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

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(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

$Assertion :$ Transverse waves are produced in a very long string fixed at one end. Only a progressive wave is observed near the free end.
$Reason :$ Energy of the reflected wave does not reach the free end.

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