E-notes Dr.

Ajit Prasad S LVibrations [10ME72]

Mechanical
Professor in Mechanical Engineering
UNIT -3 -3
UNIT PES College of Engineering
Mandya
DAMPED FREE VIBRATIONS

Damped Free Vibrations

Single Degree of Freedom Systems
Introduction:
Damping – dissipation of energy.
For a system to vibrate, it requires energy. During vibration of the system, there will be
continuous transformation of energy. Energy will be transformed from potential/strain to
kinetic and vice versa.
In case of undamped vibrations, there will not be any dissipation of energy and the
system vibrates at constant amplitude. Ie, once excited, the system vibrates at constant
amplitude for infinite period of time. But this is a purely hypothetical case. But in an
actual vibrating system, energy gets dissipated from the system in different forms and
hence the amplitude of vibration gradually dies down. Fig.1 shows typical response
curves of undamped and damped free vibrations.

Types o damping:
(i) Viscous damping
In this type of damping, the damping resistance is proportional to the relative velocity
between the vibrating system and the surroundings. For this type of damping, the
differential equation of the system becomes linear and hence the analysis becomes easier.
A schematic representation of viscous damper is shown in Fig.2.

Here, F α x& or F = cx& , where, F is damping resistance, x& is relative velocity and c is the
damping coefficient.

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(ii) Dry friction or Coulomb damping

In this type of damping, the damping resistance is independent of rubbing velocity and is
practically constant.

(iii) Structural damping

This type of damping is due to the internal friction within the structure of the material,
when it is deformed.

Spring-mass-damper system:

Fig.3 shows the schematic of a simple spring-mass-damper system, where, m is the mass
of the system, k is the stiffness of the system and c is the damping coefficient.
If x is the displacement of the system, from Newton’s second law of motion, it can be
written
m&x& = −cx& − kx
Ie m&x& + cx& + kx = 0 (1)
This is a linear differential equation of the second order and its solution can be written as
x = e st (2)
dx
Differentiating (2), = x& = se st
dt
d 2x
= &x& = s 2 e st
dt 2
Substituting in (1), ms 2 e st + cse st + ke st = 0
( )
ms 2 + cs + k e st = 0
2
Or ms + cs + k = 0 (3)
Equation (3) is called the characteristic equation of the system, which is quadratic in s.
The two values of s are given by
2
c  c  k
s1, 2 =− ±   − (4)
2m  2m  m

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The general solution for (1) may be written as

x = C1e s1t + C 2 e s2t (5)
Where, C1 and C2 are arbitrary constants, which can be determined from the initial
conditions.
2
 c  k
In equation (4), the values of s1 = s2, when   =
 2m  m
 c  k
Or,  = = ωn (6)
 2m  m
Or c = 2mω n , which is the property of the system and is called critical
damping coefficient and is represented by cc.
Ie, critical damping coefficient = cc = 2mω n
The ratio of actual damping coefficient c and critical damping coefficient cc is called
damping factor or damping ratio and is represented by ζ.
c
Ie, ζ = (7)
cc
c c c c
In equation (4), can be written as = × c = ζ .ω n
2m 2 m c c 2m
Therefore, s1, 2 = −ζ .ω n ± (ζ .ω n )2 − ω 2 n [
= − ζ ± ζ 2 − 1 ωn ] (8)

The system can be analyzed for three conditions.

(i) ζ > 1, ie, c > cc, which is called over damped system.
(ii) ζ = 1. ie, c = cc, which is called critically damped system.
(iii) ζ < 1, ie, c < cc, which is called under damped system.

Depending upon the value of ζ, value of s in equation (8), will be real and unequal, real
and equal and complex conjugate respectively.

(i) Analysis of over-damped system (ζ > 1).

In this case, values of s are real and are given by

[ ]
s1 = − ζ + ζ 2 − 1 ω n and [
s2 = − ζ − ζ 2 − 1 ω n ]
Then, the solution of the differential equation becomes
 −ζ + ζ 2 −1  ω t  −ζ − ζ 2 −1  ω t
  n   n
x = C1e + C2e (9)
This is the final solution for an over damped system and the constants C1 and C2 are
obtained by applying initial conditions. Typical response curve of an over damped system
is shown in fig.4. The amplitude decreases exponentially with time and becomes zero at t
= ∞.

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(ii) Analysis of critically damped system (ζ = 1).

In this case, based on equation (8), s1 = s2 = -ωn
The solution of the differential equation becomes
x = C1e s1t + C 2 te s2t

Ie, x = C1e −ωnt + C 2 te −ωn t

Or, x = (C1 + C 2 t )e −ωnt (10)

This is the final solution for the critically damped system and the constants C1 and C2 are
obtained by applying initial conditions. Typical response curve of the critically damped
system is shown in fig.5. In this case, the amplitude decreases at much faster rate
compared to over damped system.

