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66 CHAPTER 2 Linear Second-Order Equations
The bob passes back and forth through the equilibrium point as t increases. Specifically, it passes
through the equilibrium point exactly when y(t) = 0, which occurs at times
1 2n + 1
t = arctan + π
3 2
for n = 0,1,2,···.
Next we will pursue the effect of a driving force on the motion of the bob.
2.4.2 Forced Motion
Different driving forces will result in different motion. We will analyze the case of a periodic
driving force f (t) = A cos(ωt). Now the spring equation (2.12) is
c k A
y + y + y = cos(ωt). (2.13)
m m m
We have solved the associated homogeneous equation in all cases on c, k, and m. For the general
solution of equation (2.13), we need only a particular solution. Application of the method of
undetermined coefficients yields the particular solution
2
mA(k − mω )
Y p (t) = cos(ωt)
(k − mω ) + ω c
2 2
2 2
Aωc
+ sin(ωt).
(k − mω ) + ω c
2 2
2 2
√
It is customary to denote ω 0 = k/m to write
2
2
mA(ω − ω )
Y p (t) = 0 cos(ωt)
2
2
2 2
2 2
m (ω − ω ) + ω c
0
Aωc
+ sin(ωt).
2
2 2
2
m (ω − ω ) + ω c
2 2
0
We will analyze some specific cases to get some insight into the motion with this forcing function.
Case 1: Overdamped Forced Motion
√ √
2
Overdamping occurs when c − 4km > 0. Suppose c = 6,k = 5, m = 1 A = 6 5 and ω = 5.
If the bob is released from rest from the equilibrium position, then y(t) satisfies the initial value
problem
√ √
y + 6y + 5y = 6 5cos( 5t); y(0) = y (0) = 0
The solution is
√
5 √
−t
y(t) = (−e + e −5t ) + sin( 5t).
4
A graph of this solution is shown in Figure 2.7. As time increases, the exponential terms decay to
√
zero, and the displacement behaves increasingly like sin( 5t), oscillating up and down through
√
the equilibrium point with approximate period 2π/ 5. Contrast this with the overdamped motion
without the forcing function in which the bob began above the equilibrium point and moved with
decreasing speed down toward it but never reached it.
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October 14, 2010 14:12 THM/NEIL Page-66 27410_02_ch02_p43-76