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24 April, 21:04

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

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  1. 24 April, 21:27
    0
    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
  2. 25 April, 00:38
    0
    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
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