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 1st harmonic (fundamental frequency) in $AB$ is equal to 2nd harmonic frequency in $CD$. Then, length of $BO$ is :

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

Answer 1

Answer: (A)

$L/5$

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

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

For rotational equilibrium

$T_{1}x=T_2(L-x)\Rightarrow x=\frac{L}{5}$