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 ${}^{\prime }{x}^{\prime }$ 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, length of $BO$ is :-

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

Answer 1

Answer: (A)

According to question,
$\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}=\frac{1}{l}\sqrt{\frac{T_2}{\mu }}$
$T_2=T_1/4$
For rotational equilibrium,
$T_{1}x=T_2\left(L-x\right)\Rightarrow x=\frac{L}{5}$