Question

A uniform horizontal rod of length 0.40 m and mass 1.2 kg is supported by two identical wires as shown in figure$(g=10m/se{c}^2)$, then the distance x (in cm) from left wire, where a mass of 4.8 kg should be placed on the rod, so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone is

Answer 1

Let $n_1$ be frequency of vibration of left wire and $n_2$ that of right wire. It is given that $n_2=n_1.$ If $T_1$ and $T_2$ are tensions in left and right wire, then
$n_1=\frac{1}{2l}\sqrt{\frac{T_1}{m}}\text{and}{n}_2=\frac{2}{2l}\sqrt{\frac{T_2}{m}}$
(where $l$ is the length of identical rods)
$\therefore$ $\frac{n_2}{n_1}=2\sqrt{\frac{T_2}{T_1}}$

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
$T_1+T_2=4.8+1.2=6kgwt....\left(ii\right)$From Eqs. (i) and (ii)$T_1=4.8kg,T_2=1.2kg$



Taking moments about
$\begin{array}{c}{T}_{1}x+W(0.2-x)=T_2(0.4-x)\\ 4.8x+0.24-1.2x=0.48-1.2x\\ \text{or}4.8x=0.24\\ x=\frac{0.24}{4.8}=0.05m=5cm\end{array}$