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(A) Let the initial length and tension be $l$ and $T$ respectively. The mass per unit length $m$ remains constant.The fundamental frequency is given b

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$2:1$

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(A) Let the initial length and tension be $l$ and $T$ respectively. The mass per unit length $m$ remains constant.The fundamental frequency is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.After the changes:New length $l' = l - 0.40l = 0.6l = \frac{3}{5}l$.New tension $T' = T + 0.44T = 1.44T = \frac{144}{100}T$.The new frequency $n'$ is given by $n' = \frac{1}{2l'} \sqrt{\frac{T'}{m}}$.Taking the ratio of final to initial frequency:$\frac{n'}{n} = \frac{l}{l'} \sqrt{\frac{T'}{T}} = \frac{l}{0.6l} \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} 	imes \sqrt{1.44} = \frac{1}{0.6} 	imes 1.2 = \frac{1.2}{0.6} = 2$.Therefore,the ratio of final to initial frequency is $2:1$.

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

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