Let initially the cone be in equilibrium i.e. floating
From the diagram, it is clear that Buoyant force
(Fb)= weight of the cone.
(Fw) ρw×(πr2x)×g=ρc×(πR2H)×g ⇒2r2x=R2H (∵ρc=) ⇒x= ........(i)
From the diagram,
tanθ== ⇒x= .....(ii)
From Eqs. (i) and (ii)
= ⇒=r3⇒r=()1∕3R Now, the cone. is depressed by small distance
δ(<<) and released it executes simple harmonic motion. In simple harmonic motion, the buoyancy force acts as restoring force which is given as
Fb= density × volume × gravity
Since, the displacement is very small, the excess submerged shape of cone. can be considered as a cylinder whose volume is given by
πr2δ .
∴Fb=ρwπr2δg=ρwπ[()1∕3R]2δg Thus, force is equal to restoring force, hence
Fb=kδ=mω2δ ⇒=ω2 or