Question:
A massless rod is suspended by two identical strings. A mass is hung at distance \(x\) from left. Fundamental frequency in AB equals 2nd harmonic in CD. Find \(BO\).
A massless rod is suspended by two identical strings. A mass is hung at distance \(x\) from left. Fundamental frequency in AB equals 2nd harmonic in CD. Find \(BO\).
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Frequency ratio $\Rightarrow$ square relation gives tension ratio.
Updated On: Apr 23, 2026
- \(L/5\)
- \(4L/5\)
- \(3L/4\)
- \(L/4\)
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The Correct Option is A
Solution and Explanation
Concept:
Frequency of stretched string:
\[
f \propto \sqrt{\frac{T}{l}}
\]
Step 1: Let tensions
Taking moments about left: \[ T_2 L = mgx,\quad T_1 = mg - T_2 \] \[ T_2 = \frac{mgx}{L},\quad T_1 = mg\left(1 - \frac{x}{L}\right) \]
Step 2: Frequency relation
\[ f_1 \propto \sqrt{\frac{T_1}{L}},\quad f_2 \propto \sqrt{\frac{T_2}{L}} \] Given: \[ f_{AB} = 2f_{CD} \] \[ \sqrt{T_1} = 2\sqrt{T_2} \Rightarrow T_1 = 4T_2 \]
Step 3: Substitute
\[ mg\left(1 - \frac{x}{L}\right) = 4\cdot \frac{mgx}{L} \] \[ 1 - \frac{x}{L} = \frac{4x}{L} \Rightarrow 1 = \frac{5x}{L} \Rightarrow x = \frac{L}{5} \] Conclusion: \[ \frac{L}{5} \]
Step 1: Let tensions
Taking moments about left: \[ T_2 L = mgx,\quad T_1 = mg - T_2 \] \[ T_2 = \frac{mgx}{L},\quad T_1 = mg\left(1 - \frac{x}{L}\right) \]
Step 2: Frequency relation
\[ f_1 \propto \sqrt{\frac{T_1}{L}},\quad f_2 \propto \sqrt{\frac{T_2}{L}} \] Given: \[ f_{AB} = 2f_{CD} \] \[ \sqrt{T_1} = 2\sqrt{T_2} \Rightarrow T_1 = 4T_2 \]
Step 3: Substitute
\[ mg\left(1 - \frac{x}{L}\right) = 4\cdot \frac{mgx}{L} \] \[ 1 - \frac{x}{L} = \frac{4x}{L} \Rightarrow 1 = \frac{5x}{L} \Rightarrow x = \frac{L}{5} \] Conclusion: \[ \frac{L}{5} \]
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