Resonance and Harmonic Excitation

Chapter 2 Response to Harmonic
Excitation

Introduces the important
concept of resonance

© Eng. Vib, 3rd Ed.
1/49

@ProfAdhikari, #EG260

2.1 Harmonic Excitation of Undamped
Systems
• Consider the usual spring mass damper system
with applied force F(t)=F0cosωt
•
•

ω is the driving frequency
F0 is the magnitude of the applied force

• We take c = 0 to start with
Displacement
x
F=F0cosωt

2/49

College of Engineering

M

k

Equations of motion
Figure 2.1
• Solution is the sum of
homogenous and
particular solution
• The particular
solution assumes
form of forcing
m = −kx(t) + F0 cos(ω t)
x(t)
function (physically
2
the input wins):
 + ω n x(t) = f0 cos(ω t)
x(t)
x p (t ) = X cos(ωt )

F0
where f0 =

3/49


m

,

ωn = k

m

Substitute particular solution into
the equation of motion: x (t ) = X cos ( ωt )
p

p
x

2
ωn x p

  
 
2
−ω 2 X cosω t + ω n X cosω t = f0 cos ω t
f0
solving yields: X = 2
ωn − ω 2
Thus the particular solution has the form:

f0
x p (t ) = 2
cos(ωt )
2
ωn − ω
4/49


Add particular and homogeneous
solutions to get general solution:
x(t) =
particular
 


f0
A1 sin ω n t + A2 cos ω n t + 2
cosω t
2
 

 ω −ω
n
homogeneous

A1 and A2 are constants of integration.

5/49


Apply the initial conditions to
evaluate the constants
f0
f
cos 0 = A2 + 2 0 2 = x0
2
ωn − ω 2
ωn − ω
f
⇒ A2 = x0 − 2 0 2
ωn − ω

x(0) = A1 sin 0 + A2 cos0 +


x(0) = ω n (A1 cos0 − A2 sin 0) −

f0
sin 0 = ω n A1 = v0
2
2
ωn − ω

⇒ A1 =

v0
ωn

⇒


v0
f0 
f
x(t) = sin ω n t +  x0 − 2
cos ω n t + 2 0 2 cos ω t
2÷
ωn
ωn − ω 
ωn − ω

6/49


(2.11)

Comparison of free and forced
response
• Sum of two harmonic terms of different frequency
• Free response has amplitude and phase effected
by forcing function
• Our solution is not defined for ωn = ω because it
produces division by 0.
• If forcing frequency is close to natural frequency
the amplitude of particular solution is very large

7/49


Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5
v0=0.1m/s and x0= -0.02 m.

Displacement (x)

0.05

0

-0.05

0

2

4
6
Time (sec)

8

10

Note the obvious presence of two harmonic signals
8/49


Go to code demo

What happens when ω is near ωn?
x (t ) =

2 f0
 ω − ω   ωn + ω 
sin  n
t ÷sin 
t ÷ (2.13)
2
2
ωn − ω
 2
  2


When the drive frequency and natural frequency
are close a beating phenomena occurs

Displacement (x)

1

2 f0
 ω −ω 
sin  n
t÷
2
2
ωn − ω
 2


0.5
0

Larger
amplitude

-0.5
-1
0

9/49

5

10

15
20
Time (sec)

25

30

Called beating

What happens when ω is ωn?
x p (t ) = tX sin(ωt )
grows without bound

substitute into eq. and solve for X
X=


f
x(t) = A1 sin ω t + A2 cos ω t + 0 t sin(ω t)
2ω

f0
2ω

When the drive
frequency and natural
frequency are the
same the amplitude
of the vibration grows
without bounds. This
is known as a
resonance condition.
The most important
concept in Chapter 2!
10/49

Displacement (x)

