4—Differential Equations 72Interchangeα 1 andα 2 to getB.
The final result isx(t) =
F 0
α 1 −α 2
(
α 2 (mβ^2 −bβ)−kβ
)
eα^1 t−
(
α 1 (mβ^2 −bβ)−kβ
)
eα^2 t
k(mβ^2 −bβ+k)
+F 0
(
1
k
−
1
mβ^2 −bβ+k
e−βt
)
(4.15)
If you think this is messy and complicated, you haven’t seen messy and complicated. When it takes
20 pages to write out the equation, then you’re entitled say that it is starting to become involved.
Why not start with a simpler example, one without all the terms? The reason is that a complex
expression is often easier to analyze than a simple one. There are more things that you can do to it, and
so more opportunities for it to go wrong. The problem isn’t finished until you’ve analyzed the supposed
solution. After all, I may have made some errors in algebra along the way. Also, analyzing the solution
is the way you learn how these functionswork.
1. Everything in the solution is proportional toF 0 and that’s not surprising.
- I’ll leave it as an exercise to check the dimensions.
3. A key parameter to vary isβ. What should happen if it is either very large or very
small? In the former case the exponential function in the force drops to zero quicklyso the force jumps from zero toF 0 in a very short time — a step in the limit that
β→ 0.
4. Ifβis very small the force turns on very gradually and gently, as though you are being
very careful not to disturb the system.Take point 3 above: for largeβthe dominant terms in both numerator and denominator every-
where are themβ^2 terms. This result is then very nearly
x(t)≈
F 0
α 1 −α 2
(
α 2 (mβ^2 )
)
eα^1 t−
(
α 1 (mβ^2 )
)
eα^2 t
kmβ^2
+F 0
(
1
k
−
1
(mβ^2 )
e−βt
)
≈
F 0
k(α 1 −α 2 )
[
(α 2 eα^1 t−α 1 eα^2 t
]
+F 0
1
k
Use the notation of Eq. (4.9) and you have
x(t)≈
F 0
k
(
−γ+iω′−(−γ−iω′)
)
[(
(−γ−iω′)e(−γ+iω
′)t−(−γ+iω′)e(−γ−iω
′)t]