A cylindrical object of outer diameter 10 cm , height 20 cm , and density 8000 kg m ⁻³ is held by a vertical spring of spring constant 500 N m ⁻¹ . The object is initially in equilibrium, half dipped in water. If it is slightly depressed and released, calculate the time period of the resulting oscillations.

Step-by-step solution

Mass of the cylinder, m = r^2 h m = (0.05)^2 (0.2) 8000 = 4 kg When the cylinder is depressed by a small distance x , the restoring force is provided by the spring and the additional buoyant force. F = -kx - ( r^2 x) _w g F = -(k + r^2 _w g)x Effective spring constant, k_ eff = k + r^2 _w g Using g = 10 m s ⁻² : k_ eff = 500 + (0.05)^2 (1000)(10) = 500 + 25 N m ⁻¹ Time period, T = 2 m k_ eff T = 2 4 500 + 25 = 4 500 + 25 T 12.56 3.14 578.5 12.56 0.0737 0.93 s Answer: 0.93 s