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

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$2 : 1$

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(D) The fundamental frequency of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.Since $\mu$ remains constant,$n \propto \frac{1}{L} \sqrt{T}$.Let the initial length be $L_1$ and initial tension be $T_1$. The final length $L_2 = L_1 - 0.40L_1 = 0.60L_1$ and final tension $T_2 = T_1 + 0.44T_1 = 1.44T_1$.The ratio of the final frequency $n_2$ to the initial frequency $n_1$ is given by $\frac{n_2}{n_1} = \frac{L_1}{L_2} \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{n_2}{n_1} = \frac{L_1}{0.60L_1} \sqrt{\frac{1.44T_1}{T_1}} = \frac{1}{0.6} 	imes \sqrt{1.44} = \frac{1}{0.6} 	imes 1.2 = 2$.Thus,the ratio is $2 : 1$.

If the length of a stretched string is shortened by $40\%$ and the tension is increased by $44\%$,then the ratio of the final and initial fundamental frequencies is

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