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 . Determine the speed 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 time period of a physical pendulum is given by: T = 2 I Mgd For a closed circular wire (ring) of mass M and radius R oscillating in its own plane about a point on its circumference, the moment of inertia is: I = I_ cm + MR^2 = MR^2 + MR^2 = 2MR^2 The distance of the center of mass from the suspension point is d = R . Substituting these into the time period formula: T = 2 2MR^2 MgR = 2 2R g Given T = 2 s and g = ^2 m/s ^2 : 2 = 2 2R ^2 1 = 2R 2R = 1 m The farthest point on the wire from the point of suspension i