A wooden cylinder of mass 20 g and area of cross-section 1 cm 2 , having a piece of lead of mass 60 g attached to its bottom, floats in water. The cylinder is depressed and then released. The frequency of oscillations is N π s − 1 . Find the value of N . [Neglect the volume of water displaced by the lead piece, take g = 9.8 m/s 2 , density of water ρ w = 1 g cm - 3 ]

Step-by-step solution

Suppose that the loaded wooden block sinks upto a height h. Then, weight of water displaced by the block = weight of the block with lead If a is area of cross-section of the block and σ, the density of water, then a h ⁡ σ g = M + m g or h ⁡ = M + m a σ = 2 0 + 6 0 1 × 1 = 8 0 cm ∵ density of water = 1 g cm - 3 Now for small displacement ' x ' from equilibrium σ g a ( x + h ) − ( M + m ) g = -( M + m ) A σ g a x = -( M + m ) A A=- σ g a M + m x f = 1 2 π g h = 1 2 π 980 80 = 1.75 π s − 1