4—Differential Equations 69Push this to the extreme case where the damping vanishes: b= 0. Thenα 1 =i
√
k/mand
α 2 =−i
√
k/m. Denoteω 0 =
√
k/m.
x(t) =Aeiω^0 t+Be−iω^0 t (4.7)
You can write this in other forms using sines and cosines, see problem4.10. To determine the arbitrary
constantAandByou need two equations. They come from some additional information about the
problem, typically some initial conditions. Take a specific example in which you start from the origin
with a kick,x(0) = 0andx ̇(0) =v 0.
x(0) = 0 =A+B, x ̇(0) =v 0 =iω 0 A−iω 0 B
Solve forAandBto getA=−B=v 0 /(2iω 0 ). Then
x(t) =
v 0
2 iω 0
[
eiω^0 t−e−iω^0 t
]
=
v 0
ω 0
sinω 0 t
As a check on the algebra, use the first term in the power series expansion of the sine function to see
howxbehaves for smallt. The sine factor issinω 0 t≈ω 0 t, and thenx(t)is approximatelyv 0 t, just as
it should be. Also notice that despite all the complex numbers, the final answer is real. This is another
check on the algebra.
Damped Oscillator
If there is damping, but not too much, then theα’s have an imaginary partanda negative real part.
(Is it important whether it’s negative or not?)
α=
−b±i
√
4 km−b^2
2 m
=−
b
2 m
±iω′, where ω′=
√
k
m
−
b^2
4 m^2
(4.8)
This represents a damped oscillation and has frequency a bit lower than the one in the undamped case.
Use the same initial conditions as above and you will get similar results (letγ=b/ 2 m)
x(t) =Ae(−γ+iω
′)t+Be(−γ−iω
′)tx(0) =A+B=0, vx(0) = (−γ+iω′)A+ (−γ−iω′)B=v 0 (4.9)
The two equations for the unknownsAandBimplyB=−Aand
2 iω′A=v 0 , so x(t) =
v 0
2 iω′
e−γt
[
eiω
′t−e−iω
′t]
=v 0
ω′
e−γtsinω′t (4.10)
For small values oft, the first terms in the power series expansion of this result are
x(t) =
v 0
ω′
[1−γt+γ^2 t^2 / 2 −...][ω′t−ω′^3 t^3 /6 +...] =v 0 t−v 0 γt^2 +...
The first term is what you should expect, as the initial velocity isvx=v 0. The negative sign in the
next term says that it doesn’t move as far as it would without the damping, but analyze it further.