Question

A massless rod 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 $x^{\prime }$. Further, it is observed that the frequency of lst harmonic (fundamental frequency) in AB is equal to $2^{nd}$ harmonic frequency in CD. Then, length of BO is

• A. $L/5$
• B. $4L/5$
• C. $3L/4$
• D. $L/4$

Answer 1

The correct option is A $L/5$

Given length of OB is $x$ and length of OD is $L-x$

Let $T_1$ be tension in the wire AB and $T_2$ is the tension in the wire

First harmonic frequency in AB is $\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}$

Second harmonic frequency in CD is $\frac{1}{l}\sqrt{\frac{T_2}{\mu }}$

Equalling both the values we get $T_1=4T_2$

By Torque Equilibrium about point O we get $x{T}_1=(L-x)T_2$

Using above two equations we get $x=\frac{L}{5}$