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(iii) Analysis of under damped system (ζ < 1).

In this case, the roots are complex conjugates and are given by

[ ]
s1 = − ζ + j 1 − ζ 2 ω n

s2 = [− ζ − j 1 − ζ ]ω 2
n

The solution of the differential equation becomes

 −ζ + j 1−ζ 2  ω t  −ζ − j 1−ζ 2  ω t
x = C1e   n
+ C 2 e   n

This equation can be rewritten as

 j 1−ζ 2  ω n t
  − j 1−ζ 2   ω t
   n

x = e −ζωnt C1e  
+ C 2 e   (11)
 
Using the relationships
e iθ = cos θ + i sin θ
e − iθ = cos θ − i sin θ
Equation (11) can be written as

[ { } {
x = e −ζωnt C1 cos 1 − ζ 2 ω n t + j sin 1 − ζ 2 ω n t + C 2 cos 1 − ζ 2 ω n t − j sin 1 − ζ 2 ω n t }]
Or x = e [(C + C ){cos
−ζω n t
1 2 1−ζ 2 ω t }+ j (C − C ){sin
n 1 2 1 − ζ 2 ωnt }] (12)
In equation (12), constants (C1+C2) and j(C1-C2) are real quantities and hence, the
equation can also be written as

[{ } {
x = e −ζωnt A cos 1 − ζ 2 ω n t + B sin 1 − ζ 2 ω n t }]
Or, x = A1e −ζω n t [{sin( 1 − ζ ω t + φ )}]
2
n 1 (13)
The above equations represent oscillatory motion and the frequency of this motion is
represented by ωd = 1 − ζ 2 ωn (14)
Where, ωd is the damped natural frequency of the system. Constants A1 and Φ1 are
determined by applying initial conditions. The typical response curve of an under damped
system is shown in Fig.6.

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Applying initial conditions,

x = Xo at t = 0; and x& = 0 at t = 0, and finding constants A1 and Φ1,
equation (13) becomes
  1−ζ 2 
Xo 
x= e −ζωnt sin 1 − ζ 2 ω n t + tan −1    (15)
1−ζ 2   ζ  
   
Xo
The term e −ζωnt represents the amplitude of vibration, which is observed to decay
2
1−ζ
exponentially with time.

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LOGARITHMIC DECREMENT

Referring to Fig.7, points A & B represent two successive peak points on the response
curve of an under damped system. XA and XB represent the amplitude corresponding to
points A & B and tA & tB represents the corresponding time.

We know that the natural frequency of damped vibration = ω d = 1 − ζ 2 ω n rad/sec.

ωd
Therefore, f d = cycles/sec
2π
1 2π 2π
Hence, time period of oscillation = t B − t A = = = sec (16)
f d ωd 1 − ζ 2 ωn
From equation (15), amplitude of vibration
Xo
XA = e −ζωnt A
2
1−ζ
Xo
XB = e −ζωnt B
2
1−ζ
XA
Or, = e −ζωn (t A −t B ) = e ζωn (t B −t A )
XB
2πζ
XA 1− ζ 2
Using eqn. (16), =e
XB
XA 2πζ
Or, log e =
XB 1−ζ 2

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This is called logarithmic decrement. It is defined as the logarithmic value of the ratio of
two successive amplitudes of an under damped oscillation. It is normally denoted by δ.
XA 2πζ
Therefore, δ = log e = (17)
XB 1−ζ 2
This indicates that the ratio of any two successive amplitudes of an under damped system
is constant and is a function of damping ratio of the system.
For small values of ζ, δ ≈ 2πζ
If X0 represents the amplitude at a particular peak and Xn represents the amplitude after
X X X
‘n’ cycles, then, logarithmic decrement = δ = log e 0 = log e 1 = …… = log e n−1
X1 X2 Xn
X 0 X1 X
Adding all the terms, nδ = log e × ...... n −1
X1 X 2 Xn
1 X
Or, δ= log e 0 (18)
n Xn

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Solved problems
1) The mass of a spring-mass-dashpot system is given an initial velocity 5ωn, where
ωn is the undamped natural frequency of the system. Find the equation of motion
for the system, when (i) ζ = 2.0, (ii) ζ = 1.0, (i) ζ = 0.2.
Solution:
Case (i) For ζ = 2.0 – Over damped system
For over damped system, the response equation is given by
 −ζ + ζ 2 −1  ω t  −ζ − ζ 2 −1  ω t
  n   n
x = C1e + C2e

