2.4 Time Domain Analysis
29
x p (t) = α sin
ν t
+ β cos
ν t
, t ≥ 0,
(2.46)
where α and β are constants which need to be determined. The equation of motion
with x p (t) in place of x(t) has the form
· · ·
sin
ν t
+
· · ·
cos
ν t
=
q/m
sin(ν t),
where the square brackets depend on α and β. For x p (t) to be a particular solution,
the first and second square brackets must be equal to q/m and zero. These conditions
give a system of equations for α and β. Following are some computational details.
The equation of motion with x p (t) in place of x(t) gives
− α ν
2 cos(ν t) − β ν
2 sin(ν t) + 2 ζ ω
− α ν sin(ν t) + β ν cos(ν t)
+ ω
2
α cos(ν t) + β sin(ν t)
= (q/m) sin(ν t) or, equivalently,
− α ν
2
+ 2 ζ ω β ν + α ν
2
cos(ν t) +
− β ν
2
− 2 ζ ω α ν + β ν
2
sin(ν t) = (q/m) sin(ν t).
Since the trial x p (t) has to satisfy the equation of motion at all times, we require
− α ν
2
+ 2 ζ ω β ν + α ν
2
= 0 and
− β ν
2
− 2 ζ ω α ν + β ν
2
= q/m.
These conditions define a linear system of equations for the unknown coefficients α
and β. The solution of this system gives the particular solution of Eq. 2.44.
Similar calculations show that the particular solution for the forcing function
f (t) = q cos(ν t) has the expression
x p (t) = x st r d (ν) cos(ν t − ϕ)
(2.47)
with the notations of Eqs. 2.44 and 2.45 (see Problem 2.5).
We conclude that the particular solutions for the forcing functions f (t) =
sin(ν t) and f (t) = cos(ν t) are
f (t) = sin(ν t) ⇒ x p (t) =
r d (ν)/ω
2
sin(ν t − ϕ)
f (t) = cos(ν t) ⇒ x p (t) =
r d (ν)/ω
2
cos(ν t − ϕ)
(2.48)
since sin(ν t) = (m/q)
(q/m) sin(ν t)
and the system is linear so that the
particular solution of Eq. 2.45 can be scaled by (q/m), which gives
(m/q)
x st r d (ν) sin(ν t − ϕ)
= (m/q) (q/k) r d (ν) sin(ν t − ϕ)
= (1/ω
2 ) r d (ν) sin(ν t − ϕ).