Physics - Wave on String Question with Solution | TestHub

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 $\begin{array}{rl}\mathrm{n}& =\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ \therefore {\mathrm{n}}_1& =\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\\ {\mathrm{n}}_2& =\frac{1}{2l}\sqrt{\frac{{\mathrm{T}}_{\mathrm{new}}}{\mathrm{m}}}\\ \therefore \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}& =\frac{l\text{ aew }}{l}\times \frac{\sqrt{\mathrm{T}}}{\sqrt{{\mathrm{T}}_{\mathrm{new}}}}\\ & =\frac{\frac{3}{5}l}{l}\times \sqrt{\frac{100\mathrm{T}}{144\mathrm{T}}}\\ & =\frac{3}{5}\times \frac{10}{12}=\frac{1}{2}\\ \therefore \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}& =\frac{2}{1}\end{array}$