A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? Which pulse arrives first?(Ignore the effect of the weight of the wires on the tension in the wires).

Support of Beam \(A = \frac{2}{3} \times 1750\;N = 1166.6\;N\)

Support of Beam \(A = \frac{1}{3} \times 1750\;N = 583.3\;N\).

Mass of each wire \(m = \frac{{0.260}}{{9.8}} = 0.02653\;kg\)

Linear density \(U = \frac{{0.02653}}{{1.20}} = 0.0221\;{\rm{kg/m}}\)

The speed of transverse waves on a string depends on the tension and mass per unit length. It is given by:

\(V = \sqrt {\frac{T}{U}} \)

Here, \(T\) is tension, \(U\) is linear density.

The speed of beam A is:

\(\begin{array}{c}{V_A} = \sqrt {\frac{T}{U}} \\ = \sqrt {\frac{{1166.6}}{{0.0221}}} \\ = 229.75\;{\rm{m/s}}\end{array}\)

The speed of beam B is:

\(\begin{array}{c}{V_B} = \sqrt {\frac{T}{U}} \\ = \sqrt {\frac{{583.3}}{{0.0221}}} \\ = 162.44\;{\rm{m/s}}\end{array}\)

Now, time for beam A is:

\(\begin{array}{c}{t_A} = \frac{d}{{{V_A}}}\\ = \frac{{1.2}}{{229.75}}\\ = 0.00522306\;{\rm{s}}\end{array}\)

Time for beam B is:

\(\begin{array}{c}{t_B} = \frac{d}{{{V_B}}}\\ = \frac{{1.2}}{{162.44}}\\ = 0.00738734\;{\rm{s}}\end{array}\)

Thus, the time delay is calculated as:

\(\begin{array}{c}t = {t_B} - {t_A}\\ = 0.00738734\;{\rm{s}}\; - {\rm{0}}{\rm{.00522306}}\;{\rm{s}}\\ = 0.00216034\;s\end{array}\)

The pulse A arrives first.

Hence, the time delay is \(0.00216034\;s\) and the pulse A arrives first.