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 $x$. Further, it is observed that the frequency of $1^{st}$ harmonic (fundamental frequency) in $AB$ is equal to $2^{nd}$ harmonic frequency in $CD$. Then, the length of $BO$ is:

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

Answer 1

The correct option is A $L/5$

$\frac{1}{2\lambda }\sqrt{\frac{T_1}{m_u}}$ $=\frac{1}{\lambda }\sqrt{\frac{T_1}{m_u}}$

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

Now, $T_{1}x=T_2(L-x)$

$\Rightarrow 4T_{2}x=T_2(L-x)$

or $x=\frac{L}{5}$