13.058. In the laboratory, the speed at which sound propagates along rods made of different materials is measured. In the first experiment, it turned out that the entire path, consisting of three sequentially connected rods, is traveled by sound in $a$ s, and the path consisting of the second and third rods is traveled by sound twice as fast as the first rod alone. In another experiment, the second rod was replaced with a new one, and then the sequential connection of three rods was traveled by sound in $b$ s, and the connection of the first and second rods was twice as slow as the third rod alone. Find the speed of sound propagation in the new rod, if its length is $l$ m.
Solution.

Let $t_{1}, t_{2}, t_{3}, t_{4}$ be the times it takes for sound to travel along the 1st, 2nd, 3rd, and new rods, respectively. The fourth rod is the new one that we use to replace the second rod in the second experiment. According to the problem, in the first experiment:

$$
\left\{\begin{array}{l}
t_{1}+t_{2}+t_{3}=a, \\
2\left(t_{2}+t_{3}\right)=t_{1},
\end{array} \Rightarrow 2 t_{1}=2 a-t_{1}, t_{1}=\frac{2}{3} a\right.
$$

in the second experiment $\left\{\begin{array}{l}t_{1}+t_{4}+t_{3}=b, \\ t_{1}+t_{4}=2 t_{3},\end{array} \Rightarrow 3 t_{1}+3 t_{4}=2 b\right.$;

$t_{4}=\frac{2}{3} b-t_{1}=\frac{2}{3}(b-a)$.

Then the speed of sound in the new rod $\mathrm{v}=\frac{l}{t_{4}}=\frac{3 l}{2(b-a)}$ m/s; where $b>a$.

Answer: $\frac{3 l}{2(b-a)}$ m/s; makes sense when $b>a$.