As depicted in the figure, both strings are composed of the same...

Let the tension in the string $AB$ be $T_1$ and the tension in the string $CD$ be $T_2$.

Since the pulley $B$ is light (massless), the net force acting on it must be zero. The string $AB$ passes around pulley $B$, so there are two upward segments of this string supporting the pulley, each exerting a force $T_1$. The string $CD$ pulls downwards on the axle of the pulley with tension $T_2$.

Balancing the forces on pulley $B$, we get:
$2T_1 = T_2 \Rightarrow \dfrac{T_1}{T_2} = \dfrac{1}{2}$

The speed of a transverse wave in a string is given by $v = \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.

Since both strings are made of the same material and have the same cross-sectional area, they have the same linear mass density $\mu$.

The ratio of the transverse wave speeds is:
$\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1 / \mu}{T_2 / \mu}} = \sqrt{\dfrac{T_1}{T_2}}$

Substituting the ratio of tensions:
$\dfrac{v_1}{v_2} = \sqrt{\dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}$