Answer
The time delay is 2.33 ms. The pulse that travels along wire A arrives first.
Work Step by Step
The magnitude of the torque from the tension in each string about the center of mass must be equal.
$\tau_A = \tau_B$
$T_A~\frac{L}{3}= T_B~\frac{2L}{3}$
$T_A = 2~T_B$
The sum of the tension in each wire must be equal to the weight of the beam.
$T_A+T_B = 1750~N$
$2~T_B+T_B = 1750~N$
$T_B = 583.3~N$
We can find the tension $T_A$.
$T_A = 2~T_B = (2)(583.3~N) = 1166.7~N$
We can find the mass density in each wire.
$\mu = \frac{m}{L}$
$\mu = \frac{0.290~N}{(9.80~m/s^2)(1.25~m)}$
$\mu = 0.02367~kg/m$
We can find the time $t_A$ it takes the pulse to reach the ceiling along wire A.
$t_A = \frac{L}{v_A}$
$t_A = \frac{L}{\sqrt{\frac{T_A}{\mu}}}$
$t_A = \frac{1.25~m}{\sqrt{\frac{1166.7~N}{0.02367~kg/m}}}$
$t_A = 0.00563~s = 5.63~ms$
We can find the time $t_B$ it takes the pulse to reach the ceiling along wire B.
$t_B = \frac{L}{v_B}$
$t_B = \frac{L}{\sqrt{\frac{T_B}{\mu}}}$
$t_B = \frac{1.25~m}{\sqrt{\frac{583.3~N}{0.02367~kg/m}}}$
$t_B = 0.00796~s = 7.96~ms$
We can find the time delay.
$\Delta t = t_B-t_A$
$\Delta t = 7.96~ms-5.63~ms$
$\Delta t = 2.33~ms$
The time delay is 2.33 ms. The pulse that travels along wire A arrives first.
