Oscillations Part 2

Concept:When a floating object is displaced vertically, the extra buoyant force acts as a restoring force leading to simple harmonic motion.Explanation:Let the submerged height in equilibrium be $h$.Then $Mg = (Ah)\rho g$.When depressed by $x$, extra displaced volume is $Ax$.Extra buoyant force (restoring) is $F = - (Ax)\rho g$.Using $F = Ma$, we get $Ma = -\rho A g x$.Thus acceleration $a = -\left(\frac{\rho A g}{M}\right)x$.For SHM, $a = -\omega^2 x$.Comparing, $\omega^2 = \frac{\rho A g}{M}$, so $\omega = \sqrt{\frac{\rho A g}{M}}$.Time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{\rho A g}}$.Answer:Option C: $2\pi \sqrt{\frac{M}{\rho A g}}$.