Substituting ζ = 2.0, x = C1e [−0.27 ]ω nt + C 2 e [−3.73]ωnt (a)

Differentiating, x& = −0.27ωnC1e −0.27ω n t − 3.73ωnC2e −3.73ω n t (b)

Substituting the initial conditions
x = 0 at t = 0; and x& = 5ωn at t = 0 in (a) & (b),
0 = C1 + C2 (c)
5ωn = -0.27 ωn C1 – 3.73 ωn C2 (d)
Solving (c) & (d), C1 = 1.44 and C2 = -1.44.
Therefore, the response equation becomes
(
x = 1.44 e[−0.27 ]ω n t − e[−3.73]ω n t ) (e)
Case (ii) For ζ = 1.0 – Critically damped system
For critically damped system, the response equation is given by
x = (C1 + C 2 t )e −ωnt (f)

Differentiating, x& = −(C1 + C2t )ωne −ω n t + C2e −ω n t (g)

Substituting the initial conditions
x = 0 at t = 0; and x& = 5ωn at t = 0 in (f) & (g),
C1 = 0 and C2 = 5ωn
Substituting in (f), the response equation becomes
x = (5ωnt )e −ω n t (h)
Case (iii) For ζ = 0.2 – under damped system
For under damped system, the response equation is given by

[{
x = A1e −ζωn t sin 1 − ζ 2 ω n t + φ1 }]
Substituting ζ = 0.2, x = A1e −0.2ω n t [{sin (0.98ωnt + φ1 )}] (p)

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Differentiating,
x& = −0.2ωn A1e −0.2ω n t [{sin (0.98ωnt + φ1 )}] + 0.98ωn A1e −0.2ω n t cos(0.98ωnt + φ1 ) (q)
Substituting the initial conditions
x = 0 at t = 0; and x& = 5ωn at t = 0 in (p) & (q),
A1sinΦ1 = 0 and A1 cosΦ1 = 5.1
Solving, A1 = 5.1 and Φ1 = 0
Substituting in (p), the response equation becomes
x = 5.1e −0.2ω n t [{sin (0.98ωnt )}] (r)

2) A mass of 20kg is supported on two isolators as shown in fig.Q.2. Determine the

undamped and damped natural frequencies of the system, neglecting the mass of the
isolators.

Solution:
Equivalent stiffness and equivalent damping coefficient are calculated as
1 1 1 1 1 13
= + = + =
k eq k1 k 2 10000 3000 30000
1 1 1 1 1 4
= + = + =
C eq C1 C 2 300 100 300

k eq 30000
Undamped natural frequency = ωn = = 13 = 10.74rad / sec
m 20
10.74
fn = = 1.71cps
2π

Damped natural frequency = ω d = 1 − ζ 2 ω n

C eq 300
ζ = = 4 = 0.1745
2 k eq m 2× 30000 × 20
13

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∴ ω d = 1 − 0.1745 2 × 10.74 = 10.57 rad / sec

10.57
Or, fd = = 1.68cps
2×π

3) A gun barrel of mass 500kg has a recoil spring of stiffness 3,00,000 N/m. If the
barrel recoils 1.2 meters on firing, determine,
(a) initial velocity of the barrel
(b) critical damping coefficient of the dashpot which is engaged at the end of the
recoil stroke
(c) time required for the barrel to return to a position 50mm from the initial
position.

Solution:
(a) Strain energy stored in the spring at the end of recoil:
1 1
P = kx 2 = × 300000 × 1.2 2 = 216000 N − m
2 2
Kinetic energy lost by the gun barrel:
1 1
T = mv 2 = × 500 × v 2 = 250v 2 , where v = initial velocity of the barrel
2 2
Equating kinetic energy lost to strain energy gained, ie T = P,
250v 2 = 216000
v = 29.39m/s
(b) Critical damping coefficient = C c = 2 km = 2 300000 × 500 = 24495 N − sec/ m
(c) Time for recoiling of the gun (undamped motion):
k 300000
Undamped natural frequency = ω n = = = 24.49r / s
m 500
2π 2π
Time period = τ = = = 0.259 sec
ω n 24.29
τ 0.259
Time of recoil = = = 0.065 sec
4 4
Time taken during return stroke:
Response equation for critically damped system = x = (C1 + C 2 t )e −ωnt
Differentiating, x& = C 2 e −ωn t − (C1 + C 2 t )ω n e −ωn t
Applying initial conditions, x = 1.2, at t = 0 and x& = 0 at t = 0,
C1 = 1.2, & C2 = 29.39
Therefore, the response equation = x = (1.2 + 29.39t )e −24.49t
When x = 0.05m, by trial and error, t = 0.20 sec
Therefore, total time taken = time for recoil + time for return = 0.065 + 0.20 = 0.265 sec
The displacement – time plot is shown in the following figure.

