4—Differential Equations 90- Everything in the solution is proportional toF 0 and that’s not surprising.
- I’ll leave it as an exercise to check the dimensions.
- 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 quickly so the force jumps from zero
toF 0 in a very short time — a step in the limit thatβ→ 0. - Ifβis very small the force turns on very gradually and gently, being careful not to disturb the system.
Take point 3 above: For largeβthe dominant terms in both numerator and denominator everywhere 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
kUse the notation of Eq. ( 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]
+F 01
k=F 0 e−γt
k(2iω′)[
− 2 iγsinω′t− 2 iω′cosω′t]
+F 0
1
k=F 0 e−γt
k[
−
γ
ω′sinω′t−cosω′t]
+F 0
1
k(14)
At timet= 0this is still zero even with the approximations. That’s comforting, but if it hadn’t happened
it’s not an insurmountable disaster. This is an approximation to the exact answer after all, so it could happen