Question

3. (20 pts) A body of uniform cross-section area A, mass density p floats in a liquid of density po and at equilibrium displaces a volume V. Find the period of small oscillations about the equilibrium position To

Answer 1

using the concept of Archimede's principle of buoyancy and the idea of simple harmonic motion we can solve this as follows....

cross sectional area - A I is the it, the volume is, v'a te, where height of the body. let, at equilibrium, t height is immersed in the lituite. if mass of the body is m, then from efuilibrium At equilibrium censition, ing = (A+) f g 1-0 Here in this equation , we applied the law of Archimede's, ie the brinciple of buoyancy, that is the upward force to the body is equal to the weight of the liquide that is displaced by the body. again if we immerse the body by some more volume u, where V, da . Then the force of buoyancy is, Fauoyancy = A (x++) lo g to which is an upward force. . And downward force is mg . so, Free is the net force: Fuet = A (x++) f g -mg = Axe g + At f g - mg from O. At fog=mg so, Free = AX fog + AYPg - Afg net

So, Ines = Ans.g. thy is an upward force and we take displacement of the body & Sounward to émmerse some extar volume v so from above equation, -marx = dago 1 m dex + w x =0 [ comparing with the differential efuation of SHM] w = Ala wa ſasa now, it is given that, at efuilibrium, body Sisplaces a volume v. of réduide. So, from 0 mg = At f.g=veg 2) V= At ) ta and maulo Vlo and W = 216 » To = 21 17 = 20 2n 2n g