91影视

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.