A string vibrates with a frequency of $200 \mathrm{~Hz}$. When its length is doubled and tension is altered, it begins to vibrate with a frequency of $300 \mathrm{~Hz}$. The ratio of the new tension to the original tension is

The fundamental frequency of a vibrating string is given by $f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Let the initial frequency be $f_1 = 200 \text{ Hz}$, length be $L_1 = L$, and tension be $T_1$. Then $f_1 = \dfrac{1}{2L} \sqrt{\dfrac{T_1}{\mu}}$.

Let the new frequency be $f_2 = 300 \text{ Hz}$, new length be $L_2 = 2L$, and new tension be $T_2$. Then $f_2 = \dfrac{1}{2(2L)} \sqrt{\dfrac{T_2}{\mu}} = \dfrac{1}{4L} \sqrt{\dfrac{T_2}{\mu}}$.

Taking the ratio of the two frequencies:

$\dfrac{f_2}{f_1} = \dfrac{\dfrac{1}{4L} \sqrt{\dfrac{T_2}{\mu}}}{\dfrac{1}{2L} \sqrt{\dfrac{T_1}{\mu}}} = \dfrac{1}{2} \sqrt{\dfrac{T_2}{T_1}}$.

Substituting the given values:

$\dfrac{300}{200} = \dfrac{1}{2} \sqrt{\dfrac{T_2}{T_1}} \Rightarrow \dfrac{3}{2} = \dfrac{1}{2} \sqrt{\dfrac{T_2}{T_1}}$.

Multiplying both sides by 2:

$3 = \sqrt{\dfrac{T_2}{T_1}}$.

Squaring both sides:

$9 = \dfrac{T_2}{T_1}$.

Therefore, the ratio of the new tension to the original tension is $9:1$.

Answer: 9 : 1