To solve this problem, we need to understand the relationship between the tension in a string and the speed of a transverse wave along the string.$\\$The speed $v$ of a transverse wave on a stretched string is given by the formula:$\\ v = \sqrt{\frac{T}{\mu}} \\$where:$\\$• $T$ is the tension in the string.$\\$• $\mu$ is the linear mass density of the string.$\\$Initially, let the tension be $T_1$ and the speed be $v_1. \\ v_1 = \sqrt{\frac{T_1}{\mu}} \\$When the tension is doubled, the new tension is $T_2 = 2T_1. \\$Let the new speed be $v_2. \\ v_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{2T_1}{\mu}} \\$We can express $v_2$ in terms of $v_1$ as follows:$\\ v_2 = \sqrt{2} \cdot \sqrt{\frac{T_1}{\mu}} = \sqrt{2} \cdot v_1 \\$Therefore, the ratio of the initial and final speeds is:$\\ \frac{v_1}{v_2} = \frac{v_1}{\sqrt{2} \cdot v_1} = \frac{1}{\sqrt{2}} \\$Thus, the ratio of the initial and final speeds of the transverse wave is $1 : \sqrt{2}. \\$This corresponds to Option 3.