Question

We know that objects undergoing simple harmonic motion have an acceleration that is proportional to their position, ay = -w^2 y, and so any object with that equation will be a simple harmonic oscillator. What about an object bobbing in water? Show that it is a SHO and find its w in terms of its mass. (Look at a free body diagram of an object partially submerged.)

Answer 1

Consider that the object is floating in a water of density d and it is partially subnerged. Let us assume that the volume V of the object is submerged. Suppose y axis point upwards orthogonal to the water surface.

When the object is floating at rest, the net force it experiences is zero. If we displace the object by a small amount y from its equilibrium position, then the submerged volume will change by an amount

$\delta V= -Ay ...(1)$

where A is the cross-sectional area of the portion of the object intersecting the surface of the water. The buoyant force experienced by the displaced object will have changed by an amount equal to the change in the weight of the displaced liquid

$\delta F_B= mg =d \delta Vg =-d Agy....(2)$

Here, g is the acceleration due to gravilty. Since the net force on the object is the difference of buoyant force and the force due to gravity. But only the buoyant force changes for the displaced object, the net force on the object when it is displaced by an amount y is

Applying Newton's Second law in the y direction

$m \frac{d^2 y}{dt^2}=-d Agy...(4)$

$\frac{d^2 y}{dt^2}=-\frac{d Ag}{m}y$

$a=-\frac{d Ag}{m}y....(5)$

Clearly this is the equation of SHM as the acceleration is directly proportional to the displacement.

Comparing Eq. (5) with

$a=-\omega^2 y$

we get

$\omega^2 =\frac{d Ag}{m}$

or

$\omega=\sqrt{\frac{d Ag}{m}}$