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 oscillations frequency by 20%. The value of ratio m/m is close to:

Answer 1

Time Period of Torsional pendulum case -

A rod of mass 'm' and length '2l.'

T=2π√1/K

And the distance attached about the center l/2

Whereas

I= moment of inertia

K= torsional constant

Therefore Frequency f =K/√I

f1 = K/√(M(2L)²/12)

f2 = K/√(M(2L)²/12) + 2m (l/2)²

f2 = 0.8 f1

The value of the ratio m/M is close to = 0.35