Response Of Linear SDOF
Systems To Harmonic Excitation
Many dynamic systems are subjected to
harmonic (sinusoidal) excitation. Rotating
machinery is an example.
Objectives
• Learn how o find response to harmonic
excitation.
• Understand concept of resonance
Undamped systems
• Response = harmonic wave with
frequency ωn +harmonic wave with
frequency of excitation
• The first component is due to initial
conditions plus the force.
• The second component is due to the force
General equation for response to
force P sin(t )
x(t )  x (0) cos( nt )  [

P
2
2
m( n   )
sin(t )
x (0)
n

P
m n ( n2   2 )
] sin( nt )
Harmonic Response Of Undamped System
natural frequency=1 rad/sec, excitation frequency=2 rad/sec,
x(0)=0.01 m, xd(0)=0.01
.
0.1
x( t)
0
xd( t)
0.1
0
10
20
t
30
40
Harmonic Response Of Undamped System
natural frequency= 1 rad/sec, excitation frequency=0.95 rad/sec
t  0  0.01  250
zero initial displacement and velocity
.
2
x( t)
0
2
0
50
100
150
t
200
250
Harmonic Response Of Undamped System
natural frequency= 1 rad/sec, excitation frequency=0.9999 rad/sec
zero initial displacement and velocity
t  0  0.01  50 0
.
50
x( t)
0
50
0
100
200
300
t
400
500
Harmonic Response Of Undamped System
1 rad/sec, excitation frequency=20 rad/sec
t  0natural
 0.01 frequency=
10
zero initial displacement and velocity
.
0.01
x( t)
0
0
2
4
6
t
8
10
Observations
• Response=harmonic wave with frequency ωn
+harmonic wave with frequency of excitation
• When excitation frequency is almost equal to
natural frequency, vibration amplitude is very large.
Rapid oscillation with slowly varying amplitude.
• When excitation frequency is equal to the systems
natural frequency, vibration amplitude=  . This
condition is called resonance.
• When excitation frequency>>natural frequency,
vibration amplitude is very small. The reason is that
the system is too slow to follow the excitation.
Damped systems
x(t )  xc (t )  x p (t )
Free vibration
response (also called
transient response)
Particular
response (steadystate response)
• Use equations that we learnt in the chapter for free vibration
response for transient response. Transient dies out with time.
• Response converges to particular response with time.
Steady-state response
• We often ignore transient response and focus
on steady-state response.
• Steady-state is response is solution of
equation of motion:
x  2 n x   n2 x 
F
sin(t )
m
• From experience we know that:
x(t )  A cos(t )  B sin(t )
• Find coefficients A and B by substituting
assumed equation for response in
equation of motion and solving for these
coefficients.
• Sine-in, sine-out property: Steady-state
response to sine wave is also sine wave
with same frequency as the excitation.
Amplitude of response equal to quasistatic response times dynamic
amplification factor. This is true for any
value of damping ratio
x (t )  A sin(t   )
F
k
A
 2 2
 2
2
(1  ( ) )  4 ( )
n
n

2
n

1
   tan (
)
 2
1 ( )
n
Natural frequency 5 rad/sec,
excitation frequency, 1 rad/sec
1.041
1.5
1
0.5
x ( t  1)
p ( t  1)
0
0.5
1
 1.041 1.5
0
0
2
4
6
t
8
10
10
Natural frequency 5 rad/sec,
excitation frequency, 5 rad/sec
5
6
4
2
x ( t  5)
p ( t  5)
0
2
4
5
6
0
0
2
4
6
t
8
10
10
Natural frequency 5 rad/sec,
excitation frequency, 20 rad/sec
1
1
0.5
x ( t  20)
p ( t  20)
0
0.5
1
1
0
0
0.5
1
t
1.5
2
2
Effect of frequency and damping on amplitude
Natural freq.=5 rad/sec. Amplitude of excitation=1.
5.001
A    0.1
A    0.5
A    2
6
4
=0.1
=0.5
2
=0.5
0.046
0
0
0
5
10

15
20
20
Effect of frequency and damping on phase
angle.
Natural freq.=5 rad/sec. Amplitude of excitation=1.
0
0
    0.1
    0.5 1.57
    2
 3.14 3.14
0
0
5
10

15
20
20
Observations
•
•
•
•
•
•
Response amplitude depends only on damping
ratio, , and ratio of frequency of excitation over
natural frequency, /n.
/n small, response amplitude = F/k (quasi-static
response)
/n much greater than one, response amplitude is
close to zero
/n =1, and  small (less than 0.1), response
amplitude very large. If =0, response amplitude =

2





1

2

Response amplitude is maximum for
m
n
(resonant frequency).
To reduce response amplitude a) change /n so
that it is as far away from 1 as possible, or b)
increase damping.
Observations (continued)
• Phase angle is always negative.
• For small /n, phase angle is almost
zero.
• For /n =1, phase angle is -900.
• For large /n phase angle tends to
-1800.