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

The time required for the impluse passing through each other is approximately 0.18seconds

Explanation:

Given:

Length,L = 50m

M/L = 0.020kg/m

FA = 5.7×10^2N

FB = 2.5×10^2N

The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.

Ca(t) + CB(t) = 50

Where CA and CB are the velocities of the wire A and B

t = 50/ (CA + CB)

But C = Sqrt(FL/M)

Substituting gives:

t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))

t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))

t = 50 / (168.62 + 111.83)

t = 50/280.15

t = 0.18 seconds

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

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be