4—Differential Equations 98
To complete this idea, the external force is the sum of a lot of terms, the force betweent 1 andt 2 , that
betweent 2 andt 3 etc. The total response is the sum of all these individual responses.
x(t) =
∑
k
{F(t
k)∆tk
mω 0 sin
(
ω 0 (t−tk)
)
(t > tk)
0 (t≤tk)
For a specified timet, only the timestkbefore and up totcontribute to this sum. The impulses occurring at
the times after the timetcan’t change the value ofx(t); they haven’t happened yet. In the limit that∆tk→ 0 ,
this sum becomes an integral.
x(t) =
∫t
−∞
dt′
F(t′)
mω 0
sin
(
ω 0 (t−t′)
)
(23)
Apply this to an example. The simplest is to start at rest and begin applying a constant force from time
zero on.
Fext(t) =
{
F 0 (t > 0 )
0 (t≤ 0 )
x(t) =
∫t
0
dt′
F 0
mω 0
sin
(
ω 0 (t−t′)
)
and the last expression applies only fort > 0. It is
x(t) =
F 0
mω 02
[
1 −cos(ω 0 t)
]
As a check for the plausibility of this result, look at the special case of small times. Use the power series expansion
of the cosine, keeping a couple of terms, to get
x(t)≈
F 0
mω^20
[
1 −
(
1 −(ω 0 t)^2 / 2
)]
=
F 0
mω^20
ω^20 t^2
2
=
F 0
m
t^2
2
and this is just the result you’d get for constant accelerationF 0 /m. In this short time, the position hasn’t changed
much from zero, so the spring hasn’t had a chance to stretch very far, so it can’t apply much force, and you have
nearly constant acceleration.
For another, completely different approach to Green’s functions, see section15.5.