A massless rod of length L is suspended by tw | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The frequency of the $n^{th}$ harmonic of a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$.Frequency of the $1^{st}$ harmonic of $AB

Vedclass · Accepted Answer

$\frac{L}{5}$

Vedclass · Answer

(A) The frequency of the $n^{th}$ harmonic of a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$.Frequency of the $1^{st}$ harmonic of $AB$ is $f_{AB} = \frac{1}{2l} \sqrt{\frac{T_{AB}}{\mu}}$.Frequency of the $2^{nd}$ harmonic of $CD$ is $f_{CD} = \frac{2}{2l} \sqrt{\frac{T_{CD}}{\mu}} = \frac{1}{l} \sqrt{\frac{T_{CD}}{\mu}}$.Given that $f_{AB} = f_{CD}$,we have:$\frac{1}{2l} \sqrt{\frac{T_{AB}}{\mu}} = \frac{1}{l} \sqrt{\frac{T_{CD}}{\mu}}$$\Rightarrow \frac{T_{AB}}{4} = T_{CD} \Rightarrow T_{AB} = 4T_{CD} \quad ...(i)$For rotational equilibrium of the rod about point $O$:$T_{AB} \cdot x = T_{CD} \cdot (L - x) \quad ...(ii)$For translational equilibrium:$T_{AB} + T_{CD} = mg \quad ...(iii)$Substituting $(i)$ into $(iii)$:$4T_{CD} + T_{CD} = mg \Rightarrow 5T_{CD} = mg \Rightarrow T_{CD} = \frac{mg}{5}$.Then,$T_{AB} = 4 \left( \frac{mg}{5} \right) = \frac{4mg}{5}$.Substituting these into $(ii)$:$\left( \frac{4mg}{5} \right) x = \left( \frac{mg}{5} \right) (L - x)$$4x = L - x \Rightarrow 5x = L \Rightarrow x = \frac{L}{5}$.

Similar Questions

The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by $8 \ N$,it is in resonance with the first overtone of the closed tube. The initial tension in the wire is .... $N$.

The wavelength of light in the visible part $(\lambda_V)$ and for sound $(\lambda_S)$ are related as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module