A massless rod of length L is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to ‘x’. Further it is observed that the frequency of 1st harmonic in AB is equal to 2nd harmonic frequency in CD. ‘x’ is 

Frequency of Ist harmonic of AB = $\frac{1}{2\mathcal{l}}\sqrt{\frac{T_{AB}}{m}}$

Frequency of 2nd harmonic of CD =$\frac{1}{\mathcal{l}}\sqrt{\frac{T_{CD}}{m}}$

Given that the two frequencies are equal. 

$$\begin{gathered} \therefore \frac{1}{2\mathcal{l}}\sqrt{\frac{T_{AB}}{m}}=\frac{1}{\mathcal{l}}\sqrt{\frac{T_{CD}}{m}} \Rightarrow \frac{T_{AB}}{4}=T_{CD} \\ \Rightarrow T_{AB}=4T_{CD} .........\left(i\right) \end{gathered}$$

For rotational equilibrium of massless rod, taking torque about point O. 

                         T_AB × x = T_CD (L – x)                  ... (ii) 

For translational equilibrium,

                       T_AB + T_CD = mg                          ... (iii) 

On solving, (i) and (iii), we get

              $T_{CD}=\frac{mg}{5} \therefore T_{AB}=\frac{4mg}{5}$

Substituting these values in (ii), we get 

       $\frac{4mg}{5}\times x=\frac{mg}{5}(L-x) \Rightarrow x=\frac{L}{5}$