If the length of stretched string is reduced by $40 \%$ and tension is increased by $44 \%$ then the ratio…
If the length of stretched string is reduced by $40 \%$ and tension is increased by $44 \%$ then the ratio of final to initial frequencies of stretched string is
$2: 1$
$3:2$
$3:4$
$1:3$
Solution
Let the initial length and tension be $l$ and $\mathrm{T}$ respectively.
After shortening,
The new length $l_{\text {new }}=l-\frac{40}{100} l=\frac{3}{5} l$
After increase in tension, the new tension $\mathrm{T}_{\text {new }}=\mathrm{T}+\frac{44}{100} \mathrm{~T}=\frac{144 \mathrm{~T}}{100}$
Fundamental frequency of a vibrating string is given by
$\begin{aligned}
\mathrm{n} & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
\therefore \quad \mathrm{n}_1 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
\mathrm{n}_2 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{\mathrm{new}}}{\mathrm{m}}} \\
\therefore \quad \frac{\mathrm{n}_1}{\mathrm{n}_2} & =\frac{l \text { aew }}{l} \times \frac{\sqrt{\mathrm{T}}}{\sqrt{\mathrm{T}_{\mathrm{new}}}} \\
& =\frac{\frac{3}{5} l}{l} \times \sqrt{\frac{100 \mathrm{~T}}{144 \mathrm{~T}}} \\
& =\frac{3}{5} \times \frac{10}{12}=\frac{1}{2} \\
\therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1} & =\frac{2}{1}
\end{aligned}$