A rod of mass M and length 2 L is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass m , are attached at a 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

Step-by-step solution

Let C be the torsional constant of the wire. f = 1 2 π C M 2 L 2 12 = 1 2 π 3 C M .   L 2 After masses are attached, f ′ = 1 2 π   C M . 2 L 2 12 + m L 2 4 × 2 ⇒ 0.8 f = 1 2 π   C M 3 + m 2 L 2 ⇒ 0.64 × 3 C M = C M 3 + m 2 0.64 M + 0.64 × 3 2 m = M ⇒ m M = 0.37