JEE Mains · Physics · STD 11 - 13. oscillations
A rod of mass \(‘M’\) and length \(‘2L’\) is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of \(‘m’\) are attached at distance \(‘L/2’\) from its centre on both sides, it reduces the oscillation frequency by \(20\%\). The value of ratio \(m/M\) is close to
- A \(0.77\)
- B \(0.57\)
- C \(0.37\)
- D \(0.17\)
Answer & Solution
Correct Answer
(C) \(0.37\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}\right)}} \& 0.8 \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\right)}}\)…
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