$C _1=\sqrt{\frac{ T }{\mu}}, C _2=\sqrt{\frac{ T }{4 \mu}}=\frac{ C _1}{2}$

For node at $O$ :

$L =\frac{ n \lambda_1}{2}, 2 L =\frac{ m \lambda_2}{2} \text { (n, } m \text { are integers) }$

$\lambda_1=\frac{2 L }{ n }, \lambda_2=\frac{4 L }{ m }$

$\frac{ C _1}{\lambda_1}=\frac{ C _2}{\lambda_2}$

$\Rightarrow \frac{ C _1}{\frac{2 L }{ n }}=\frac{\frac{ C _1}{2}}{\frac{4 L }{ m }}$

$\Rightarrow 4 n = m$

For minimum frequency, $n =1, m =4$

$\therefore v_{\min }=\frac{ C _1 \times 1}{2 L }=\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}=v_0$

The string will look like

(image)

Total no. of nodes $=6$ including the end nodes

For antinode at $O$ :

$L =(2 n +1) \frac{\lambda_1}{4} ; 2 L =(2 n +1) \frac{\lambda_2}{4} \quad \text { (n, } m \text { are integers) }$

$\lambda_1=\frac{4 L }{(2 n +1)} ; \lambda_2=\frac{8 L }{(2 m +1)}$

$\frac{ C _1}{\lambda_1}=\frac{ C _2}{\lambda_2}$

$\frac{ C _1}{ C _2}=\frac{\lambda_1}{\lambda_2}$

$2=\frac{\frac{4 L }{(2 n +1)}}{\frac{8 L }{(2 m +1)}}$

$4=\frac{(2 m +1)}{(2 n +1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }$