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IV. Exercises · Q3

Q.A piece of wood of mass mm is floating erect in a liquid whose density is ρ\rho. If it is slightly pressed down and released, then it executes simple harmonic motion. Show that its time period of oscillation is T=2πmAρgT = 2\pi\sqrt{\dfrac{m}{A\rho g}}

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Step 1. Let the wood of mass mm and cross-sectional area AA float upright in a liquid of density ρ\rho. At equilibrium, by Archimedes' principle, the buoyant force (weight of displaced liquid) balances the weight: mg=Ax0ρgmg=A x_0\rho g, where x0x_0 is the equilibrium submerged depth.

Step 2. If the wood is pushed down a further small depth xx and released, the total submerged depth becomes x0+xx_0+x, and the buoyant force becomes A(x0+x)ρgA(x_0+x)\rho g.

Step 3. The net restoring force is this buoyant force minus the weight: F=mg−A(x0+x)ρg=[mg−Ax0ρg]−Aρg x=0−Aρg xF=mg-A(x_0+x)\rho g=[mg-Ax_0\rho g]-A\rho g\,x=0-A\rho g\,x (using the equilibrium condition from Step 1), so F=−Aρg xF=-A\rho g\,x -- a restoring force directly proportional to displacement xx, exactly like a spring. …

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