A sonometer string vibrates with a frequency of 400 Hz . When the length of the string is halved and the tension is altered, it begins to vibrate with a frequency of 200 Hz . The ratio of the new tension to the original tension in the string is:

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

Let the initial frequency be $f_1 = 400$ Hz, length be $L_1 = L$, and tension be $T_1 = T$. Then $f_1 = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$.

Let the final frequency be $f_2 = 200$ Hz, length be $L_2 = L/2$, and tension be $T_2 = T'$. Then $f_2 = \dfrac{1}{2(L/2)} \sqrt{\dfrac{T'}{\mu}} = \dfrac{1}{L} \sqrt{\dfrac{T'}{\mu}}$.

Taking the ratio of the two frequencies:

$\dfrac{f_2}{f_1} = \dfrac{\dfrac{1}{L} \sqrt{\dfrac{T'}{\mu}}}{\dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}} = 2 \sqrt{\dfrac{T'}{T}}$.

Substituting the given values:

$\dfrac{200}{400} = 2 \sqrt{\dfrac{T'}{T}} \Rightarrow \dfrac{1}{2} = 2 \sqrt{\dfrac{T'}{T}} \Rightarrow \sqrt{\dfrac{T'}{T}} = \dfrac{1}{4}$.

Squaring both sides:

$\dfrac{T'}{T} = \left( \dfrac{1}{4} \right)^2 = \dfrac{1}{16}$.

Thus, the ratio of the new tension to the original tension is $1:16$.

Answer: $1: 16$