DOM Unit 4 (12 ME 63)

Chapter 2: Damped free vibrations

• Damping is the resistance offered by a body to the
motion of a vibrating system.
• The resistance may be applied by a liquid or solid,
internally or externally.
• Due to damping, the amplitude of vibrations will be
controlled so that failure of mechanical systems due to
resonance is avoided.
• Within the scope of our syllabus, only damping due to a
fluid dashpot (viscous damping) is discussed.
 Types of Damping
The following are the common types of damping;
(i) Viscous damping: (Due to a viscous medium such as a fluid
dashpot)
(ii) Eddy current damping: (Due to generation of eddy currents
which sets up a magnetic field which in turn opposes the motion
of the body)
(iii) Coulomb damping: (Due to dry friction between sliding
surfaces)
(iv) Solid or structural or Hysteresis damping: (Due to internal
friction of the molecules)
(v) Slip or interfacial damping: Due to microscopic slip due to
fluctuating loads in machine parts in contact.
 Derivation of differential equation for a single degree of
freedom damped free vibrations (Viscous damping)

k C

m
x
Damped free vibration

• Consider a spring of stiffness ‘k’ & a mass ‘m’ system

constrained to move in a vertical direction.
• Let the damping coefficient of the fluid dashpot be ‘c’
(Force per unit velocity , N/m/sec or N-sec/m)
 Damped free vibrations (Viscous damping)

.. .
mx cx kx

Free Body diagrams

When the mass is displaced from its equilibrium position by a distance ‘ x’ & released,
after a time ' t ', for equilibrium, Inertia force + Damper force + Spring force = 0
Here, inertia force = mass  acceleration  mx
Damper force = damping coefficient ' c '( force per unit velocity )  velocity  cx
Spring force = stiffness ' k '( force per unit deflection)  deflection  kx
c k
i.e. mx + cx + kx = 0 or x +   x +   x = 0
m m
Let the solution of the second order differential equation
c k
x +   x +   x = 0 (i) is of the form x = Ae1t  Be  2t where;
m m
A & B are some constants & 1 &  2 are the roots of the auxiliary equation
2
c k c 1  c  k
        0
2
i.e. 1,2      
m m 2m 2  2m   m 
2
 c  k
The ratio of   to   represents the degree of damping provided
 2m  m
in the system & its square root is known as damping factor 
2 2
 c   c 
     
i.e.         c  c
2 m 2 m k
 n   k =mn 
2

k n2 2mn 2 km  m 

 
m
Rearranging the terms, Damping coefficient c = 2ζmωn = 2ζ km
When ζ = 1, the damping is known as critical. The corresponding value of
damping coefficient is known as critical damping coefficient, demoted by cc .
i.e. cc  2mn  2 km
c Actual damping coefficient
Hence damping ratio ζ = =
cc Critical damping coefficient
Thus, when;
 =1, the system is said to be critically damped
 <1, the system is said to be under-damped
  1, the system is said to be over-damped
Equation (i) can also be written as x + 2ζωn x + ωn2 x = 0 (ii)

and 1,2   ωn   ωn 
2
 
 ωn2     2  1 n
The exact solution of the equation (ii) depends on whether the roots
α1,2 are real or imaginary.
Rearranging the terms, Damping coefficient c = 2ζmωn = 2ζ km
When ζ = 1, the damping is known as critical. The corresponding value of
damping coefficient is known as critical damping coefficient, demoted by cc .
i.e. cc  2mn  2 km
c Actual damping coefficient
Hence damping ratio ζ = =
cc Critical damping coefficient
Thus, when;
 =1, the system is said to be critically damped
 <1, the system is said to be under-damped
  1, the system is said to be over-damped
Equation (i) can also be written as x + 2ζωn x + ωn2 x = 0 (ii)

and 1,2   ωn   ωn 
2
 
 ωn2     2  1 n
The exact solution of the equation (ii) depends on whether the roots
α1,2 are real or imaginary.
 (i) ζ > 1, i.e. the system is over-damped.

The roots are real & distinct. The solution is x  Ae

    1 t  Be    1 t
2
n
2
n

The motion is aperiodic & hence no vibrations are possible. The system returns to equilibrium
with passage of considerable time.

Displacement


Time
Over-damped system

(ii) ζ = 1, i.e. the system is critically damped.

