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A rod of length 'L' and negligible mass is suspended by two identical strings $A B$ and $C D$ as shown in the figure. $A$ mass ' $M$ ' is suspended from point ' $O$ ' which is at a distance ' $x$ ' from $\mathrm{B}$. If the frequency of the first harmonic of $\mathrm{AB}$ is equal to the frequency of the second harmonic of $C D$, then the value of ' $x$ ' is
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Verified Answer
The correct answer is:
$\frac{L}{5}$
Frequency of Harmenic motion is given by:
$$
\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}
$$
Frequency of first Harmonic of $\mathrm{AB}$ is given by:
$$
\mathrm{f}_{\mathrm{A}}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}
$$
Frequency of second Harmonic of $\mathrm{CD}$ is given by:
$$
\mathrm{f}_{\mathrm{C}}=\frac{2}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}
$$
Since frequencies are equal.
$$
\begin{aligned}
& \mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \\
& \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}} \\
& \frac{1}{4}\left(\mathrm{~T}_{\mathrm{A}}\right)=\mathrm{T}_{\mathrm{C}} \\
& \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{C}}
\end{aligned}
$$
Since the system is in equilibrium the torque about point $\mathrm{O}$ is same.
$$
\begin{aligned}
& \mathrm{T}_{\mathrm{A}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{~T}_{\mathrm{C}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \\
& \Rightarrow 5 \mathrm{x}=\mathrm{L} \\
& \mathrm{x}=\frac{\mathrm{L}}{5}
\end{aligned}
$$
$$
\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}
$$
Frequency of first Harmonic of $\mathrm{AB}$ is given by:
$$
\mathrm{f}_{\mathrm{A}}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}
$$
Frequency of second Harmonic of $\mathrm{CD}$ is given by:
$$
\mathrm{f}_{\mathrm{C}}=\frac{2}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}
$$
Since frequencies are equal.
$$
\begin{aligned}
& \mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \\
& \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}} \\
& \frac{1}{4}\left(\mathrm{~T}_{\mathrm{A}}\right)=\mathrm{T}_{\mathrm{C}} \\
& \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{C}}
\end{aligned}
$$
Since the system is in equilibrium the torque about point $\mathrm{O}$ is same.
$$
\begin{aligned}
& \mathrm{T}_{\mathrm{A}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{~T}_{\mathrm{C}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \\
& \Rightarrow 5 \mathrm{x}=\mathrm{L} \\
& \mathrm{x}=\frac{\mathrm{L}}{5}
\end{aligned}
$$
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