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(A) The fundamental frequency 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 $

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$2: 1$

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(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.Initial frequency $n_1 = \frac{1}{2l} \sqrt{\frac{T}{m}}$.New length $l' = l - 0.40l = 0.6l$.New tension $T' = T + 0.44T = 1.44T$.New frequency $n_2 = \frac{1}{2l'} \sqrt{\frac{T'}{m}}$.Taking the ratio of the final to initial frequency: $\frac{n_2}{n_1} = \frac{l}{l'} \sqrt{\frac{T'}{T}}$.Substituting the values: $\frac{n_2}{n_1} = \frac{l}{0.6l} \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} 	imes \sqrt{1.44} = \frac{1.2}{0.6} = 2$.Therefore,the ratio $n_2 : n_1 = 2 : 1$.

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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