The roots are real & equal. 1   2  n The solution is x  ( A  Bt )e nt
The motion is aperiodic & hence no vibrations are possible.The system returns to equilibrium
within shortest possible time.
Displacement



Time
Criticaly-damped system
(iii) ζ < 1, i.e. the system is under-damped.The roots are imaginary.

x  Ae
   i 1  t
2
n
 Be
   i 1  t
2

 e  t  Aei 1  t  Be  i 1  t 

n
n
2
n
2
n
 
i 1 2 n t
Using Euler's theorem, e  cos( 1   2 nt )  i sin( 1   2 nt )
 i 1 2 n t
e  cos( 1   2 nt )  i sin( 1   2 nt )
 x  e nt ( A  B ) cos( 1   2 nt )  i ( A  B ) sin( 1   2 nt ) 
 
Let ( A  B )  X sin  & i ( A  B )  X cos  , &
1   2 n  d , frequency of damped vibrations
Then, x  e nt  X sin  cos(d t )  X cos  sin(d t ) 
 x  e-nt  X sin(d t   )  where X &  are constants which can be
determined from initial conditions.
The solution x  e-nt  X sin(d t   )  consists of three terms
 X , which is a constant,
 e-nt which decreases with time
 sin(d t   ) which represents a periodic motion
Thus the resulting motion is oscillatory with decreasing amplitudes
2
with frequency d  n 1   2 & time period Td 
d
 Logarithmic Decrement
• It is defined as the natural logarithm of the ratio of any two
successive amplitudes on the same side of the mean position in
an under-damped system. It is denoted by d .
• The ratio of any two successive amplitudes in an under-damped
system is always a constant.
Td Td Td
Displacement

x1
x2
x3
O

Time
 Logarithmic Decrement
Td Td Td

Displacement
x1
x2
x3
O

Time

The displacement of an under-damped system is x  e-nt  X sin(d t   ) 

whose peak amplitude is Xe-nt when sin(d t   ) is equal to unity. (or one)
Hence maximum amplitude x  Xe-nt at any given time 't'
Let x1 be the maximum amplitude when time is t1 & x2 when time is t2
x1 Xe-nt1
 x1  Xe-nt1 & x2  Xe-nt2   -nt2  en (t2 t1 )
x2 Xe
 Logarithmic Decrement
2 2
But (t2  t1 )  Td  Time period of damped oscillations  
d n 1   2
 2  2 2
n  
x  2 
1  1 2 x x x 1 2
 1 e  n
e . Similarly, 2  3   n e
x2 x3 x4 xn 1
Taking natural logarithm on both sides, we get logarthmic decrement d
 xn  2
d = ln   If  is very small, δ = 2πζ
 xn 1  1  2

Also, if the initial amplitude x1 & amplitude after n cycles xn+1 are given,
1
n
 x1   x1 x2 x3 xn   x1  x1  x1  n
          
 xn 1   x2 x3 x4 xn 1   x2  x2  xn 1 
1  x1 
 δ = ln  
n  xn+1 
 Use of critical damping
• When a critically damped system is displaced from
equilibrium, it returns to its original position in a shortest
possible time without oscillations.
• This feature of critical damping is used in dashpots with
recoil springs of large gun barrels so that after firing, the
gun returns to its original position and ready for the next
shot without delay.
• Critical damping is also employed in telescopic shock
absorbers of vehicles, automatic spring door closers, etc.
 Numerical problem 1
A gun barrel is so designed that, on firing, the barrel
recoils against a spring. A dashpot, at the end of the recoil,
allows the barrel to come back to its initial position within
the minimum time without any oscillation. The gun barrel
has a mass of 500 kg and a recoil spring of 300 N/mm.
The barrel recoils 1 m on firing. Determine;
(i) The initial recoil velocity of the gun barrel
(ii) The critical damping coefficient of the dashpot
engaged at the end of the recoil stroke.
Data : m  500 kg , k  300 N / mm  300 103 N / m, x  1 m
Solution :
(i) Initial velocity of recoil : As the dashpot is engaged only at
the end of recoil, K.Eof the barrel = Work done on the spring
1 2 1 2 1 2 1 2
i.e. mx  kx    500  x     300  10  1 
3

2 2 2  2 
 Initial velocity of recoil x  24.5 m / sec
(ii) Critical damping coefficient :
cc  2mn or 2 km  2 300 103  500  24495 N - sec / m
 Numerical problem 2
For the system shown in fig, the characteristic of the
dashpot is such that when a constant force of 49 N is
applied to the piston, its velocity is found to be constant at
0.12 m/sec. State whether the motion is periodic or not.

