Question

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. 0.17
• B. 0.37
• C. 0.57
• D. 0.77

Answer 1

Answer: (B)

0.37

Answer 2

Frequency of torsonal oscillations is given by $f = \frac{k}{\sqrt{I}}$
$f_{1} = \frac{k}{\sqrt{\frac{M\left(2L\right)^{2}}{12}}}$
$f_{2} = \frac{k}{\sqrt{\frac{M\left(2L\right)^{2}}{12} +2m \left(\frac{L}{2}\right)^{2}}}$
$f_{2} = 0.8 f_{1}$
$\frac{m}{M} = 0.375$