Page 83 - Advanced engineering mathematics

2.4 Spring Motion 63

√
Clearly, λ 2 < 0. Since m and k are positive, c − 4km < c ,so c − 4km < c and λ 1 < 0.
2
2
2
Therefore,
lim y(t) = 0
t→∞
2
regardless of initial conditions. In the case that that c − 4km > 0, the motion simply decays to
zero as time increases. This case is called overdamping.

EXAMPLE 2.18 Overdamping
−t
Let c = 6,k = 5, and m = 1. Now the general solution is y(t) = c 1 e + c 2 e −5t . Suppose the bob
was initially drawn upward 4 feet from equilibrium and released downward with a speed of 2
feet per second. Then y(0) =−4 and y (0) = 2, and we obtain

1
y(t) = e −t −9 + e −4t .
2
Figure 2.4 is a graph of this solution. Keep in mind here that down is the positive direction. Since
−9 + e −4t < 0for t > 0, then y(t)< 0, and the bob always remains above the equilibrium point.
−t
Its velocity y (t)=e (9−5e −4t )/2 decreases to zero as t increases, so the bob moves downward

toward equilibrium with decreasing velocity, approaching arbitrarily close to but never reaching
this position and never coming completely to rest.

2
Case 2: c − 4km = 0
In this case, the general solution of the spring equation is
y(t) = (c 1 + c 2 t)e −ct/2m .

This case is called critical damping. While y(t) → 0as t →∞, as with overdamping, now the
bob can pass through the critical point, as the following example shows.

t
0 1 2 3 4
0

–1

–2

–3

–4

FIGURE 2.4 Overdamped, unforced motion in
Example 2.18.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 14:12 THM/NEIL Page-63 27410_02_ch02_p43-76