HW 18
March 6, 2013
3.8.14. Find the velocity of the steady state response given by Eq. (10).
Then show that the velocity is a maximum when ω = ω0 .
Solution: Define Γ = γ 2 /mk. Then the velocity V (ω) is given by
V (ω) = R(ω)ω =
k 1−
F0 ω
2
ω2
ω02
+
2
Γ ωω2
0
.
To show that the maximum velocity occurs when ω = ω0 , it suffices to
differentiate V (ω) and set the resulting expression equal to zero, i.e.
F0
1−
V 0 (ω) =
ω2
ω02
2
+
2
Γ ωω2
0
k
1+
1 − ω2
F0
=
k 1−
ω2
0
ω2
ω02
2
+
− F0 ω2
1−
2
ω2
ω02
h
2
ω2
ω02
1−
2
+ Γ ωω2
i
−2 ωω2 + 2Γ ωω2
0
0
3/2
0
ω
ω02
2
Γ ωω2
0
3/2
= 0.
The unique nonnegative solution is ω = ω0 .
3.8.17 Consider a vibrating system described by the initial value problem
1
u00 + u0 + 2u = 2 cos ωt,
4
u(0) = 0, u0 (0) = 2.
1
(a) Determine the steady state part of the solution of this problem.
Posit a solution of the form u(t) = A cos ωt + B sin ωt and substitute into
the i.v.p. to see
− Aω 2 cos ωt − Bω 2 sin ωt
1
+
[−Aω sin ωtBω cos ωt]
4
+ 2 [A cos ωt + B sin ωt]
= 2 cos ωt.
Therefore, we need to solve
Bω
+ 2A = 2
4
Aω
−Bω 2 −
+ 2B = 0.
4
−Aω 2 +
This system has the solution
2(2 − ω 2 )
ω 2 /16 + (2 − ω 2 )2
ω/2
B =
.
2
ω /16 + (2 − ω 2 )2
A =
Therefore, the steady state is
ω/2
2(2 − ω 2 )
u(t) = 2
cos
ωt
+
sin ωt.
ω /16 + (2 − ω 2 )2
ω 2 /16 + (2 − ω 2 )2
(b) The amplitude of the steady state is
s
p
A(ω) = A(ω)2 + B(ω)2 = A(ω) 1 +
= q
2
ω2
16(2 − ω 2 )2
2
ω2
16
+ (2 − ω 2 )2
.
(c) Plot A(ω). This function starts at A(0) = 1 and then increases until
1
2
, whereupon it decreases to zero in the limit x → ∞.
it reaches ωmax
= 2 − 32
3.8.18. Consider the forced but undamped system described by the i.v.p.
u00 + u = 3 cos ωt, u(0) = u0 (0) = 0.
(a) Find the solution u(t) for ω 6= 1.
The homogeneous solutions to u00 + u = 0 take the form
C1 sin t + C2 cos t.
Now posit a particular solution to the inhomogenous problem of the form
u(t) = A sin ωt + B cos ωt. Substitute into the i.v.p. to deduce
−Aω 2 + A = 0
−Bω 2 + B = 3.
As ω 6= 1, conclude A = 0 and B =
3
.
1−ω 2
u(t) = C1 sin t + C2 cos t +
Our solution now looks like
3
cos ωt.
1 − ω2
The initial conditions ensure u(0) = C2 +
Conclude with
3
1−ω 2
= 0 and u0 (0) = C1 = 0.
1+ω
1−ω
3
6
sin
u(t) =
(cos ωt − cos t) = −
t sin
t .
1 − ω2
1 − ω2
2
2
6
(b) The amplitude A(ω) = |1−ω
2 | has a singularity at ω = 1.
(c) Plot A(ω). Straightforward.
3