Oscillations

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Question : 18
Total: 25
A cylindrical piece of cork of base area A and height h floats in a liquid of density ρ1. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T=2π
hρ
ρ1g

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:  
Let X be the equilibrium position of a cylinder floating in a given liquid.
A= area of cross-section of the cylindrical piece of cork.
h= height of the cylindrical piece of cork.
ρ= density of the material of the cylindrical cork.
ρe= density of the liquid in which it floats.
l= length of the cylindrical piece of cork dipping in the liquid upto point P in position X.
W= weight of the cylindrical cork.
W1= weight of the liquid displaced by the cork.
V= its volume, m= mass of cork =Ahρ.
V=Ah,
W=mg=(Vρ)g=(Ahρ)g
W1= Area of cross-section of cork × length of cylinder dipping in liquid × density of liquid ×g=Alρ1g
∴ According to the law of flotation, W=W1orAhρg=Alρ1g
or h=
lρ1
ρ
.
.
.(i)

Let the cylinder be pushed into the liquid through a distance PQ=y. Then the restoring force acting on the cylinder is given by
F= weight of the liquid displaced by portion PQ of the cylindrical cork
=(Ay)ρ1×g=(Ag )y...(ii)
or F=ky
where k=Aρ1g is the force constant.
If a be the acceleration produced in the cylindrical piece of cork, then
a=
F
mass of cork

=
F
m
=
Aρ1gy
Ahρ

or a=
ρ1g
hρ
y
.
.
.(iii)

Now as the acceleration of the cylindrical cork is directly proportional to its displacement from equilibrium position and acts towards the equilibrium position, so the motion of the bob is simple harmonic having time period (T) given by
T=2π
Displacement
Acceleration

=2π
y
ρ1g
hρ
y
or T=2π
hρ
ρ1g
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