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2.4 Spring Motion 65

EXAMPLE 2.20 Underdamping
Let c = k = 2 and m = 1. The general solution is
−t
y(t) = e [c 1 cos(t) + c 2 sin(t)].
Suppose the bob is driven downward from a point three feet above equilibrium with an initial
speed of two feet per second. Then y(0) =−3, and y (0) = 2. The solution is

−t
y(t) =−e (3cos(t) + sin(t)).
The behavior of this solution is visualized more easily if we write it in phase angle form. Choose
C and δ so that
3cos(t) + sin(t) = C cos(t + δ).
For this, we need
3cos(t) + sin(t) = C cos(t)cos(δ) − C sin(t)sin(δ).
Then

C cos(δ) = 3 and C sin(δ) =−1,
so
C sin(δ) 1
= tan(δ) =− .
C cos(δ) 3
Now
1 1

δ = arctan − =−arctan .
3 3
To solve for C, write
2
2
2
2
2
2
C cos (δ) + C sin (δ) = C = 3 + 1 = 10.
√
Then C = 10, and the solution can be written in phase angle form as
√
−t
y(t) = 10e cos(t − arctan(1/3)).
√
−t
The graph is a cosine curve with decaying amplitude squeezed between graphs of y = 10e and
√
−t
y =− 10e . Figure 2.6 shows y(t) and these two exponential functions as reference curves.
0.4
0.2

0
2 3 4 5 6
t
–0.2

–0.4

FIGURE 2.6 Underdamped, unforced motion in
Example 2.20.

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October 14, 2010 14:12 THM/NEIL Page-65 27410_02_ch02_p43-76