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Vibration and Noise 354

Dr J Muiyser

Lecture 2: Damped free vibration1

What is damping?
Damping is a reality of the world in which we live. All free vibration stops after a certain
amount of time as a result of various factors such as friction, aerodynamic forces etc. that
dissipates the energy in the system.
One way to model damping is as a viscous damper which is easily understood as that
which occurs inside a shock absorber where a viscous fluid (oil) is forced through a small
opening and friction causes the energy to be converted to heat.

The equation of motion for a damped SDOF system

Consider the damped SDOF system:

x(t)

c
fc

m
mg
k fk
N1 N2

The damping force is proportional to the velocity of the element:

f c (t) = c ẋ(t) (1)

where c is the viscous damping coefficient in N s/m.

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Lecture notes adapted from: Inman, DJ; Engineering Vibration; Fourth Edition; pp 33-43 & 70-75 and Van
Niekerk, JL; Vibration and Noise 354 class notes; 2016

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 The equation of motion of the system is:

− f k (t) − f c (t) = mẍ(t) (2)

mẍ(t) + c ẋ(t) + k x(t) = 0 (3)

Solution to the equation of motion of a damped SDOF system

As before, choose the general solution of the form:

x(t) = aeλt (4)

Substitution into the equation of motion and simplification yields:

mλ2 + cλ + k = 0 (5)
The two solutions to this quadratic equation are:

c 1 p
λ1,2 = − ± c 2 − 4km (6)
2m 2m
The value of the discriminant, c 2 − 4km will determine whether the roots are distinct real
numbers, a complex conjugate pair or equal real numbers. The critical damping coefficient is
defined by:
p
ccr = 2mωn = 2 km (7)
This leads to the definition of the damping ratio:
c c c
ζ= = = p (8)
ccr 2mωn 2 km
The roots are then:
p
λ1,2 = −ζωn ± ωn ζ2 − 1 (9)

Underdamped motion

If the damping ratio is less than 1 (0 < ζ < 1) the discriminant will be negative, resulting in
a complex conjugate pair of roots. If the discriminant is negative:
p p
ζ2 − 1 = 1 − ζ2 j (10)
And the roots are:
p
λ1,2 = −ζωn ± ωn 1 − ζ2 j (11)
As with the undamped case, the solution is then:
 p 2 p 2 ‹
x(t) = e−ζωn t a1 e j 1−ζ ωn t + a2 e− j 1−ζ ωn t (12)

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 Using the Euler relations this can be written as:

x(t) = Ae−ζωn t sin(ωd t + φ) (13)

where A and φ are once again constants of integration dependent on the initial conditions
and ωd is now the damped natural frequency. These are given by:
p
ω d = ω n 1 − ζ2 (14)

s
(v0 + ζωn x 0 )2 + (x 0 ωd )2
A= (15)
ω2d
x 0 ωd
φ = tan−1 (16)
v0 + ζωn x 0

Overdamped motion

When the damping ratio is larger than 1 (ζ > 1) the discriminant is positive, resulting in
distinct real roots:
p
λ1,2 = −ζωn ± ωn ζ2 − 1 (17)
The solution is then:
 p p ‹
−ζωn t − ζ2 −1ωn t ζ2 −1ωn t
x(t) = e a1 e + a2 e (18)

This represents a nonoscillatory response.

Critically damped motion

When the damping ratio is equal to 1 (ζ = 1) the discriminant is equal to zero. The roots are
then equal and real:

λ1,2 = −ωn (19)

The solution is then:

x(t) = (a1 + a2 t)e−ωn t (20)

This represents the smallest amount of damping that will provide a nonoscillatory re-
sponse.

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4
Measuring the damping ratio
One can see that underdamped motion is contained within the decay envelope defined by
±Ae−ζωn t . One can use the displacement response of the system to determine the viscous
damping ratio.
The logarithmic decrement is given by:

x(t)
δ = ln (21)
x(t + T )
where x(t) and x(t + T ) are measurements taken at two adjacent peaks of the displace-
ment response. The damping ratio is then given by:

δ
ζ= p (22)
4π2 + δ2
The accuracy of the measurement may be increased by taking a measurement over a larger
number of peaks:

1 x(t)
δ= ln (23)
n x(t + nT )

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 Matlab code for free response

1 clear
2

3 % System p a r a m e t e r s
4 m = 1; % Mass [ kg ]
5 k = 10; % S t i f f n e s s [N/m]
6 ccr = 2 ∗ sqrt ( k ∗ m ) ; % C r i t i c a l damping c o e f f i c i e n t [Nm/ s ]
7 wn = s q r t ( k / m ) ; % N a t u r a l f r e q u e n c y [ rad / s ]
8

9 % I n i t i a l conditions
10 x0 = 1 ; % I n i t i a l d i s p l a c e m e n t [m]
11 v0 = 0 ; % I n i t i a l v e l o c i t y [m]
12

