A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end as shown in figure. Mass per unit length of the strings are $\mu$ and 9$\mu$  and the tension is T.
Choose the correct statement (s) :

When A is a node:
Suppose  $n_1$  and $n_2$   are the complete loops formed on left and right side of point A. Then  $f_1=f_2$
Or   $n_1\left(\frac{v_1}{2L}\right)=n_2\left(\frac{v_2}{2L}\right)$
or    $\frac{n_1}{n_2}\left(\frac{v_2}{v_1}\right)=\sqrt{\frac{{\mu }_1}{{\mu }_2}}=\frac{1}{3},\frac{2}{6},\frac{3}{9},\mathrm{........},etc$            $(asv\propto \frac{1}{\sqrt{\mu }})$
  Possible frequencies are,
$\frac{v_1}{2L},2\left(\frac{v_1}{2L}\right),\frac{3v_1}{2L},\mathrm{.....},etc$                  $(v_1=\sqrt{\frac{T}{\mu }})$        
Or  $\frac{1}{2L}\sqrt{\frac{T}{\mu }},\frac{1}{L}\sqrt{\frac{T}{\mu }},\frac{3}{2L}\sqrt{\frac{T}{2\mu }},\mathrm{........},etc$
Where A is an antinode:
Suppose  $n_1$ and $n_2$  are complete loops on left and right side of point A.
 $$\begin{gathered} n_1\frac{{\lambda }_1}{2}+\frac{{\lambda }_1}{4}=L\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{f}_1=\frac{v_1}{L}(\frac{n_1}{2}+\frac{1}{4}) \\ {n}_2\frac{{\lambda }_2}{2}+\frac{{\lambda }_2}{4}=L\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{f}_2=\frac{v_2}{L}(\frac{n_2}{2}+\frac{1}{4}) \end{gathered}$$
 Substituting, $f_1=f_2$  we get
 $\frac{2n_1+1}{2n_2+1}=\frac{1}{3}$
  For  $n_1=1,n_2=4$
 $n_1=2,n_2=7$
  $n_1=3,n_2=10$etc
…             …           …
Therefore, the possible frequencies are
 $\frac{v_1}{L}(\frac{1}{2}+\frac{1}{4}),\frac{v_1}{L}(\frac{2}{2}+\frac{1}{4}),\frac{v_1}{L}(\frac{3}{2}+\frac{1}{4}),\mathrm{.......},etc$
 Or  $\frac{3}{4L}\sqrt{\frac{T}{\mu }},\frac{5}{4L}\sqrt{\frac{T}{\mu }},\frac{7}{4L}\sqrt{\frac{T}{\mu }},\mathrm{.....},etc$
 
$\frac{1}{4L}\sqrt{\frac{T}{\mu }}$