Question

A rod $AB$ of length $L$ is hung from two identical wires $1$ and $2$. A block of mass $m$ is hung at point $O$ of the rod as shown in Fig. The value of $x$ so that a tuning fork excites the fundamental mode in wire $1$ and the second harmonic in wire $2$ is

• A. $\frac{L}{5}$
• B. $\frac{L}{4}$
• C. $\frac{L}{3}$
• D. $\frac{2L}{3}$

Answer 1

The correct option is A $\frac{L}{5}$


Let $T_1$ and $T_2$ be the tensions in wires 1 and 2 respectively.
Let $\mu$ be the mass per unit length of each wire and
$l$ be the length of each wire.

Given that tuning fork is in resonance with wires, hence
$v=\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}=\frac{2}{2l}\sqrt{\frac{T_2}{\mu }}$
$\frac{T_1}{T_2}=4$

For rotational equilibrium of the rod about $O$, we have
$T_1\times AO=T_2\times BO$
$T_1\times x=T_2(L-x)$,
$\frac{T_1}{T_2}=\frac{(L-x)}{x}$.
$4=\frac{(L-x)}{x}$
$5x=L$
$x=\frac{L}{5}$.
Hence the correct choice is (a).