Q. 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 :

Solution:

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 $

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