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

Vedclass · Accepted Answer

$\frac{9}{4}$

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

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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