Solutions 2.6-Page 167
Problem 17
The problem gives the parameters for a forced mass-spring-dashpot system with equation
mx ′′ + cx ′ + kx = F0 cos ωt . Investigate the possibility of practical resonance of this
system. In particular, find the amplitude C (ω ) of steady periodic forced oscillations with
frequency ω . Sketch the graph of C (ω ) and find the practical resonance frequency ω
(if any).
m = 1, c = 6, k = 45, F0 = 50
The amplitude of steady periodic forced oscillation is given by eq.21 on pg.165.
C (ω ) =
C (ω ) =
F0
(k − mω 2 ) 2 + (cω ) 2
=
50
(45 − (1)ϖ 2 ) 2 + (6ω ) 2
=
50
(ω 4 − 90ω 2 + 2025) + 36ω 2
50
ω 4 − 54ω 2 + 2025
The maximum amplitude (practical resonance) occurs when C ′(ω ) = 0 .
C ′(ω ) = 50(−1 / 2)(ω 4 − 54ω 2 + 2025) −3 / 2 (4ω 3 − 108ω ) =
Thus
ω = 27 = 3 3 = 5.196 rad/s
− 100ω (ω 2 − 27)
=0
(ω 4 − 54ω 2 + 2025) 3 / 2
Problem 25
Derive the steady periodic solution of
mx ′′ + cx ′ + kx = F0 sin ωt ,
In particular, show that it is what one would expect-the same as the formula in (20) with
the same values of C and ω , except with sin(ωt − α ) in place of cos(ωt − α ) .
The assumed form of the particular solution will be x sp = A cos ωt + B sin ωt .
As stated on pg.165, the solution of a damped system, whether underdamped, critically
damped, or overdamped will not contain any of the terms in the assumed particular
solution. The necessary derivatives for substitution into the differential equation are as
follows:
x ′sp = ωB cos ωt − ωA sin ωt
x ′sp′ = −ω 2 A cos ωt − ω 2 B sin ωt
Substituting into the differential equation yields:
m(−ω 2 A cos ωt − ω 2 B sin ωt )m + c(ωB cos ωt − ωA sin ωt ) + k ( A cos ωt + B sin ωt ) = F0 sin ωt
∴
(1) cωB − mω 2 A + kA = (k − mω 2 ) A + cωB = 0
(2) kB − mω 2 B − cωA = −cωA + (k − mω 2 ) B = F0
− cωB
(k − mω 2 )
From (2) A =
This value is substituted in (2)
c 2ω 2 B
+ (k − mω 2 ) B = F0
2
(k − mω )
((k − mω 2 ) 2 + (cω ) 2 ) B
= F0
(k − mω 2 )
∴A=
(k − mω 2 ) F0
− cωF0
and
B
=
(k − mω 2 ) 2 + (cω ) 2
(k − mω 2 ) 2 + (cω ) 2
There is trigonometric identity that states
A
~
~
~
~
A cos x + B sin x = A 2 + B 2 sin( x ± δ ) , tan δ = ±
. Thus the steady periodic solution
B
can be written in the form C sin(ωt − α ) instead of C cos(ωt − α ) where ω would be the
same in both cases.
In this case,
C=

− cωF0
A + B = 
2 2
2
 (k − mω ) + (cω )
C=
(cωF0 ) 2 + (k − mω 2 ) 2 F0
=
((k − mω 2 ) 2 + (cω ) 2 ) 2
2
2
2
2
 
(k − mω 2 ) F0
 + 
2 2
2
  (k − mω ) + (cω )
F0 ((k − mω 2 ) 2 + (cω ) 2 )
=
((k − mω 2 ) 2 + (cω ) 2 ) 2
2
Therefore C is the same in both cases also.



2
F0
(k − mω 2 ) 2 + (cω ) 2
Problem 30
Problems 29 and 30 deal further with the car of Example 5. Its upward displacement
function satisfies the equation mx ′′ + cx ′ + kx = cy ′ + ky when the shock absorber is
connected (so that c > 0). With y = a sin ωt for the road surface, this differential
equation becomes
mx ′′ + cx ′ + kx = E 0 cos ωt + F0 sin ωt
where E 0 = cωa and F0 = ka .
Figure 2.6.12 shows the graph of the amplitude function C (ω ) using the numerical data
given in Example 5 (including c = 3000 N*s/m). It indicates that, as the car accelerates
gradually from rest, it initially oscillates with amplitude slightly over 5 cm. Maximum
resonance vibrations with amplitude about 14 cm occur around 32 mi/h, but then subside
to more tolerable levels at high speeds. Verify these graphically based conclusions by
analyzing the function C (ω ) . In particular, find the practical resonance frequency and
the corresponding amplitude.
The result of Problem 29 gives the amplitude function.
a k 2 + (cω ) 2
(0.05) (70000) 2 + (3000ω ) 2
C (ω ) =
=
(k − mω 2 ) 2 + (cω ) 2
((7 x 10 4 ) − 800ω 2 ) 2 + (3000ω ) 2
after simplifying
C (ω ) =
0.25 9ω 2 + 4900
16ω 4 − 2575 ω 2 + 122500
The maximum amplitude (practical resonance) occurs when C ′(ω ) = 0 .
C ′(ω ) = (.25)(1 / 2)(9ω 2 + 4900) −1 / 2 (18ω )(16ω 4 − 2575ω 2 + 122500) −1 / 2
+ (.25)(−1 / 2)(9ω 2 + 4900)1 / 2 (16ω 4 − 2575ω 2 + 122500) −3 / 2 (64ω 3 − 10300ω )
after simplifying
C ′(ω ) =
− 36ω (ω 4 + 1088.89ω 2 − 95277.8)
(
(9ω 2 + 4900) 16ω 4 − 2575ω 2 + 122500 3 / 2
)
Practical resonance will occur at the roots of the equation ω 4 + 1088.89ω 2 − 95277.8 .
The equation can be rewritten in the form (ω 2 ) 2 + 1088.89(ω 2 ) − 95277.8 . Now the
quadratic equation can be used to find the roots.
− 1088.89 ± (1088.89) 2 − 4(−95277.8) − 1088.89 ± 1251.72
=
= −544.445 ± 625.858
2
2
ω 2 = 81.4129 , − 1170.3
∴
ω2 =
ω = ± 81.4129 , ± − 1170.3
ω = ±9.02 , ± 34.2i
Only the real values of the practical resonance are of any use. You cannot have an
imaginary circular frequency.
Therefore ω = ±9.02
Substituting ω = ±9.02 into C (ω ) will give the amplitude at practical resonance. The
result is:
C (±9.02) = .136 m = 13.6 cm