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madreJ [45]
3 years ago
11

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?
Physics
2 answers:
Elodia [21]3 years ago
7 0

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

<u>Now, the velocity of the wave pulse in the stretched spring is given as:</u>

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

<u>Now the time taken to travel this distance by pulse in wire A:</u>

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.

Virty [35]3 years ago
4 0

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

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