InterviewSolution
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A cylindrical log of wood of height H and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.T = \(2\pi\sqrt\frac{m}{Aρg}\)Where m is mass of the body and is density of the liquid. |
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Answer» Given, m = mass of cylinder h = height of cylinder h1 = length of cylinder dipping in liquid in equilibrium position ρ = density of liquid A = area of cross section of cylinder mg = buoyant force = weight of water displaced by body = ρ(Ah1)g …(i) log is pressed gently through small distance x vertically and released. FB = ρA(h1 + x)g ∴ Net restoring force, F = Buoyant Force – weight = ρA(h1 + x)g − mg = ρA(h1 + x)g − ρ(Ah1 )g [from (ii)] = (Aρg)x ∴ F and x are in opposite direction. F = −(Aρg)x a = \(\frac{-(Aρg)}{m}x\) …(ii) for standard SHM a = w2x …(iii) ∴ by (ii) & (iii) w2 = \(\frac{Aρg}{m}\) or w = \(\sqrt{\frac{Aρg}{m}}\) ∴ T = \(2\pi\sqrt{\frac{m}{Aρg}}\) |
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