Simple Harmonic Motion

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/ballon.png



F is the NET upthrust by the air
Fcosθ is balanced by the tension in the string
so Fsinθ is the net force acting on the balloon

for small θ, sinθ=θ so
mα= Fsinθ = F(-θ)
since the force points to the opposite direction of the angular displacement measured from the vertical (the force is restoring and always points to the equilibrium position)
Hence

α= - (F/m)θ


mass of the helium gas=(ρ1)V
where V is the volume of the balloon

mass of air displaced=(ρ2)V

so F=(ρ2-ρ1)V

Hence
α= - [(ρ2-ρ1)V/((ρ1)V)]θ
= - ((ρ2-ρ1)/(ρ1)θ

2010-03-04 16:23:19 補充：
some mistake in the previous answer, see this
http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-03-04-16-21-19.png

If you don't want to use the idea in rotational dynamics, you may use physics8801's method:
me a=αL

these are the two only methods

2010-03-05 16:20:54 補充：
franng2008
exactly!
that's how i think of the solution

This is the same as the simple harmonic motion of simple pendulum with all the forces are in opposite direction.

Let U be the upthrust acting on the balloon, m is the mass of the balloon, d1 be the density of helium, and d2 be the density of air.

Hence, by Archimedes' Principle, upthrust U = weight of air diaplced
U = volume of air displaced x density of air x g, where g is the acceleration due to gravity
i.e. U = (m/d1).(d2.g)
Net upward force acting on the balloon, F = U - mg = (m/d1).(d2.g) - mg
F = (d2/d1 - 1).mg ------------------ (1)

Since F is vertically upward, now resolve F into two components, one is along the string direction (=F.cos(theta) ), and the other is perpendicular to the string direction (= F.sin(theta) ).
Consider the component perpendicualr to the string direction,

F.sin(theta) = -m.a
where a is the acceleration of the balloon. The -ve sign indicates that the acceleration is always pointing towards the equilibrium position, opposite to the displacement.

But a = L.(alpha), where (alpha) is the angular acceleration
hence, F.sin(theta) = -m.L.(alpha)
Substitute F from equation (1),
[(d2/d1 - 1).mg].sin(theta) = -m.L.(alpha)
when (theta) is small, sin(theta) = (theta) approximately

therefore, [(d2/d1 - 1)].g.(theta) = -L.(alpha)
(alpha) = - [(d2/d1 - 1).g/L].(theta)

since the angular acceleration is proportional to the angular displacement (alpha), and in a direction always opposite to it, the motion is simple harmonic.

[Note: in fact, if you turn the whole balloon arrangement upside down, this is just similar to a simple pendulum, with the weight of the pendulum bob be replaced by the net force F].