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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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$2: 1$
Fundamental frequency in case of string is
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n \propto \frac{\sqrt{T}}{1} \Rightarrow \frac{n^{\prime}}{n}=\sqrt{\frac{T^{\prime}}{T}} \times \frac{1}{r}$
putting $T^{\prime}=T+0.44 T=\frac{144}{100} T$ and $\left.l^{\prime}=|-0.4|=\frac{3}{5} \right\rvert\,$
We get $\frac{n^{\prime}}{n}=\frac{2}{1}$.
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n \propto \frac{\sqrt{T}}{1} \Rightarrow \frac{n^{\prime}}{n}=\sqrt{\frac{T^{\prime}}{T}} \times \frac{1}{r}$
putting $T^{\prime}=T+0.44 T=\frac{144}{100} T$ and $\left.l^{\prime}=|-0.4|=\frac{3}{5} \right\rvert\,$
We get $\frac{n^{\prime}}{n}=\frac{2}{1}$.
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