4—Differential Equations 79The way to make use of this picture is to take a sequence of contiguous steps. One step follows
immediately after the preceding one. If two such impulses are two steps
F 0 =
{
F(t 0 ) (t 0 < t < t 1 )
0 (elsewhere) and F^1 =
{
F(t 1 ) (t 1 < t < t 2 )
0 (elsewhere)m ̈x+kx=F 0 +F 1 (4.33)
then ifx 0 is the solution to Eq. (4.30) with only theF 0 on its right, andx 1 is the solution with only
F 1 , then the full solution to Eq. (4.33) is the sum,x 0 +x 1.
Think of a general forcing functionFx,ext(t) in the way that you would set up an integral.
Approximate it as a sequence of very short steps as in the picture. Betweentkandtk+1the force is
essentiallyF(tk). The response ofmto this piece of the total force is then Eq. (4.32).
xk(t) =
{F(t
k)∆tk
mω 0 sin(
ω 0 (t−tk)
)
(t > tk)
0 (t≤tk)
where∆tk=tk+1−tk.
F
t 1 t 2 t^5
+ + + +
t
To complete this idea, the external force is the sum of a lot of terms, the force betweent 1 and
t 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 timestk before 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′)
)
(4.34)
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) =
∫t0dt′
F 0
mω 0
sin