A massless rod of length $L$ 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 in $AB$ is equal to $2^{nd}$ harmonic frequency in $CD$. $‘x’$ is
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Frequency of $1$ st harmonic of $AB$
$=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}$
Frequency of $2$ nd harmonic of $\mathrm{CD}$
$=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}$
Given that the two frequencies are equal.
$\therefore \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}$
$\Rightarrow \frac{\mathrm{T}_{\mathrm{AB}}}{4}=\mathrm{T}_{\mathrm{CD}} \Rightarrow \mathrm{T}_{\mathrm{AB}}=4 \mathrm{T}_{\mathrm{CD}}$ $...(i)$
For rotational equilibrium of massless rod, taking torque about point $O.$
$\mathrm{T}_{\mathrm{AB}} \times \mathrm{x}=\mathrm{T}_{\mathrm{CD}}(\mathrm{L}-\mathrm{x})$ $...(ii)$
For translational equilibrium,
$\mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=\mathrm{mg}$ $..(iii)$
On solving, $(i)$ $\&(iii)$ we get, $\mathrm{T}_{\mathrm{CD}}-\frac{\mathrm{mg}}{5}$
$\therefore \mathrm{T}_{\mathrm{AB}}=\frac{4 \mathrm{mg}}{5}$
Substituting these values in $(ii)$ we get
$\frac{4 m g}{5} \times x=\frac{m g}{5}(L-x)$
$\Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{L}}{5}$
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