James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 17.3 Applications of Second-Order Differential Equations 1171
ExamplE 2 Suppose that the spring of Example 1 is immersed in a fluid with damping
constant c − 40. Find the position of the mass at any time t if it starts from the
equilibrium position and is given a push to start it with an initial velocity of 0.6 mys.
SOLUTION From Example 1, the mass is m − 2 and the spring constant is k − 128, so
the differential equation (3) becomes
2 d 2 x
dt 2
dx
1 40
dt 1 128x − 0
or
d 2 x
dt 2
dx
1 20
dt 1 64x − 0
The auxiliary equation is r 2 1 20r 1 64 − sr 1 4dsr 1 16d − 0 with roots 24 and
216, so the motion is overdamped and the solution is
Figure 6 shows the graph of the position
function for the overdamped motion in
Example 2.
0.03
xstd − c 1 e 24t 1 c 2 e 216t
We are given that xs0d − 0, so c 1 1 c 2 − 0. Differentiating, we get
x9std − 24c 1 e 24t 2 16c 2 e 216t
so x9s0d − 24c 1 2 16c 2 − 0.6
0 1.5
FIGURE 6
Since c 2 − 2c 1 , this gives 12c 1 − 0.6 or c 1 − 0.05. Therefore
x − 0.05se 24t 2 e 216t d
■
Forced Vibrations
Suppose that, in addition to the restoring force and the damping force, the motion of the
spring is affected by an external force Fstd. Then Newton’s Second Law gives
m d 2 x
dt 2
− restoring force 1 damping force 1 external force
− 2kx 2 c dx
dt 1 Fstd
Thus, instead of the homogeneous equation (3), the motion of the spring is now governed
by the following nonhomogeneous differential equation:
5 m d 2 x
dt 2
1 c dx
dt
1 kx − Fstd
The motion of the spring can be determined by the methods of Section 17.2.
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