Two uniform strings of mass per unit length \(\mu\) and \(4 \mu\), and length \(L\) and \(2 L\), respectively, are joined at point \(\mathrm{O}\), and tied at two fixed ends \(\mathrm{P}\) and \(\mathrm{Q}\), as shown in the figure. The strings are under a uniform tension \(T\). If we define the frequency \(v_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\), which of the following statement(s) is(are) correct?

\(\mathrm{C}_1=\sqrt{\frac{\mathrm{T}}{\mu}}, \mathrm{C}_2=\sqrt{\frac{\mathrm{T}}{4 \mu}}=\frac{\mathrm{C}_1}{2}\)
\(\underline{\text { For node at } \mathrm{O} \text { : }}\)
\(\mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}\) ( \(\mathrm{n}, \mathrm{m}\) are integers)
\(\begin{aligned} & \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{n}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{4 \mathrm{~L}}{m}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m}\end{aligned}\)
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 like

Total no. of nodes \(=6\) including the end nodes \(\underline{\text { For antinode at } \mathrm{O} \text { : }}\)
\(\mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad\) \(\text { (n, m are integers) } \)
\( \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \)
\( \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \)
\( \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \)
\( 2=\frac{\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)}}{\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)}} \)
\( 4=\frac{(2 \mathrm{~m}+1)}{(2 \mathrm{n}+1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow\) \(\text { This node is not possible }\)