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(D) The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,a

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$2.25$

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(D) The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.From this formula,we have $n \propto \frac{\sqrt{T}}{l}$.Let the initial frequency be $n_1$ and the final frequency be $n_2 = 2n_1$.Let the initial length be $l_1$ and the final length be $l_2 = \frac{3}{4}l_1$.Using the proportionality $n \propto \frac{\sqrt{T}}{l}$,we can write $\frac{n_2}{n_1} = \frac{\sqrt{T_2}/l_2}{\sqrt{T_1}/l_1} = \sqrt{\frac{T_2}{T_1}} \cdot \frac{l_1}{l_2}$.Substituting the given values: $2 = \sqrt{\frac{T_2}{T_1}} \cdot \frac{l_1}{(3/4)l_1} = \sqrt{\frac{T_2}{T_1}} \cdot \frac{4}{3}$.Squaring both sides: $4 = \frac{T_2}{T_1} \cdot \frac{16}{9}$.Therefore,$\frac{T_2}{T_1} = 4 \cdot \frac{9}{16} = \frac{9}{4} = 2.25$.

In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is:

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