1. Tardigrade
  2. Question
  3. Physics
  4. 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 :- <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-o31nmtsxavbwhdp2.jpg />

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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 :-
Question

NTA AbhyasNTA Abhyas 2020

Solution:

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