Page 82 - Advanced engineering mathematics

62 CHAPTER 2 Linear Second-Order Equations

(a) Unstretched (b) Static (c) System in
equilibrium motion

d

m y 0

y(t)

m

FIGURE 2.3 Spring at natural and
equilibrium lengths and in motion.

Assuming that the mass is constant, Newton’s second law of motion gives us

my =−ky − cy + f (t)

or
c k 1

y + y + y = f (t). (2.12)
m m m
This is the spring equation. Solutions give the displacement of the bob as a function of time and
enable us to analyze the motion under various conditions.

2.4.1 Unforced Motion

The motion is unforced if f (t) = 0. Now the spring equation is homogeneous, and the
characteristic equation has roots
c 1 √
2
λ =− ± c − 4km.
2m 2m
As we might expect, the solution for the displacement, and hence the motion of the bob, depends
on the mass, the amount of damping, and the stiffness of the spring.

2
Case 1: c − 4km > 0
Now the roots of the characteristic equation are real and distinct:

c 1 √ c 1 √
2
2
λ 1 =− + c − 4km and λ 2 =− − c − 4km.
2m 2m 2m 2m
The general solution of the spring equation in this case is
λ 2 t
y(t) = c 1 e λ 1 t + c 2 e .

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October 14, 2010 14:12 THM/NEIL Page-62 27410_02_ch02_p43-76