Physics - Sound Wave Question with Solution | TestHub

The fundamental frequency of open tube is ${\mathrm{n}}_1=\frac{\mathrm{v}}{2{\ell }_1}$ When tube is dipped in water, one-fourth of it is in water and three-fourth is in air. Hence, it becomes a tube closed at one end with length ${\ell }_2=\frac{3}{4}{\ell }_1$ The fundamental frequency of closed tube is $\begin{array}{rl}& {\mathrm{n}}_2=\frac{\mathrm{v}}{4{\ell }_2}\\ & \therefore \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{1}{4{\ell }_2}\times 2{\ell }_1=\frac{{\ell }_1}{2{\ell }_2}=\frac{4}{2\times 3}\left[\because \frac{{\ell }_1}{{\ell }_2}=\frac{4}{3}\right]\\ & \therefore {\mathrm{n}}_2=\frac{2}{3}{\mathrm{n}}_1=\frac{2}{3}\mathrm{n}\end{array}$