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 . Compute the acceleration of the particle on the wire farthest from the point of suspension when it is at an extreme position. Take g = ^2 m/s ^2 .

Step-by-step solution

For a closed circular wire (ring) of radius R suspended from a point on its circumference and oscillating in its own plane, the moment of inertia about the point of suspension is I = mR^2 + mR^2 = 2mR^2 . The distance from the point of suspension to the center of mass is d = R . The time period of this physical pendulum is given by: T = 2 I mgd = 2 2mR^2 mgR = 2 2R g Given T = 2 s and g = ^2 m/s ^2 , substituting these values yields: 2 = 2 2R ^2 1 = 2R 2R = 1 m The farthest point on the wire from the point of suspe