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JEE Advanced 2016 Paper 2 · Physics · Simple Harmonic Motion · easy

A uniform cylinder of mass 0.50 kg and cross-sectional area 50 cm^2 floats upright in water of density 1000 kg m^-3, partly submerged, while being…

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A uniform cylinder of mass 0.50 kg0.50\ \mathrm{kg} and cross-sectional area 50 cm250\ \mathrm{cm^2} floats upright in water of density 1000 kg m−31000\ \mathrm{kg\,m^{-3}}, partly submerged, while being held from a fixed support above by a light vertical spring of force constant 51 N m−151\ \mathrm{N\,m^{-1}}. It is pressed slightly down and released, and executes small vertical oscillations without rotation. Take g=9.8 m s−2g=9.8\ \mathrm{m\,s^{-2}} and neglect damping. The period of oscillation is closest to
  1. A. 0.62 s0.62\ \mathrm{s}
  2. B. 0.36 s0.36\ \mathrm{s}
  3. C. 0.44 s0.44\ \mathrm{s}
  4. D. 0.27 s0.27\ \mathrm{s}
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Answer: (C) 0.44 s0.44\ \mathrm{s}

Let the cylinder be displaced downward by xx from equilibrium. The extra displaced volume is AxAx, so the buoyant force rises by ρgAx\rho gAx and acts upwards. The spring, which is already stretched, also exerts an extra upward force kxkx. Both are restoring and add, giving mx¨=−(k+ρgA)xm\ddot{x}=-(k+\rho gA)x and an effective stiffness keff=k+ρgAk_{eff}=k+\rho gA. Here ρgA=1000×9.8×5×10−3=49 N m−1\rho gA=1000\times 9.8\times 5\times10^{-3}=49\ \mathrm{N\,m^{-1}}, so keff=100 N m−1k_{eff}=100\ \mathrm{N\,m^{-1}}. The period is T=2πm/keff=2π0.50/100=0.444 sT=2\pi\sqrt{m/k_{eff}}=2\pi\sqrt{0.50/100}=0.444\ \mathrm{s}. Using the spring alone gives 0.62 s0.62\ \mathrm{s}, and doubling the buoyancy term gives 0.36 s0.36\ \mathrm{s}.
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