A body of mass 2 kg floats vertically in water. Its cross-sectional area at the water surface remains approximately constant at 0.01 m² during small vertical oscillations. Take water density 1000 kg/m³ and g = 10 m/s². If ω² = N s⁻², find N.
Physics · Simple Harmonic Motion · time period of linear shm
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Question

A body of mass 2 kg floats vertically in water. Its cross-sectional area at the water surface remains approximately constant at 0.01 m² during small vertical oscillations. Take water density 1000 kg/m³ and g = 10 m/s². If ω² = N s⁻², find N.

Physics · Simple Harmonic Motion · time period of linear shm

G1 Physics · Accepted Answer

For a small downward displacement x, the extra displaced volume is Ax. Thus the extra buoyant force is F = ρgAx. Hence the effective spring constant is kₑ = ρgA. Substitute ρ = 1000, g = 10, A = 0.01. Then, kₑ = 100 N/m. Therefore, ω² = kₑ/m = 100/2 = 50 s⁻². Hence, N = 50.