A uniform horizontal rod of length 0.40 m and mass 1.2 kg is supported by two identical wires as shown in figure.  Where should a mass of 4.8 kg be placed on the rod, so that the same tuning fork may excit the wire on left into its fundamental vibrations and that on right into its first overtone? $g=10m/s^2$   (In cm from left wire)

Let $n_1$  be frequency of vibration of left wire and  $n_2$ that of second wire.  Then given $n_2=n_1.$   If  $T_1$ and  $T_2$ are tensions in first and second wire, then
$n_1=\frac{1}{2l}\sqrt{\frac{T_1}{m}}$ and  $n_2=\frac{2}{2l}\sqrt{\frac{T_2}{m}}$
Therefore,  $\frac{n_2}{n_1}=2\sqrt{\frac{T_2}{T_1}}$
As  $n_1=n_2$ or  $\sqrt{T_1}=2\sqrt{T_2}$
$\therefore \sqrt{\frac{T_1}{T_2}}=2$ or  $T_1=4T_2$  …..(i)
For vertical equilibrium of rod
$T_1+T_2=4.8+1.2=6$ kg wt
From Eqs. (i) and (ii)       ……(ii)
$T_1=4.8$ kg, $T_2=1.2$  kg

Taking moments about  $A$
$$\begin{gathered} T_{1}x+W(0.2-x)=T_2(0.4-x) \\ 4.8x+0.24-1.2x=0.48-1.2x \end{gathered}$$
or   $4.8x=0.24$
$x=\frac{0.24}{4.8}=0.05m=5cm$