Answer: 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
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