5

0

-5
0

5

10

15
Time (sec)


20

25

30

Example 2.1.1: Compute and plot the response for
m=10 kg, k=1000 N/m, x0=0,v0=0.2 m/s, F=23 N, ω =2 ω n.
k
1000 N/m
ωn =
=
= 10 rad/s, ω = 2ω n = 20 rad/s
m
10 kg
F 23 N
v0 0.2 m/s
f0 = =
= 2.3 N/kg,
=
= 0.02 m
m 10 kg
ω n 10 rad/s
f0
2.3 N/kg
=
= −7.9667 ×10 −3 m
2
ω n − ω 2 (10 2 − 20 2 ) rad 2 / s2
Equation (2.11) then yields:
x(t) = 0.02sin10t + 7.9667 ×10−3 (cos10t − cos20t)
11/49


Example 2.1.2 Given zero initial conditions a
harmonic input of 10 Hz with 20 N magnitude and k= 2000 N/m,
and measured response amplitude of 0.1m, compute the mass
of the system.

f0
cos20π t − cosω n t ) for zero initial conditions
2
2 (
ωn − ω
 ωn − ω   ωn + ω 
2 f0
trig identity ⇒ x(t) = 2
sin 
t ÷sin 
t÷
2
 2
  2

ωn − ω

 
x(t) =

0.1 m

⇒

2 f0
2(20 / m)
= 0.1 ⇒
= 0.1
2
2
2
2000 − (20π )
ωn − ω
m
m = 0.45 kg

12/49

(

)


Example 2.1.3 Design a rectangular
mount for a security camera.

0.02 x 0.02 m
in cross section

Compute l > 0.5 m so that the mount keeps the camera from
vibrating more then 0.01 m of maximum amplitude under a wind
load of 15 N at 10 Hz. The mass of the camera is 3 kg.
13/49


Solution:Modeling the mount and camera as a
beam with a tip mass, and the wind as harmonic, the
equation of motion becomes:
3EI
m +
x

From strength of materials:



3

x(t) = F0 cosω t

bh 3
I=
12

Thus the frequency expression is:

3Ebh 3 Ebh 3
2
ωn =
=
3
12m 4m3

Here we are interested computing  that will make the
amplitude less then 0.01m:

(a)
2 f0

< 0.01 ⇒ 
2
ωn − ω 2
 (b)



14/49

− 0.01 <

2 f0
2
, for ωn − ω 2 < 0
2
ωn − ω 2

2 f0
2
< 0.01, for ωn − ω 2 > 0
2
ωn − ω 2


Case (a) (assume aluminum for the material):
2 f0
Ebh3
2
−0.01 < 2
⇒ 2 f 0 < 0.01ω 2 − 0.01ωn ⇒ 0.01ω 2 − 2 f 0 > 0.01
ωn − ω 2
4m3
Ebh3
⇒  > 0.01
= 0.321 ⇒  > 0.6848 m
2
4m(0.01ω − 2 f 0 )
3

Case (b):
2 f0
Ebh3
2
2
2
< 0.01 ⇒ 2 f 0 < 0.01ωn − 0.01ω ⇒ 2 f 0 + 0.01ω < 0.01
2
2
ωn − ω
4m3
Ebh3
⇒  < 0.01
= 0.191 ⇒  < 0.576 m
2
4m(2 f 0 + 0.01ω )
3

15/49


Remembering the constraint that the length
must be at least 0.5 m, (a) and (b) yield

0.5 <  < 0.576, or  > 0.6848 m
Less material is usually desired, so chose case
b, say  = 0.55 m.
To check, note that
3Ebh3
2
2
2
ωn − ω =
− ( 20π ) = 1742 > 0
3
12m

Thus the case a condition is met.
Next check the mass of the designed beam to
insure it does not change the frequency. Note
it is less then m.
16/49

m = ρ bh
= (2.7 × 103 )(0.55)(0.02)(0.02)
= 0.594 kg


A harmonic force may also be represented by
sine or a complex exponential. How does this
change the solution?
m + kx(t) = F0 sin ω t or  + ω n x(t) = f 0 sin ω t
x(t)
x(t) 2