245 N/m C

15 kg
Data : m  15 kg , k  245 N / m
Solution :
(i) Damping coefficient (c) :
 Damping force   49 
c    408.33 N - sec / m
 Velocity   0.12 
(ii) Critical damping coefficient cc :
cc  2 km  2 245 15  121.24 N - sec / m
 c   408.33 
 Damping factor         3.37
 cc   121.24 
As the damping ratio is more than unity, the motion
is over damped & hence aperiodic.
 Numerical Problem 3
A mechanical vibrating system has mass of 10 kg and
stiffness of springs 5 N/mm along with a dashpot which
exerts a force of 40 N when the mass has a velocity 1m/sec.
Determine;
(i) Critical damping coefficient
(ii) Damping factor
(iii) Frequency of damped vibrations
(iv) Logarithmic decrement
(v) Ratio of any two successive amplitudes
 k C

m
x
Damped free vibration

Data : m  10 kg , k  5 N / mm  5000 N / m, c  40 N / m / sec

k 5000
Solution : Natural frequency n    22.36 rad / sec
m 10
(i) Critical damping coefficient : cc  2mn  2 10  22.36  447.2 N - sec / m
c 40
(ii) Damping factor :  =   0.0894 (dimensionless number)
cc 447.2
(iii) Damped frequency : d  n 1   2  22.36 1  0.08942  22.27 rad / sec
2 2  0.0894
(iv) Logarithmic decrement : d    0.566 (dimensionless)
1  2
1  0.0894 2

(v) Ratio of successive amplitudes = ed  e0.566  1.76

 Numerical Problem 4
The mass of a single degree damped vibrating system is 7.5
kg and it makes 24 oscillations in 14 seconds. The
amplitude of vibration reduces to 0.25 of its initial value
after 5 oscillations. Determine;
(i) Logarithmic decrement
(ii) Damping factor
(iii)Stiffness of the spring
  No of oscillations   24 
Data : m  7.5 kg , f d        1.714 Hz
 Time taken   14 
Amplitude after 5 cycles i.e. x6  0.25  Initial amplitude i.e. x1
Solution :
1  x  1  x1 
(i) Logarithimic decrem ent d  ln  1   ln    0.2773
5  x6  5  0.25 x1 
(ii) Damping factor :
 2  d
Log decrement d     Damping ratio  =
 1  2   2
 d 2
  4
0.2773
=  0.044
4  0.2773
2 2

(iii) Stiffness of the spring : Damped frequency f d  f n 1   2

1.714  f n 1  0.0442  f n  1.716 But n  2 f n  10.78 rad / sec
 Stiffness of the spring k  mn2  7.5(10.78) 2  871.88 N / m
 Numerical Problem 5
A body of mass 10 kg is suspended freely from a spring of
stiffness 2 N/mm. A damper having a resistance of 5 N at a
velocity of 0.1 m/sec is connected between the mass and
the fixed end of the spring. Determine;
(i) Damping factor & log decrement
(ii) Ratio of successive amplitudes
(iii) Amplitude of body after 10 cycles if the initial
amplitude is 15 mm.
  5 
Data : m  10 kg , k  2 N / mm  2000 N / m, c     50 N  sec/ m
 0.1 
Initial amplitude i.e. x1  15mm, No of oscillations , n  10
Solution :
(i) Damping factor & logarithimic decrement :
critical damping coefficient cc  2 km  2 2000 10  282.84 N - sec / m
c   50 
Damping factor       0.177
 cc   282.4 
 2   2  0.177 
Log decrement d     = 1.13
 1     1  0.177 2
2

 
 xn  d
(ii) Ratio of successive amplitudes =    e  e1.13
 3.095
 xn 1 
(iii) Amplitude after 10 cycles : Let x1 be the initial amplitude &
10
 x1   xn 
x11 be the amplitude after 10 oscillations. Then     
 x11   xn 1 
 15 
    (3.095)10  x11 = 1.86 × 10 -4 mm
 x11 
 Torsional Systems with viscous damping
GJ 1 GJ
(1) circular frequency n   fn  where,
IL 2 IL
G  Rigidity modulus of shaft material=80 GPa for steel
d4
d J  Polar moment of inertia of shaft 
L 32
I  Mass moment of inertia of disk, L  Length of shaft
Oil
  n   2 
(2) Logarthmic decrement d  ln     
  n 1   1   
2

I
d
 Damping ratio  =
4 2  d 2
 (3) Frequency of damped vibration d  n 1   2 rad / sec

