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JEE Main 2025 · Physics · Simple Harmonic Motion · medium

A uniform cylindrical log of length 20 cm and density 600 kg m^-3 floats upright in water of density 1000 kg m^-3. It is pushed slightly down and…

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A uniform cylindrical log of length 20 cm20\ \mathrm{cm} and density 600 kg m−3600\ \mathrm{kg\,m^{-3}} floats upright in water of density 1000 kg m−31000\ \mathrm{kg\,m^{-3}}. It is pushed slightly down and released. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}} and π2=10\pi^2=10, its time period of vertical oscillation is nearly
  1. A. 0.35 s0.35\ \mathrm{s}
  2. B. 0.69 s0.69\ \mathrm{s}
  3. C. 0.89 s0.89\ \mathrm{s}
  4. D. 1.40 s1.40\ \mathrm{s}
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Answer: (B) 0.69 s0.69\ \mathrm{s}

If the log (cross-section AA, length LL) is displaced by xx downward, the extra buoyant force ρwAg x\rho_w A g\,x acts as a restoring force, so mx¨=−ρwAg xm\ddot x = -\rho_w A g\,x with m=ρlALm = \rho_l A L. Hence ω2=ρwgρlL=1000×10600×0.2=83.3 s−2\omega^2 = \dfrac{\rho_w g}{\rho_l L} = \dfrac{1000\times10}{600\times0.2} = 83.3\ \mathrm{s^{-2}}. Then T=2πωT = \dfrac{2\pi}{\omega}, so T2=4π283.3=4083.3=0.48T^2 = \dfrac{4\pi^2}{83.3} = \dfrac{40}{83.3} = 0.48 and T≈0.69 sT\approx0.69\ \mathrm{s}. Equivalently T=2πh/gT=2\pi\sqrt{h/g} with submerged length h=12 cmh=12\ \mathrm{cm}; using the full length 2020 cm gives the trap 0.890.89 s.
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