(2.18)

The particular solution then becomes a sine:
x p (t ) = X sin ωt

(2.19)

Substitution of (2.19) into (2.18) yields:
x p (t ) =

f0
sin ωt
2
2
ωn − ω

Solving for the homogenous solution and evaluating the constants yields

v
f0 
f
ω
x(t ) = x0 cos ωnt +  0 −
sin ωnt + 2 0 2 sin ωt (2.25)
÷
ωn ωn ωn2 − ω 2 
ωn − ω

17/49


Section 2.2 Harmonic Excitation of
Damped Systems

Extending resonance and response
calculation to damped systems

18/49


2.2 Harmonic excitation of
damped systems
m + cx (t) + kx(t) = F0 cos ω t
x(t) 
 + 2ζω n x (t) + ω x(t) = f0 cosω t

x(t)
x p (t) = cos(ω)
X   t −
θ

2
n

now includes a phase shift

Displacement
x
F=F0cosωt

k

M
c

19/49


Let xp have the form:

x p (t) = As cos ω t + Bs sin ω t
 Bs 
2
2
−1
X = As + Bs , θ = tan  ÷
 As 

x p = −ω As sin ω t + ω Bs cosω t

p = −ω As cosω t − ω Bs sin ω t
x
2

2

Note that we are using the rectangular form, but
we could use one of the other forms of the solution.
20/49


Substitute into the equations of motion
(−ω As + 2ζωnω Bs + ω As − f 0 ) cos ωt
2

2
n

+ ( −ω Bs + 2ζωnω As + ω Bs ) sin ωt = 0
2

2
n

for all time. Specifically for t = 0, 2π / ω ⇒
(ω − ω ) As + (2ζωnω ) Bs = f 0
2
n

2

(−2ζωnω ) As + (ω − ω ) Bs = 0
2
n

21/49

2


Write as a matrix equation:
2
(ωn − ω 2 ) 2ζωnω   As   f0 

= 
2
2 
 −2ζωnω (ωn − ω )   Bs   0 

Solving for As and Bs:

(ω − ω ) f0
As = 2
2
(ωn − ω ) + (2ζωnω )
2
n
2 2

2

2ζωnω f0
Bs = 2
(ωn − ω 2 )2 + (2ζωnω )2
22/49


Substitute the values of As and Bs into xp:
 2ζω nω 
x p (t) =
cos(ω t − tan  2
)
2÷
2
2 2
2
 ωn − ω 
(ω n − ω ) + (2ζω nω )
 


 


f0

−1

θ

X

Add homogeneous and particular to get total solution:

x(t) = Ae−ζω nt sin(ω d t + φ ) + cos(ω)
X   t −
θ

 


homogeneous or transient solution

particular or
steady state solution

Note: that A and φ will not have the same values as in Ch 1,
for the free response. Also as t gets large, transient dies out.
23/49


Things to notice about damped forced
response
• If ζ = 0, undamped equations result
• Steady state solution prevails for large t
• Often we ignore the transient term (how
large is ζ, how long is t?)
• Coefficients of transient terms (constants
of integration) are effected by the initial
conditions AND the forcing function
• For underdamped systems at resonance,
the amplitude is finite.
24/49


Example 2.2.1: ωn = 10 rad/s, ω = 5 rad/s, ζ = 0.01, F0= 1000 N, m = 100
kg, and the initial conditions x0 = 0.05 m and v0 = 0. Compare A and φ
for forced and unforced case:
Using the equations on slide 23:

