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BASIC HOMOGENEOUS
Recall our classical spring problem deals with modeling the position (dis- placement) of a spring as it vibrates/oscillates. We start with a spring hanging from some surface, and we hang an object of mass m on it, which causes it to elongate L units. This is what we call the equilibrium/original state or position. Letting y(t) be the function which represents the displace- ment from this equilibrium state at time t, we will use the convention that upwards movements/forces are negative quantities, while downward ones are positive. So:
y(t) > 0 _ spring is stretched down y(t) = 0 _ spring is at equilibrium (no displacement) y(t) < 0 _ spring is compressed up
We used Newton’s Law, F (t) = ma(t), to get a DE for y(t), since acceler- ation is just y′′(t). To do this, we added up all of the forces acting on the object at time t. These consisted of gravity (Fg ), resistance/drag (Fr (t)), and spring force (Fs(t)). That is:
my′′(t) = F (t) = Fg + Fr(t) + Fs(t)
my′′(t) = mg − γy′(t) − k(L + y(t))
my′′(t) = −γy′(t) − ky(t) (since mg = kL)
So the our spring problem is modeled by the homogeneous DE:
my′′(t) + γy′(t) + ky(t) = 0, y(0) = y 0 , y′(0) = y 0 ′
where m = mass of the object, γ is the damping/resistance constant, and k = mg L is the spring constant.
ADDITIONAL WEIGHTS AND NONHOMOGENEOUS PROBLEMS
We can modify the original homogeneous problem described on the first page by introducing additional forces in 2 ways.
- ADDITIONAL WEIGHT - MODIFIED HOMOGENEOUS EQUATION Suppose that we have the setup described on page 1 for some spring prob- lem. But now the spring will be set into motion by throwing some additional weight, W , onto the object which is already hanging at equilibrium on the end of the spring. Let mw := W g be the mass of this additional weight.
In keeping with the spirit of the description of the homogeneous DE on the first page, we should expect to be able to model this problem by an equation:
M y′′(t) + Γy′(t) + Ky(t) = 0, y(0) = Y 0 , y′(0) = Y 0 ′
So, let’s figure out which pieces change, and how they do.
- Mass: The new mass on the end of the spring is now M = m + mw
- Damping: The new damping constant, Γ, is probably not drastically different, so there’s no need to change it (Γ = γ).
- Spring: The new spring constant, K, is also unchanged since it’s just a proportion of elongation and weight. (K = k)
- Elongation: The new elongation, L˜, is going to be different now since the additional weight would certainly increase the amount it would stretch before coming to rest. But exploiting the fact that L˜ = M K^ × g, you can figure out the new elongation amount.
- Note that y(t) will NOW measure displacement from an elongation of L˜, instead of L. This shifts things by L˜ − L units. So, Y 0 = −( L˜ − L) to reflect that you’re starting above the new equilibrium state.
If you don’t want to change your reference point, you can use the original setup, where y(t) measures displacement from an elongation of L by using the equivalent equation:
M y′′(t) + γy′(t) + ky(t) = W, y(0) = y 0 , y′(0) = y′ 0
Of course this is now non-homogeneous, but it may be easier than moving around your reference point. Note that the M here is M = m + mw
