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(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.From this formula,we observe that $n \propto \fra

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$2:1$

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(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.From this formula,we observe that $n \propto \frac{\sqrt{T}}{l}$.Therefore,the ratio of the final frequency $n'$ to the initial frequency $n$ is $\frac{n'}{n} = \sqrt{\frac{T'}{T}} 	imes \frac{l}{l'}$.Given that the length is shortened by $40\%$,the new length is $l' = l - 0.4l = 0.6l = \frac{3}{5}l$.Given that the tension is increased by $44\%$,the new tension is $T' = T + 0.44T = 1.44T = \frac{144}{100}T$.Substituting these values into the ratio formula:$\frac{n'}{n} = \sqrt{\frac{1.44T}{T}} 	imes \frac{l}{0.6l} = \sqrt{1.44} 	imes \frac{1}{0.6} = 1.2 	imes \frac{1}{0.6} = 2$.Thus,the ratio of the final to the initial fundamental frequency 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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