Question

A cuboidal piece of wood has dimensions a, b and c. Its relative density is d. It is floating in large body of water such that side a is vertical. It is pushed down a bit and released. The time period of SHM executed by it is

Answer 1

If we displace the object by a small amount 'y' from its equilibrium position,

Then the submerged volume =

V =cross sectional area X y = bcy

Now we can calculate restore force, which is buoyant force =

u=Volumex density of water xg = bcy1000g

By Newton's second theorem,

F=mA (A = acceleration )

-1000gbcy = mA

-1000gbcy = 1000dabc* A

-(g/ad) y =A

A = -(g/ad) y

But   equation of the simple harmonic motion ,

A = -w^2 * y

So our w^2 = g/ad

Then w = sqrt(g/ad)

Time period = 2* pi/w

T =2*Pi*sqrt(ad/g)