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4) A 25 kg mass is resting on a spring of 4900 N/m and dashpot of 147 N-se/m in

parallel. If a velocity of 0.10 m/sec is applied to the mass at the rest position, what
will be its displacement from the equilibrium position at the end of first second?

Solution:

The above figure shows the arrangement of the system.

Critical damping coefficient = cc = 2mω n

k 4900
Where ω n = = = 14r / s
m 25
Therefore, cc = 2 × 25 × 14 = 700 N − sec/ m
c 147
Since C< Cc, the system is under damped and ζ = = = 0.21
cc 700

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[{ (
Hence, the response equation is x = A1e −ζω n t sin 1 − ζ 2 ωnt + φ1 )}]
[{ (
Substituting ζ and ωn, x = A1e −0.21×14t sin 1 − 0.212 14t + φ1 )}]
x = A1e −2.94t [{sin (13.7t + φ1 )}]

Differentiating, x& = −2.94 A1e −2.94t [{sin (13.7t + φ1 )}] + 13.7 A1e −2.94t cos(13.7t + φ1 )
Applying the initial conditions, x = 0, at t = 0 and x& = 0.10m / s at t = 0
Φ1 = 0
0.10 = −2.94 A1 [{sin (φ1 )}] + 13.7 A1 cos(φ1 )
Since, Φ1 = 0, 0.10 = 13.7 A1; A1 = 0.0073
Displacement at the end of 1 second = x = 0.0073e −2.94 [{sin (13.7 )}] = 3.5 × 10 −4 m

5) A rail road bumper is designed as a spring in parallel with a viscous damper.

What is the bumper’s damping coefficient such that the system has a damping ratio
of 1.25, when the bumper is engaged by a rail car of 20000 kg mass. The stiffness of
the spring is 2E5 N/m. If the rail car engages the bumper, while traveling at a speed
of 20m/s, what is the maximum deflection of the bumper?

k
m

Solution: Data = m = 20000 kg; k = 200000 N/m; ζ = 1.25

Critical damping coefficient =
cc = 2 × m × k = 2 × 20000 × 200000 = 1.24 × 105 N − sec/ m

Damping coefficient C = ζ × CC = 1.25 × 1.24 × 105 = 1.58 × 105 N − sec/ m

k 200000
Undamped natural frequency = ωn = = = 3.16r / s
m 20000
Since ζ = 1.25 , the system is over damped.
For over damped system, the response equation is given by
 −ζ + ζ 2 −1  ω t  −ζ − ζ 2 −1  ω t
x = C1e   n
+ C 2 e   n

Substituting ζ = 1.25, x = C1e [−0.5]ωnt + C2 e [−2.0 ]ωnt (a)

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Differentiating, x& = −0.5ω n C1e −0.5ωnt − 2.0ω n C 2 e −2.0ωnt (b)

Substituting the initial conditions
x = 0 at t = 0; and x& = 20m / s at t = 0 in (a) & (b),
0 = C1 + C2 (c)
20 = -0.5 ωn C1 – 2.0 ωn C2 (d)
Solving (c) & (d), C1 = 4.21and C2 = -4.21
Therefore, the response equation becomes
( )
x = 4.21 e[−1.58×t ] − e[−6.32×t ] m (e)
The time at which, maximum deflection occurs is obtained by equating velocity equation
to zero.
Ie, x& = −0.5ωnC1e −0.5ωnt − 2.0ωnC2e −2.0ωnt = 0

Ie, − 6.65e −1.58t + 26.61e −6.32t = 0

Solving the above equation, t = 0.292 secs.
Therefore, maximum deflection at t = 0.292secs,
( )
Substituting in (e), x = 4.21 e[−1.58×0.292 ] − e[−6.32×0.292 ] m , = 1.99m.
6) A disc of a torsional pendulum has a moment of inertia of 6E-2 kg-m2 and is
immersed in a viscous fluid. The shaft attached to it is 0.4m long and 0.1m in
diameter. When the pendulum is oscillating, the observed amplitudes on the same
side of the mean position for successive cycles are 90, 60 and 40. Determine (i)
logarithmic decrement (ii) damping torque per unit velocity and (iii) the periodic
time of vibration. Assume G = 4.4E10 N/m2, for the shaft material.