13 t = 0:0.01:10; % Time a r r a y [ s ]
14

15 % The undamped c a s e
16 A = s q r t ( wn^2 ∗ x0^2 + v0^2) / wn; % eqn 1 . 9
17 p h i = atan ( wn ∗ x0 / v0 ) ; % eqn 1 . 9
18

19 x1 = A ∗ s i n ( wn ∗ t + p h i ) ; % eqn 1 . 3
20

21 % The underdamped c a s e
22 z = 0.1; % Damping r a t i o
23 c = z ∗ ccr ; % Damping c o e f f i c i e n t [Nm/ s ]
24 wd = wn ∗ s q r t ( 1 − z^2 ) ; % Damped n a t u r a l f r e q u e n c y [ rad / s ]
25

26 A = s q r t ( ( ( v0 + z ∗ wn ∗ x0 )^2 + ( x0 ∗ wd )^2 ) / wd^2 ) ; %

eqn 1.38
27 p h i = atan ( x0 ∗ wd / ( v0 + z ∗ wn ∗ x0 ) ) ; %
eqn 1.38
28

29 x2 = A ∗ exp ( −z ∗ wn ∗ t ) . ∗ s i n ( wd ∗ t + p h i ) ; % eqn 1.36

30

31 % The overdamped c a s e
32 z = 2 ; % Damping r a t i o
33

34 a1 = ( −v0 + ( −z + s q r t ( z^2 − 1 ) ) ∗ wn ∗ x0 ) / ( 2 ∗ wn ∗ s q r t
( z^2 − 1 ) ) ; % eqn 1.42
35 a2 = ( v0 + ( z + s q r t ( z^2 − 1 ) ) ∗ wn ∗ x0 ) / ( 2 ∗ wn ∗ s q r t (
z^2 − 1 ) ) ; % eqn 1.43
36

37 x3 = exp ( −z ∗ wn ∗ t ) . ∗ ( a1 ∗ exp ( −wn ∗ s q r t ( z^2 − 1) ∗ t ) +

a2 ∗ exp ( wn ∗ s q r t ( z^2 − 1 ) ∗ t ) ) ; % eqn 1.41
38

39 % The c r i t i c a l l y damped c a s e
40 z = 1 ; % Damping r a t i o
41

42 a1 = x0 ; % eqn 1.46

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43 a2 = v0 + wn ∗ x0 ; % eqn 1.46
44

45 x4 = ( a1 + a2 ∗ t ) . ∗ exp ( −wn ∗ t ) ; % eqn 1.45

46

47 % P l o t t i n g the curves
48 f i g u r e ( ’ D e f a u l t A x e s F o n t S i z e ’ , 12 , ’ P o s i t i o n ’ , [100 ,100 ,800 ,500]) ;
hold on ;
49

50 l1 = plot ( t , x1 , ’ LineWidth ’ , 1) ; L1 = ’ Undamped ’ ;

51 l2 = plot ( t , x2 , ’ LineWidth ’ , 1) ; L2 = ’ Underdamped ’ ;
52 l3 = plot ( t , x3 , ’ LineWidth ’ , 1) ; L3 = ’ Overdamped ’ ;
53 l4 = plot ( t , x4 , ’ LineWidth ’ , 1) ; L4 = ’ C r i t i c a l l y damped ’ ;
54

55 grid () ;
56

57 l = legend ( [ l 1 ; l 2 ; l 3 ; l 4 ] , L1 , L2 , L3 , L4 ) ;
58 s e t ( l , ’ F o n t S i z e ’ , 11) ;
59

60 x l a b e l ( ’ Time [ s ] ’ )
61 y l a b e l ( ’ D i s p l a c e m e n t [m] ’ )
62

63 p r i n t ( ’ damping ’ , ’−dpng ’ )

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Example problem: Damped free response
Consider the damped SDOF system:

x(t)

Given the following parameters:

m 1 kg
k 20 N/m
c 1 N s/m

1. Is the system over-, under- or critically damped?

2. What is the maximum displacement if x 0 = 0.1 m and v0 = 2 m/s?

Solution

The equation of motion is:

mẍ + c ẋ + k x = 0 (24)
The damping ratio is:
c
ζ= p = 0.1118 (25)
2 km
The system is therefore underdamped. As such the solution is:

x(t) = Ae−ζωn t sin(ωd t + φ) (26)

where:
r
k
ωn = = 4.4721 rad/s (27)
m
p
ωd = ωn 1 − ζ2 = 4.4441 rad/s (28)

s
(v0 + ζωn x 0 )2 + (x 0 ωd )2
A= = 0.4720 m (29)
ω2d
x 0 ωd
φ = tan−1 = 0.2135 (30)
v0 + ζωn x 0
Find the first point where ẋ = 0:

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ẋ(t) = −ζωn Ae−ζωn t sin(ωd t + φ) + ωd Ae−ζωn t cos(ωd t + φ) = 0 (31)

ωd
tan(ωd t + φ) = (32)
ζωn
ωd
tan−1 −φ
ζωn
t= = 0.2802 s (33)
ωd

x(0.2802) = 0.408 m (34)