Wire B: Vb = √(Tb/μb) = √(2.50×10²/0.02)

Va = 112m/s

At time t has elapsed, the wave in Wire A would have traveled a distance

xa = Va ×t = 169t

And the wave in Wire B would have traveled a distance

xb = Vb ×t = 112t

Since both waves are traveling along equal lengths but in opposite directions. At time t the wave in A has traveled a distance xa. At this same time the wave in B has traveled a distance xa in the opposite direction. This distance xb is equal to the distance wave A still has to travel to the other end of the of wire A. This distance is equal to 50 – xa. So

xb = 50 – xb

112t = 50 – 169t

112t + 169t = 50

281t = 50

t = 50/281= 0.178s

t = the time required for the waves to pass each other = 0.178s