Professor

Applications of Second-Order Differential Equations
Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration of springs and electric circuits. Vibrating Springs We consider the motion of an object with mass m at the end of a spring that is either vertical (as in Figure 1) or horizontal on a levelsurface (as in Figure 2). In Section 6.5 we discussed Hooke’s Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x :
m equilibrium position
0

restoring force

kx

x x

m

where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to airresistance or friction) then, by Newton’s Second Law (force equals mass times acceleration), we have
1

FIGURE 1 equilibrium position m
0 x x

m

d 2x dt 2

kx

or

m

d 2x dt 2

kx

0 k 0

This is a second-order linear differential equation. Its auxiliary equation is mr 2 i, where with roots r sk m. Thus, the general solution is xt which can also be written as xt where A cos A cos tc1 cos t c2 sin t

FIGURE 2

sk m (frequency) sc 2 1 c1 A c2 2
(amplitude)

sin

c2 A

is the phase angle

(See Exercise 17.) This type of motion is called simple harmonic motion.
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 thenreleased 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

k 0.2

25.6 2

so k 25.6 0.2 128. Using this value of the spring constant k, together with m in Equation 1, we have 2 d 2x dt 2 128x 0

As in the earlier general discussion, the solution of this equation is
2

xt

c1 cos 8t

c2 sin 8t
12 ■ APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS

We are given the initial condition that x 0 0.2. But, from Equation 2, x 0 Therefore, c1 0.2. Differentiating Equation 2, we get x t 8c1 sin 8t 8c2 cos 8t

c1.

Since the initial velocity is given as x 0 xt Damped Vibrations

0, we have c2
1 5

0 and so the solution is

cos 8t

We next consider the motion of a spring thatis 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 hasbeen confirmed, at least approximately, by some physical experiments.) Thus
m

damping force
FIGURE 3

c

dx dt

where c is a positive constant, called the damping constant. Thus, in this case, Newton’s Second Law gives m or
3

d 2x dt 2

restoring force

damping force

kx

c

dx dt

m

d 2x dt 2

c

dx dt

kx

0

Equation 3 is a second-order linear differentialequation and its auxiliary equation is mr 2 cr k 0. The roots are
4

r1

c

sc 2 2m

4mk

r2

c

sc 2 2m

4mk

We need to discuss three cases.
4 mk 0 (overdamping) In this case r1 and r 2 are distinct real roots and
■

CASE I

c2

x

x

c1 e r1 t

c2 e r2 t

0

t

x

Since c, m, and k are all positive, we have sc 2 4mk c, so the roots r1 and r 2 given byEquations 4 must both be negative. This shows that x l 0 as t l . Typical graphs of x as a function of t are shown in Figure 4. Notice that oscillations do not occur. (It’s possible for the mass to pass through the equilibrium position once, but only once.) This is because c 2 4mk means that there is a strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass. c...