Physics - Wave on String Question with Solution | TestHub

${\mathrm{C}}_1=\sqrt{\frac{\mathrm{T}}{\mu }},{\mathrm{C}}_2=\sqrt{\frac{\mathrm{T}}{4\mu }}=\frac{{\mathrm{C}}_1}{2}$ $\underline{\text{ For node at }\mathrm{O}\text{ : }}$ $\mathrm{L}=\frac{\mathrm{n}{\lambda }_1}{2},2\mathrm{L}=\frac{\mathrm{m}{\lambda }_2}{2}$ ( $\mathrm{n},\mathrm{m}$ are integers) $\begin{array}{rl}& {\lambda }_1=\frac{2\mathrm{L}}{\mathrm{n}},{\lambda }_2=\frac{4\mathrm{L}}{\mathrm{m}}\\ & \frac{{\mathrm{C}}_1}{{\lambda }_1}=\frac{{\mathrm{C}}_2}{{\lambda }_2}\\ & \Rightarrow \frac{{\mathrm{C}}_1}{\frac{2\mathrm{L}}{n}}=\frac{\frac{{\mathrm{C}}_1}{2}}{\frac{4\mathrm{L}}{m}}\\ & \Rightarrow 4\mathrm{n}=\mathrm{m}\end{array}$ For minimum frequency, $\mathrm{n}=1,\mathrm{m}=4$ $\therefore v_{\min }=\frac{{\mathrm{C}}_1\times 1}{2\mathrm{L}}=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{T}}{\mu }}=v_0$ The string will look likeTotal no. of nodes $=6$ including the end nodes $\underline{\text{ For antinode at }\mathrm{O}\text{ : }}$ $\begin{array}{rl}& \mathrm{L}=(2\mathrm{n}+1)\frac{{\lambda }_1}{4};2\mathrm{L}=(2\mathrm{n}+1)\frac{{\lambda }_2}{4}\text{ (n, m are integers) }\\ & {\lambda }_1=\frac{4\mathrm{L}}{(2\mathrm{n}+1)};{\lambda }_2=\frac{8\mathrm{L}}{(2\mathrm{m}+1)}\\ & \frac{{\mathrm{C}}_1}{{\lambda }_1}=\frac{{\mathrm{C}}_2}{{\lambda }_2}\\ & \frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}=\frac{{\lambda }_1}{{\lambda }_2}\\ & 2=\frac{\frac{4\mathrm{L}}{(2\mathrm{n}+1)}}{\frac{8\mathrm{L}}{(2\mathrm{m}+1)}}\\ & 4=\frac{(2\mathrm{m}+1)}{(2\mathrm{n}+1)}\Rightarrow \text{ even }=\frac{\text{ odd }}{\text{ odd }}\Rightarrow \text{ This node is not possible }\end{array}$