Question

A half immersed sphere with radius R and relative density 0.5 is in equilibrium with water. The sphere is pushed down lightly into the water and released. What's the oscillatory frequency of the sphere? Density of water is `1.0 g//cm^3`.

Answer 1

At equilibrium,we can say,the buoyancy force of water acting on half the volume of the sphere was balancing its weight.

So,when it will be pushed downward,amount of buoyancy force will be increased than its weight,so it will be forced to move upwards,due to this unbalanced force.

But on going to its position of equilibrium,it won't be able to stop due to inertia of motion,so it will try to go up further beyond its point of equilibrium,but then its weight will increase than its buoyancy force and will come down again,thus an oscillation will go on.

So,on pushing it fully by the length of its radius,net force acting on it is `(4/3 pi R^3 -4/3 pi R^3*0.5)*g=2/3 pi R^3g`

Now,this force is acting upwards,against the direction of movement.

So,this will be performing S.H.M.

We know,equation of S.H.M is `F=-momega^2 *x` (where, `x` is the amount of displacement)

Comparing this equation,we can rearrange as `F= - (2/3 pi R^2g) *R` (we could have directly written this expression to prove its motion to be S.H.M without the previous discussion)

So,`m*omega ^2 = 2/3 pi R^2g`

or, `omega^2=g/R` (as `m= 4/3 pi R^3 *0.5`)

or, `omega =sqrt(g/R)`

so, frequency =(`nu`) = `1/(2pi) *sqrt(g/R)`