2.3 Response to Harmonic Excitation
Many systems in mechanical engineering are subject to harmonic excitation. Sources of harmonic
excitation include:



automobile engines
aircraft propellers + turbine engines
unbalanced tires, gears, linkages, etc.
Studying harmonic excitation will lead us to one of the most important phenomena in vibration:
resonance.
Let us begin by studying harmonic excitation of an undamped system.
Harmonic Excitation of Undamped Systems
k
m
f(t)
Assume the mass/spring system is excited by the force f, where
𝑓(𝑡) = 𝐹 cos 𝜔𝑡
Excitation frequency = ω = property of the excitation
Natural frequency
= ωn = property of mass/spring system
Doing a force balance on the system leads to
𝑚𝑥̈ + 𝑘𝑥 = 𝐹 cos 𝜔𝑡
This is a second-order, non-homogeneous differential equation. The solution will be a combination
of the homogeneous solution (from earlier) and a particular solution.
Let us assume that the particular solution takes the same form as the forcing function.
𝑥𝑝 = 𝑋 cos 𝜔𝑡
𝑥̇ 𝑝 = −𝜔𝑋 sin 𝜔𝑡
𝑥̈ 𝑝 = −𝜔2 𝑋 cos 𝜔𝑡
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Substitute this into the equation of motion
−𝑚𝜔2 𝑋 cos 𝜔𝑡 + 𝑘𝑋 cos 𝜔𝑡 = 𝐹 cos 𝜔𝑡
−𝑚𝜔2 𝑋 + 𝑘𝑋 = 𝐹
Divide through by m
𝑘
𝐹
𝑋=
𝑚
𝑚
𝐹
2
2
(𝜔𝑛 − 𝜔 )𝑋 =
𝑚
−𝜔2 𝑋 +
Finally, solving for X
𝑋=
𝐹
𝑚(𝜔𝑛2 − 𝜔 2 )
𝑥𝑝 =
𝐹 cos 𝜔𝑡
𝑚(𝜔𝑛2 − 𝜔 2 )
Thus, the particular solution is
Now, let us rewrite the homogeneous part of the solution as
𝑥ℎ = 𝑎1 sin 𝜔𝑛 𝑡 + 𝑎2 cos 𝜔𝑛 𝑡
in the same way we did for Homework problem 1.7. Thus, the total solution is
𝑥 = 𝑥𝑝 + 𝑥ℎ
𝐹 cos 𝜔𝑡
=
+ 𝑎1 sin 𝜔𝑛 𝑡 + 𝑎2 cos 𝜔𝑛 𝑡
𝑚(𝜔𝑛2 − 𝜔 2 )
Taking the time derivative, we can find the velocity
𝑣=
−𝜔𝐹 sin 𝜔𝑡
+ 𝑎1 𝜔𝑛 cos 𝜔𝑛 𝑡 − 𝑎2 𝜔𝑛 sin 𝜔𝑛 𝑡
𝑚(𝜔𝑛2 − 𝜔 2 )
To find a1 and a2, we need to enforce the initial conditions.
𝑥(0) = 𝑥0 =
𝐹
+ 𝑎2
− 𝜔2)
𝑚(𝜔𝑛2
Thus,
𝑎2 = 𝑥0 −
𝐹
− 𝜔2)
𝑚(𝜔𝑛2
2
also
𝑣(0) = 𝑣0 = 𝑎1 𝜔𝑛
thus
𝑎1 =
𝑣0
𝜔𝑛
The complete solution is then
𝑥=
𝐹
𝑣0
(cos
𝜔𝑡
−
cos
𝜔
𝑡)
+
sin 𝜔𝑛 𝑡 + 𝑥0 cos 𝜔𝑛 𝑡
𝑛
𝑚(𝜔𝑛2 − 𝜔 2 )
𝜔𝑛
Forced Response
Initial conditions
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Cases to Consider
1. The natural frequency is much greater than the excitation frequency.
0.4
0.2
x( t )
0
0.2
0.4
0
2
4
6
t
1/f
8
10
1/f n
2. The excitation frequency is much greater than the natural frequency.
0.4
0.2
x( t )
0
0.2
0.4
0
2
1/f
4
6
t
8
10
1/f n
Both cases behave the same, even though the cause of the behavior is quite different!
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3. The excitation frequency is close to (but not equal to) the natural frequency
4
2
x( t )
0
2
4
0
5
10
15
20
t
Here we see the phenomenon of “beating”. The angular frequency of the beats is seen to be
𝜔𝑏𝑒𝑎𝑡 =
|𝜔𝑛 − 𝜔|
2
𝑓𝑏𝑒𝑎𝑡 =
|𝜔𝑛 − 𝜔|
4𝜋
so that the frequency is
Why is this? Let us assume for the moment that both initial conditions are zero. Then we are left
with only the forced response:
𝑥=
𝐹
(cos 𝜔𝑡 − cos 𝜔𝑛 𝑡)
− 𝜔2)
𝑚(𝜔𝑛2
Using a half-angle identity, we can write this as
𝑥=
2𝐹
𝜔𝑛 − 𝜔
𝜔𝑛 + 𝜔
sin (
𝑡) sin (
𝑡)
2
−𝜔 )
2
2
𝑚(𝜔𝑛2
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The first sine function provides a low-frequency “envelope” that contains the second, highfrequency sine function. The high-frequency component vibrates at the average of the excitation
and natural frequencies.
4. The natural frequency is the same as the excitation frequency
In the preceding equation, we note that as the two frequencies approach each other, the beat
frequency becomes very low (the beats last longer) and the amplitude becomes large.
|𝑥| =
2𝐹
→∞
𝑚(𝜔𝑛2 − 𝜔 2 )
In the limit, as ω → ωn, the beat period is of infinite length and the magnitude increases without
bound (or at least until the destruction of the system)!
20
10
x( t )
0
10
20
0
2
4
6
8
10
t
This phenomenon is known as “resonance”.
If we drive a system at its natural frequency, we say we are driving it at resonance.
Good resonance phenomena
 wind-driven musical instruments
 acoustic compressors
 thermoacoustic devices
Bad resonance phenomena
 wind effects on buildings, bridges
 just about anything not related to music!
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