A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On  slightly pressing it downwards and releasing, it starts executing simple harmonic  motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is  close to (density of water ρ=103kg/m3)

The restoring force responsible for executing simple harmonic motion in this case  is the buoyant force. equate , buoyant force= FB=ρVg and restoring force F=kx .here ρ=density;V=volume; g=acceleration due to gravity;k=force constant; x=displacementComparing the two equations,  kx=ρVg   ----(1)Here volume= V=Ax, ----(2)    here A=area of cross section, x=displacementSubstituting equation (2) in (1) this we get  kx=ρAxgHence,  k=ρAg---(3)For a body executing simple harmonic motion, its angular frequency =  ω=km Substituting k=ρAg in  ω=km, we get ω=ρAgm .Now, the mass of water in the bottle  m=ρAh  ω=ρAgρAh ω=ghIn the question, the volume of water is given as 500 ml. Using the conversion that  1l=10−3 m3,  we get  1ml=10−6 m3,Hence, volume of 500ml of water is  5×10−4 m3 Again using  V=Ah,  5×10−4 =AhTherefore  h=5×10−4 A=5×10−4 πr2. Given the radius if the base is 5 cm,h=5×10−4 π×5×10−22 = 1 5π h=115.70.Substituting this in equation (1), ω=gh=9.8115.70ω=12.4 rad/s