A cone with half the density of water is floating in water as shown in figure. It is depressed down by a small distance δ(≪H) and released. The frequency of simple harmonic oscillations of the cone is
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×(
1
3
πr2x)×g=ρc×(
1
3
πR2H)×g ⇒2r2x=R2H (∵ρc=
ρw
2
) ⇒x=
R2H
2r2
........(i) From the diagram, tanθ=
r
x
=
R
H
⇒x=
rH
R
.....(ii) From Eqs. (i) and (ii)
R2H
2r2
=
rH
R
⇒
R3
2
=r3⇒r=(
1
2
)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
2
)1∕3R]2δg Thus, force is equal to restoring force, hence Fb=kδ=mω2δ