15—Fourier Analysis 377The explicit values ofω±are
ω+=
−ib+
√
−b^2 + 4km
2 m
and ω−=
−ib−
√
−b^2 + 4km
2 m
Let ω′=
√
−b^2 + 4km
2 m
and γ=
b
2 m
The difference that appears in the preceding equation is then
ω+−ω−= (ω′−iγ)−(−ω′−iγ) = 2ω′
Eq. (15.16) is then
∫∞
−∞dω
2 π
. e
−iω(t−t′)−mω^2 −ibω+k
=−i
[
e−i(ω
′−iγ)(t−t′)− 2 mω′
+
e−i(−ω
′−iγ)(t−t′)+2mω′
]
=
−i
2 mω′
e−γ(t−t
′)[
−e−iω
′(t−t′)+e+iω
′(t−t′)]=
1
mω′
e−γ(t−t
′)
sin(
ω′(t−t′)
)
Put this back into Eq. (15.14) and you havex(t) =
∫t−∞dt′F 0 (t′)G(t−t′), where G(t−t′) =
1
mω′
e−γ(t−t
′)
sin(
ω′(t−t′)
)
(15.17)
If you eliminate the damping term, settingb= 0, this is exactly the same as Eq. (4.34). The integral
stops att′=tbecause the Green’s function vanishes beyond there. The motion at timetis determined
by the force that was applied in the past, not the future.
Example
Apply a constant external force to a damped harmonic oscillator, starting it at timet= 0and keeping
it on. What is the resulting motion?
F 0 (t) =
{
0 (t <0)
F 1 (t >0)
whereF 1 is a constant. The equation (15.17) says that the solution is (t > 0 )
x(t) =
∫t−∞dt′F 0 (t′)G(t−t′) =
∫t0dt′F 1 G(t−t′)
=F 1
∫t0dt′
1
mω′
e−γ(t−t
′)
sin(
ω′(t−t′)
)
=
F 1
2 imω′
∫t0dt′e−γ(t−t
′)[
eiω
′(t−t′)−e−iω
′(t−t′)]=
F 1
2 imω′
[
1
γ−iω′
e(−γ+iω
′)(t−t′)
−1
γ+iω′
e(−γ−iω
′)(t−t′)]t′=tt′=0=F 1
2 imω′
[
2 iω′
γ^2 +ω′^2
−
e−γt
γ^2 +ω′^2
[
2 iγsinω′t+ 2iω′cosω′t
]
]
=
F 1
m(γ^2 +ω′^2 )
[
1 −e−γt
[
cosω′t+
γ
ω′
sinω′t
]]
Check the answer. Ift= 0it is correct;x(0) = 0as it should.
Ift→∞,x(t)goes toF 1 /
(
m(γ^2 +ω′^2 )
)
; isthiscorrect? Check it out! And maybe simplify the result
in the process. Is the small time behavior correct?