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 node 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 m be the mass per unit length of each wire and let L be the length of each wire.

Given

$v=\frac{1}{2l}\sqrt{\frac{T_1}{m}}=\frac{2}{2l}\sqrt{\frac{T_2}{m}}$

Which gives $\frac{T_1}{T_2}=4$

For rotational equilibrium of the rod about O, we have

$T_1\times AO=T_2\times BO$ or $T_1\times x=T_2(L-x),$

Which gives $\frac{T_1}{T_2}=\frac{(L-x)}{x}.$ But $\frac{T_1}{T_2}=4.$ Hence $4=\frac{(L-x)}{x}$

Which gives $x=\frac{L}{5}.$