TS EAMCET · Physics · Oscillations
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 
- A \(\frac{L}{5}\)
- B \(\frac{2 L}{7}\)
- C \(\frac{3 L}{10}\)
- D \(\frac{L}{9}\)
Answer & Solution
Correct Answer
(A) \(\frac{L}{5}\)
Step-by-step Solution
Detailed explanation
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: \[ \mathrm{f}_{\mathrm{A}}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}} \]…
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