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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%

0.775

0.175

0.375

0.575

Solution:

Frequency of torsional oscillations is given by f = k√I
If f1 = k√[M(2L)²/12]
f2 = k√[M(2L)²/12 + 2m(L/2)²]
f2 = 0.8f1
Solving for m/M, we get m/M = 0.375