X = 0.133, θ = −0.013
x (t ) = Ae −0.1t sin(9.99t + φ ) + 0.133cos(5t − 0.013)
Differentiating yields:

v(t ) = − 0.01Ae −0.1t sin(9.999t + φ )
+ 9.999 Ae −0.1t cos(9.999t + φ )
− 0.665sin(5t − 0.013)
applying the intial conditions :
A = − 0.083 (0.05), φ = 1.55 (1.561)
25/49


x0 0 5
( )= .
0
= sn + . 3c s−. 1)
Ai φ 0 3o( 0 3
1
0
v0 0 −. 1 sn
( )= = 0 Ai φ
0
+ . 9Aoφ 0 6 sn . 1
9 9 c s + . 5i 0 3
9
6
0

The numbers in ( ) are
those obtained by
incorrectly using the free
response values

Proceeding with ignoring the transient
• Always check to make sure the
transient is not significant
• For example, transients are very
important in earthquakes
• However, in many machine
applications transients may be
ignored
26/49


Proceeding with ignoring the transient
Magnitude:

X=

f0
2
(ωn − ω 2 )2 + (2ζωnω )2

Frequency ratio:

ω
r=
ωn

Non dimensional
Form:

(2.39)

2
Xk Xωn
1
=
=
F0
f0
(1 − r 2 )2 + (2ζ r )2

Phase:
27/49

(

θ = tan −1 2ζ r

1 − r2

)

(2.40)

Magnitude plot

•
•

•

Resonance is close to
r=1
For ζ = 0, r =1 defines
resonance
As ζ grows
resonance moves r
<1, and X decreases
The exact value of r,
can be found from
differentiating the
magnitude

X=

30
20
10
0
-10
-20
0

28/49

(1 − r 2 ) 2 + (2ζ r )2

40

X (dB)

•

1

0.5


1
r

1.5

Fig 2.7

2

Phase plot

•

•

Resonance occurs
at Φ = π/2
The phase changes
more rapidly when
the damping is
small
From low to high
values of r the
phase always
changes by 1800 or
π radians

(

1− r2

)

3.5

ζ =0.01
ζ =0.1
ζ =0.3
ζ =0.5
ζ =1

3
2.5

Phase (rad)

•

θ = tan −1 2ζ r

2
1.5
1
0.5
0

0.5


1

1.5

r
29/49

0

Fig 2.7

2

Example 2.2.3 Compute max peak
by differentiating:
d  Xk  d 
1

÷= 
dr  F0  dr  (1 − r 2 )2 + (2ζ r ) 2



÷= 0 ⇒
÷


rpeak = 1 − 2ζ < 1 ⇒ ζ < 1 / 2

(2.41)

 Xk 
1

÷ =
F0  max 2ζ 1 − ζ 2


(2.42)

2

30/49


Effect of Damping on Peak Value
30

• The top plot shows how
the peak value becomes
very large when the
damping level is small
• The lower plot shows
how the frequency at
which the peak value
occurs reduces with
increased damping
• Note that the peak value
is only defined for
values ζ<0.707

25

Xk
(dB)
F0

20
15
10
5
0

0

0.2

0.4

0

0.2

0.4

ζ

0.6

0.8

0.6

0.8

1
0.8

rpeak

0.6
0.4
0.2
0

31/49


ζ

Fig 2.9

Section 2.3 Alternative
Representations
• A variety methods for solving differential
equations
• So far, we used the method of undetermined
coefficients
• Now we look at 2 alternatives:
a frequency response approach
a transform approach
• These also give us some insight and additional
useful tools.

32/49


2.3.2 Complex response method
Ae

jω t

= t + (Asin ω t) j
A cosω  
   
real part

(2.47)

imaginary part

jω t

m + cx (t) + kx(t) = F0 e
x(t) 


(2.48)

harmonic input

Real part of this complex solution corresponds to
the physical solution
35/49


Choose complex exponential as a
solution
x p (t) = Xe

jω t

(2.49)

(−ω m + cjω + k)Xe
2

jω t

= F0 e

jω t

F0
X=
= H ( jω )F0
2
(k − mω ) + (cω ) j
1
H ( jω ) =
2
(k − mωj
) + (cω
)


the frequency response function

Note: These are all complex functions
36/49


(2.50)
(2.51)