(4) Damping coefficent C    2 I n Nm / rad / sec
 Numerical Problem 6
The disc of a torsional pendulum has a moment of inertia of 0.06 kg-
m2 and is immersed in a viscous fluid. The brass shaft (G = 44 GPa)
attached to it is of 100 mm diameter and 400 mm long. When the
pendulum is vibrating, the amplitudes on the same side for
successive cycles are 90, 60 & 40. Determine;
(i) Logarithmic decrement
(ii) Damping torque at unit velocity
(iii) Periodic time of vibration.
What would be the frequency if the disc is removed from the viscous
fluid?
 Data : I  0.06 kg - m 2 ,L = 400 mm = 0.4 m,
G = 44 GPa = 44  10 9 N / m 2
d4
d = 100 mm = 0.1m  J =  9.82  106 m 4 ,
32
x1  90 , x2  60 , x3  40
L d
(1) Logarthmic decrement :
Oil 9 6
d  ln   or ln    0.4055
6 4
I d 0.4055
 =   0.0644
4  d
2 2
4  0.4055
2 2

GJ
 Also Circular frequency n 
 IL
(44 109 )  (9.82 106 )
n   4243 rad / sec
0.06  0.4
(2) Damping torque at unit velocity (c) :
Damping coefficent c    2 I n
 c  0.0644  2  0.06  4243  32.8 Nm / rad / sec
2
(3) Periodic time of vibration : Td 
d
2 2
Td    1.487 × 10 -3 sec
n 1   2 4243 1  0.06442
Frequency when damper is rem oved (  = 0) :
d  n 1  02  n  4243 rad / sec
 n   4243 
 fn      675.3 Hz
 2   2 
 Numerical Problem 7
A thin plate of mass m and area A
is attached to a spring and oscillates
Fluid in a viscous fluid as shown in fig.
Spring
If f n is the frequency of oscillation
of the system in air  damping neglected 
and f d is the frequency of oscillation
A in the fluid, prove that;
2 m
 f n2  f d2
A
where the damping force is 2 AV ,
V being the velocity.
 Damping force 2 AV
Data : Damping coefficient  = =2 A
Velocity V
Proof : We know that damped frequency f d = f n (1   2 )
2
f
Squaring both sides, f d2 = f n2 (1   2 )  (1   2 )  d2
fn
 f 2
  f 2
 f 2

i.e.   1  2   
2 d n
2
d
  ( f n
2
 f d
2
)   2 2
fn (i )
 fn   fn 
c 2 A 2 A A
But      Substituting in eqn (i ),
cc 2mn 2m(2 f n ) 2 mf n
2
 A   A 
2
2πm
( fn  fd )  
2 2
  f 2
    η = (f 2
- f d )
2

 2 mf n   2 m 
n n
A
 Numerical Problem 8
A pendulum is pivoted at point O as shown in fig.
Neglecting the mass of the rod, determine the damped
natural frequency of the pendulum for small oscillations.

k
o a
b
L c

m
 k 

o a
b
L  c

Solution : Taking moments about O, for equilibrium,

 
I o  cb b  (ka )a  mgl  0 i.e. ml 2  cb 2  (ka 2  mgl )  0
 cb 2   ka 2 + mgl 
 θ +  2 θ +  2 θ = 0
 ml   ml 
Comparing with the standard equation for damped system,
 cb 2
  ka 2
+ mgl 
  2n  n2  0, we have 2ζωn =  2  & ωn = 
2
2 
 ml   ml 
Also we know that damped Frequency for a general damped system is
d =n 1   2  d2  n2 (1   2 )  n2  (n ) 2 (i )
 cb 2   ka 2
+ mgl 
Here, ζωn =  2 
& ωn = 
2
2  Substituting in (i)
 2ml   ml 
2 2
 ka  mgl   cb 
2 2
 ka + mgl   cb 
2 2
 
2
d 2  2 
 ωd =  2 - 2 
 ml   2 ml   ml   2ml 
 Numerical Problem 9
Determine the damped natural frequency for the system
shown in fig. Also, if m=1.5 kg, k=4900 N/m, a=6cm & b=14
cm, determine the value of damping coefficient ‘c’ for
which the system is critically damped.

k
a
O m
c
b
 k
a
O 
m
c

Solution : Taking moments about O, for equilibrium,

 
I o  ca a  (ka )a  0 i.e. mb 2  ca 2  ka 2  0
 ca 2   ka 2 
θ+ 2 
θ+ 2 
θ=0
 mb   mb 
Comparing with the standard equation for damped system,
 ca 2
  ka 2

