3.10 Exercise 3: Oscillations of a Buoyant Object
39
3.10.9 Analytical Solution
Equations (3.18) and (3.19) can be combined to yield a single equation with the
argument z obj . This equation reads:
dw obj
dt
=
d
2
z obj
dt 2 = −g
(ρ obj − ρ amb )
ρ obj
(3.22)
For convenience, we defin z
∗
as the distance of the object from its equilibrium
density level. The buoyancy force is proportional to the distance from this equilibrium level, and, with aid of (3.17), we can write the latter equation as:
d
2
z
∗
dt 2 = −N
2
z
∗
(3.23)
Accordingly, we seek a function whose second temporal derivative gives the
same function times a constant N
2
and a sign reversal. Only one particular type
of functions does this – the sinusoidal function – and the solution is:
z
∗
(t) = z
∗
o cos(N t)
(3.24)
where z
∗
o is the initial distance from the equilibrium level. The period of this wave is
related to the stability frequency as T = 2π/N . We yield T = 628.3 s = 10.47 mins
for settings of Exercise 3. The prediction is close to this analytical result. The
reader is encouraged to test the solution (3.24) for other choices of stability frequency N , object densities or initial displacement distances. Vertical speed evolves
according to:
w obj =
dz
∗
dt
= z
∗
o N sin(N t) = w o sin(N t)
where w o = z
∗
o N is the maximum speed that the object attains as it crosses its density equilibrium level. In our example, z
∗
o = 30 m and N = 0.01 s
−1
give a maximum
vertical speed of around 30 cm/s.
3.10.10 Inclusion of Friction
Under the assumption that the object is subject to friction in proportion to its speed,
Eq. (3.18) can be expanded as:
dw obj
dt
= −g
(ρ obj − ρ amb )
ρ obj
− Rw obj
(3.25)