Suspended from a nail in a wall, a closed circular wire experiences small oscillations with an amplitude of 2^ and a time period of 2 s . Evaluate the acceleration of the particle on the wire farthest from the point of suspension as it passes through its mean position. Take g = ^2 m/s ^2 .

Step-by-step solution

The moment of inertia of the closed circular wire (ring) about the point of suspension on its circumference is I = I_ cm + M R^2 = 2 M R^2 . The distance of the center of mass from the point of suspension is d = R . The time period of the physical pendulum is given by: T = 2 I M g d = 2 2 M R^2 M g R = 2 2 R g Given T = 2 s and g = ^2 m/s ^2 , we have: 2 = 2 2 R ^2 1 = 2 R 2 R = 1 m The distance of the farthest point on the wire from the point of suspension is L = 2 R = 1 m . The amplitude of oscillation is ₀ = 2^