(2.52)

Using complex arithmetic:
X=

F0
(k − mω 2 )2 + (cω ) 2

e

− jθ

 cω 
θ = tan 
2 ÷
 k − mω 
F0
j (ω t −θ )
x p (t ) =
e
2 2
2
(k − mω ) + (cω )
−1

Has real part = to previous solution
37/49


(2.53)

(2.54)

(2.55)

Comments:
• It is the real part of this complex
solution that is physical
• The approach is useful in more
complicated problems

38/49


Example 2.3.1: Use the frequency response
approach to compute the particular solution of an
undamped system
The equation of motion is written as
2
m + kx(t) = F0 e jωt ⇒  + ω n x(t) = f0 e jωt
x(t)
x(t)

Let x p (t) = Xe jω t
2
⇒ ( −ω 2 + ω n ) Xe jωt = f0 e jωt

f0
⇒X= 2
ωn − ω 2 )
(

39/49


2.3.3 Transfer Function Method
The Laplace Transform
• Changes ODE into algebraic equation
• Solve algebraic equation then compute
the inverse transform
• Rule and table based in many cases
• Is used extensively in control analysis to
examine the response
• Related to the frequency response
function
40/49


The Laplace Transform approach:
• See appendix B and section 3.4 for details
• Transforms the time variable into an algebraic,
complex variable
• Transforms differential equations into an
algebraic equation
• Related to the frequency response method
∞

X(s) = L(x(t)) = ∫ x(t)e dt
0

41/49


− st

Take the transform of the equation of motion:

m + cx + kx = F0 cosω t ⇒
x 
F0 s
2
(ms + cs + k)X(s) = 2
s +ω2
Now solve algebraic equation in s for X(s)

F0 s
X (s) =
(ms 2 + cs + k )(s 2 + ω 2 )
To get the time response this must be “inverse
transformed”
42/49


Transfer Function Method
With zero initial conditions:
2
(ms + cs + k ) X (s) = F (s) ⇒
The transfer
X ( s)
1
function
= H ( s) = 2
F ( s)
ms + cs + k
(2.59)
1
H ( jω ) =
2
(2.60)
k − mω + cω j
= frequency response function
43/49


Example 2.3.2 Compute forced response of the
suspension system shown using the Laplace transform
Summing moments about the shaft:

Jθ (t) + kθ (t) = aF sin ω t
0

Taking the Laplace transform:
ω
s2 + ω 2
1
⇒ X ( s) = aω F0 2
( s + ω 2 ) ( Js 2 + k )
Js 2 X (s) + kX (s) = aF0

Taking the inverse Laplace transform:



1
÷
θ (t ) = aω F0 L  2
2
2
 ( s + ω ) ( Js + k ) ÷



aω F
1 1
1
k
⇒ θ (t ) =
sin ωt − sin ωnt ÷, ωn =
2 
J ω 2 − ωn  ω
ωn
J

−1

44/49


Notes on Phase for Homogeneous
and Particular Solutions
• Equation (2.37) gives the full solution for a
harmonically driven underdamped SDOF
oscillator to be

x(t) = Ae
sin ( ω d t + θ ) + X cos ( ω t − φ )
   




−ζω nt

homogeneous solution

particular solution

How do we interpret these phase angles?
Why is one added and the other subtracted?
45/49


Non-Zero initial conditions
x0 ≠ 0 v0 ≠ 0

sin (ω d t + θ )

xoω d
θ = tan
vo + ζω n xo
−1

46/49


Zero initial displacement

47/49


Zero initial velocity

48/49


Phase on Particular Solution

• Simple “atan” gives -π/2 < Φ < π/2
• Four-quadrant “atan2” gives 0 < Φ < π
49/49


Resonance and Harmonic Excitation

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