Question

1. Ablockofmass M = 4kg, cross-sectional area A = 1 m^2, and height h = 60 cm sits in a liquid with density ρ = 12 kg/m^3.

a) Find the depth d the block will sit in the water when in equilibrium.
b) At time t=0, the block is released from being completely submerged in the liquid, calculate the amplitude and frequency of oscillation.

Answer 1

When in equilibrium, the object is floating at rest and the net force on the block is zero. The buoyancy force on the block due to the displaced liquid, acts upward and the gravitational force due to the weight of the block acts downward. And so, if a depth d of the block sits in the water, and the block has a horizontal cross sectional area A, then, the mass of liquid displaced by the block when floating is
   
And so, the buyoancy force upward is
   
And the mass of the block is M, so, the weight of the block acting downward is
   
And so, we have at the equilibrium
   
  
So, putting the given values, we get
   
  

b)

When the block is completely submerged in the liquid, the buoyancy force due to the displaced liquid is higher than that in equilibrium. And so, there is a net force on the block. And if the block is displaced a distance y to into the water then the diagram is given by
  

So, the extra force acting upward will provide a restoring on the block. And this extra force is given by
   
And so, the equation of motion of the block is
  


And so, we have the angular frequency
   
So, the frequency is
  
Now putting the given values, we get
  
  
And as at time t = 0, the block was completely submerged ( the height of the block is h) , so, the block was displace from the equilibrium at time t = 0,
  
As the block is released from the rest, so, this is the amplitude of oscillation. And as h = 60 cm = 0.60 m, so, the amplitude of the oscillation is