I think I'm lost in the ugly algebra, but I want to make sure. A uniform rectangular plate is suspended at point P (top center of the rectangle), and swings in the plane of the paper about an axis through P. At what other point between P and O (center of the rectangle), along PO, could the plate be suspended to have the same period of oscillation as it has around P. O is the cm of the plate. The rectangle has a width and b length. The answer is (a^2b)/(3(a^2 + b^2)) T = period. I = moment of inertia M = mass 1) T = 2pi(I/mgd)^(1/2) I = (1/12)M(a^2 + b^2) + M(b/2)^2 d = b/2 2) T = 2pi[(8b^2 + 2a^2)/(12mgb)]^(1/2) I then substitute (b/2 - x) for d in equation 1 and set it equal to equation 2 and solve for x, but I'm not getting the correct answer. Is this the correct approach? Thanks. The perpetually confused discoverer02.