  2n  n2  0, we have 2ζωn =  2  & ωn = 
2
2 
 mb   mb 
Also we know that damped Frequency for a general damped system is
d =n 1   2  d2  n2 (1   2 )  n2  (n ) 2 (i )
 ca 2   ka 2 
Here, ζωn =  2 
& ωn = 
2
2 
Substituting in (i)
 2mb   mb 
2
 ka   ca 
2
2 2
a k  c 
d   2   
2
2 
 ω d =  - 
 mb   2 mb  b   
m 2 m 
Substituting the values, m = 1.5 kg, k = 4900 N / m, a = 6 cm, b = 14 cm
 ka 2   4900  62 
  2  
2
2 
 24.5 rad / sec
 1.5 14 
n
 mb 
 ca 2 
Also, n   2 
For critical damping,  =1 & c  cc
 2mb 
 n  2mb 2   24.5  2 1.5 142 
 cc   2  2   400 N - sec / m
 a   6 
 Numerical Problem 10
Determine the damped natural frequency for the system
shown in fig. Also, if m=1.5 kg, k=4900 N/m, a=10 cm &
b=13 cm, determine the critical damping coefficient ‘c’ for
the system.

c
k
O m
a
b
 c
k
O m 
a
b

Solution : Taking moments about O, for equilibrium,

 
I o  ca a  (kb )b  0 i.e. ma 2  ca 2  kb 2  0

c  kb 2 
 θ +  θ +  2 
θ=0
m  ma 
Comparing with the standard equation for damped system,
 
c  kb 2

  2n  n   0, we have 2ζωn =   & ωn = 
2 2
2 
m  ma 
Also we know that damped Frequency for a general damped system is
d =n 1   2  d2  n2 (1   2 )  n2  (n ) 2 (i )
 c   kb 2

Here, ζωn =   & ωn = 
2
2 
Substituting in (i)
 2m   ma 
     c 
2 2
 
2 2
kb c kb
d   2   
2
  ωd =  2  
- 
 ma   2 m   ma   2m 
Substituting the values, m = 1.5 kg, k = 4900 N / m, a = 10 cm, b = 13 cm
 ka 2   4900 132 
  2  
2
2 
 74.3 rad / sec
 1.5 10 
n
 mb 
 c 
Also, n    For critical damping,  =1 & c  cc
 2m 
 cc  n  2m  74.3  2 1.5  222.9 N - sec / m
 Numerical Problem 6

A machine of mass 20 kg is mounted on springs of

total stiffness 10 N/mm & total damping coefficient
is 0.15 N/mm/sec. If the system is initially at rest
and a velocity of 100 mm/sec is imparted to the
mass, determine;
(i) Displacement & velocity of mass as function of
time
(ii) Displacement and velocity after one second.
Data : m  20 kg , k  10 N / mm  10 103 N / m
c  0.15 N / mm / sec  150 N / m / sec, x0  0, x0  100 mm / sec
Solution :
10 103
Undamped natural frequency n   22.36 rad / sec
20
c 150 150
Damping factor      0.168
cc 2mn 2  20  22.36
 ζωn  0.168  22.36  3.76
Damped natural frequency
d  n (1   2 )  22.36 (1  0.1682 ) = 22.04 rad / sec
As ζ < 1, i.e. the system is under-damped.
For under - damped system, the displacement is given by
x  Xe-nt sin(d t   )  (i )
Differentiating w.r.t time, we get velocity x
x  X e-nt d cos(d t   )  n e-nt sin(d t   ) (ii )
It is given that system is initially at rest.
i.e. when t  0, x  0 Substituting in Eqn (i ),
0  X sin    = 0 (As X  0)
Also initial velocity is given as 100 mm/sec.
i.e. when t  0, x  100 mm / sec, put in Eqn (ii )
100  X  22.04 cos(0)  (3.76) sin(0) 
 X = 4.537 mm
The displacement at any time t is given by;
x  4.537e-3.76t sin(22.04t ) (iii )
 Displacement at time t = 1 sec is,
x(1)  4.537e-3.76 sin(22.04)  0.04 mm
Velocity at any time t is given by;
x  4.537e-3.76t  22.04 cos(22.04t )  3.76e-3.76t sin(22.04t )  (iv)
 The velcity at at time t = 1 sec is,
x(1)  4.537e-3.76  22.04 cos(22.04)  3.76e-3.76 sin(22.04) 
x(1)  0.1056  20.43  0.033  2.155 mm / sec