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

Section 17.3 Applications of Second-Order Differential Equations 1169
ExamplE 1 A spring with a mass of 2 kg has natural length 0.5 m. A force of 25.6 N is
required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length
of 0.7 m and then released with initial velocity 0, find the position of the mass at any
time t.
SOLUTION From Hooke’s Law, the force required to stretch the spring is
ks0.2d − 25.6
so k − 25.6y0.2 − 128. Using this value of the spring constant k, together with m − 2
in Equation 1, we have
2 d 2 x
dt 2 1 128x − 0
As in the earlier general discussion, the solution of this equation is
2 xstd − c 1 cos 8t 1 c 2 sin 8t
We are given the initial condition that xs0d − 0.2. But, from Equation 2, xs0d − c 1 .
Therefore c 1 − 0.2. Differentiating Equation 2, we get
x9std − 28c 1 sin 8t 1 8c 2 cos 8t
Since the initial velocity is given as x9s0d − 0, we have c 2 − 0 and so the solution is
xstd − 0.2 cos 8t
■
m
FIGURE 3
Damped Vibrations
We next consider the motion of a spring that is subject to a frictional force (in the case of
the horizontal spring of Figure 2) or a damping force (in the case where a vertical spring
moves through a fluid as in Figure 3). An example is the damping force supplied by a
shock absorber in a car or a bicycle.
We assume that the damping force is proportional to the velocity of the mass and acts
in the direction opposite to the motion. (This has been confirmed, at least approximately,
by some physical experiments.) Thus
damping force − 2c dx
dt
where c is a positive constant, called the damping constant. Thus, in this case, Newton’s
Second Law gives
m d 2 x
dt 2
dx
− restoring force 1 damping force − 2kx 2 c
dt
Schwinn Cycling and Fitness
or
3 m d 2 x
dt 2
1 c dx
dt 1 kx − 0
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