Shaft dia. = d = 0.1m

Shaft length = l = 0.4m
Moment of inertia of disc = J = 0.06 kg-m2.
Modulus of rigidity = G = 4.4E10 N/m2

Solution:
The above figure shows the arrangement of the system.
9 6
(i) Logarithmic decrement = δ = log e = log e = 0.405
6 4
(ii) The damping torque per unit velocity = damping coefficient of the system ‘C’.

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2πζ
We know that logarithmic decrement = δ = , rearranging which, we get
1−ζ 2
δ 0.405
Damping factor ζ = = = 0.0645
4π 2 + δ 2 4π + 0.405 2
2

C
Also, ζ = , where, critical damping coefficient = C C = 2 k t J
CC
GI p G πd 4 4.4 × 1010 π × 0.14
Torsional stiffness = k t = = × = × = 1.08 × 10 6 N − m / rad
l l 32 0 .4 32
Critical damping coefficient = C C = 2 k t J = 2 1.08 × 10 6 × 0.06 = 509 N − m / rad
Damping coefficient of the system = C = C C × ζ = 509 × 0.0645 = 32.8 N − m / rad
1 2π 2π
(iii) Periodic time of vibration = τ = = =
f d ωd ωn 1 − ζ 2
kt 1.08 × 10 6
Where, undamped natural frequency = ω n = = = 4242.6rad / sec
J 0.06
2π
Therefore, τ = = 0.00148 sec
4242.6 × 1 − 0.0645 2

7) A mass of 1 kg is to be supported on a spring having a stiffness of 9800 N/m. The

damping coefficient is 5.9 N-sec/m. Determine the natural frequency of the system.
Find also the logarithmic decrement and the amplitude after three cycles if the
initial displacement is 0.003m.

Solution:
k 9800
Undamped natural frequency = ω n = = = 99r / s
m 1

Damped natural frequency = ω d = 1 − ζ 2 ω n

Critical damping coefficient = cc = 2 × m × ω n = 2 × 1 × 99 = 198 N − sec/ m
c 5 .9
Damping factor = ζ = = = 0.03
cc 198
Hence damped natural frequency = ω d = 1 − ζ 2 ω n = 1 − 0.013 2 × 99 = 98.99rad / sec

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2πζ 2 × π × 0.03
Logarithmic decrement = δ = = = 0.188
1−ζ 2 1 − 0.03 2
1 X
Also, δ = log e 0 ; if x0 = 0.003,
n Xn
1 X 1 0.003
then, after 3 cycles, δ = log e 0 ie,0.188 = × log e
n Xn 3 X3
0.003
ie, X 3 = 3×0.188 = 1.71 × 10 −3 m
e

8) The damped vibration record of a spring-mass-dashpot system shows the

following data.
Amplitude on second cycle = 0.012m; Amplitude on third cycle = 0.0105m;
Spring constant k = 7840 N/m; Mass m = 2kg. Determine the damping constant,
assuming it to be viscous.
Solution:
X2 0.012
Here, δ = log e = log e = 0.133
X3 0.0105
2πζ δ 0.133
Also, δ = , rearranging, ζ = = = 0.021
1−ζ 2 4π 2 + δ 2 4π 2 + 0.133 2

Critical damping coefficient = cc = 2 × m × k = 2 × 2 × 7840 = 250.4 N − sec/ m

Damping coefficient C = ζ × C C = 0.021 × 250.4 = 5.26 N − sec/ m

9) A mass of 2kg is supported on an isolator having a spring scale of 2940 N/m and
viscous damping. If the amplitude of free vibration of the mass falls to one half its
original value in 1.5 seconds, determine the damping coefficient of the isolator.

Solution:
k 2940
Undamped natural frequency = ω n = = = 38.34r / s
m 2

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Critical damping coefficient = cc = 2 × m × ω n = 2 × 2 × 38.34 = 153.4 N − sec/ m

[{ (
Response equation of under damped system = x = A1e −ζωnt sin 1 − ζ 2 ω n t + φ1 )}]
Here, amplitude of vibration = A1e −ζωnt
X0
If amplitude = X0 at t = 0, then, at t = 1.5 sec, amplitude =
2
Ie, A1e −ζωn ×0 = X 0 or A1 = X 0

X0 X 1
Also, A1e −ζωn ×1.5 = or X 0 × e −ζ ×38.34×1.5 = 0 or e −ζ ×38.34×1.5 =
2 2 2
Ie, e ζ ×38.34×1.5 = 2 , taking log, ζ × 38.34 × 1.5 = 0.69 ∴ ζ = 0.012
Damping coefficient C = ζ × CC = 0.012 × 153.4 = 1.84 N − sec/ m

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