Q.
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, length of $BO$ is
A) $\frac{L}{5}$
B) $\frac{4L}{5}$
C) $\frac{3L}{4}$
D) $\frac{L}{4}$
Explanation
(a) According to question,
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$\frac{1}{2l}\sqrt{\frac{T_{1}}{\mu}} = \frac{1}{l}\sqrt{\frac{T_{2}}{\mu}}$
$T_{2} = T_{1}/4$
For rotational equilibrium,
$T_{1}x = T_{2}(L - x) \Rightarrow x - L/5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.