FORCED VIBRATIONS
INTRODUCTION
When a mechanical system undergoes free vibrations, an initial force (causing some
displacement) is impressed upon the system, and the system is allowed to vibrate under the
influence of inherent elastic properties. The system however, comes to rest, depending upon
the amount of damping in the system.
In engineering situations, there are instances where in an external energy source
causes vibrations continuously acting on the system. Then the system is said to undergo
forced vibrations, as it vibrates due to the influence of external energy source. The external
energy source may be an externally impressed force or displacement excitation impressed
upon the system. The excitation may be periodic, impulsive or random in nature. Periodic
excitations may be harmonic or non harmonic but periodic. The amplitude of vibrations
remains almost constant. Machine tools, internal combustion engines, air compressors, etc
are few examples that undergo forced vibration.
3.2 FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC
EXCITATION
Consider a spring mass damper system as shown in Figure 3.1 excited by a sinusoidal forcing
function F=Fo Sint
k
Cx∙
kx
C
x
F
m
F = F0 Sin ωt
F
Fo
Figure 3.1
Let the force acts vertically upwards as shown in FBD. Then the Governing Differential
Equation (GDE) can be written as
mx = - Kx - C x + F
m x + C x + Kx = F --------------- (3.1)
is a linear non homogeneous II order differential equation whose solution is in two
parts.
1. Complementary Function or Transient Response
Consider the homogenous differential equation,
x = 0 which is incidentally the GDE of a single DOF spring mass damperm +xC + Kx
system. It has been shown in earlier discussions that for different conditions of damping,
the response decays with time. Thus the response is transient in nature and therefore
termed as transient response.
For an under damped system the complementary function or transient response.
ω t
xc = X 1 ē
ζ n
ω t
n
xc = X1 ēζ
Sin (ωdt + )
(A Sin ωdt + B Cos ωdt)
------------ (3.2)
2. Particular Integral or Steady State Response
This response neither builds up nor decays with time. It is steady state harmonic
oscillation having frequency equal to that of excitation. It can be determined as
follows.
Consider non-homogenous differential equation
mx + Cx + Kx = Fo Sin ωt
----------- (3.3)
The particular integral or steady state response is a steady state oscillation of the same
frequency ω as that of external excitation and the displacement vector lags the force vector
by some angle.
Let x = X Sin (ωt - ) be the trial solution
X: Amplitude of oscillation
: Phase of the displacement with respect to the exciting force (angle by which the
displacement vector lags the force vector).
Velocity = x
x = ωX. Cos (ωt - )
= ωX Sin [90 + (ωt - )]
Acceleration
x = - ω2 X. Sin (ωt - ), substitute these values in GDE, (equation 3.1)
We get
-m ω2 X Sin (ωt - ) + Cω Sin [90 + (ωt - )]
+ KX Sin (ωt - ) = Fo Sin ωt
2
m ω X Sin (ωt - ) - Cω Sin [90 + (ωt - )]
- KX Sin (ωt - ) + Fo Sin ωt = 0
---------- (3.4)
The four terms in the above equation represent both in magnitude and direction, the
four forces namely: inertia force, damping force, spring force and impressed force, taken in
order, acting on the system and their sum is equal to zero. Thus they satisfy the D’Alemberts
principle. F = 0. Now, if vector representation as shown in Figure 3.2, is employed to
denote these forces the force polygon shown in Figure 3.3 should close.
Represent the force vectors and draw the force polygon as given below.
Figure 3.2
A
X
B
) (ωt - )
O
Reference axis
Figure 3.3
Impressed force: Fo Sin ωt: acts at an angle ωt from the reference axis.
Displacement vector X: Lags the force vector by an angle  and hence shown at (ωt - ) from
the reference axis.
Spring force: - KX Sin (ωt - ): which means that the vector – KX acting at (ωt - ) or KX
acting in opposite direction to (ωt - ) = at [90 + (ωt - )]
Damping force: - CωX [Sin (90 + (ωt - )]
- CωX acting at 90 + (ωt - ) or CωX acting in opposite direction to [90 +
(ωt - )]
2
Inertia force: mω X Sin (ωt - )
Vector mω2X acting at (ωt - )
From the force polygon, in figure 3.3
Consider the triangle OAB.
OA2 = OB2 + BA2
= (CωX)2 + (KX-mω2X)2
Fo2 = X2 (Cω)2 + x2 (K-mω2)2
Fo2 = X2 [(K-mω2)2 + (Cω)2]
F0
X=
--------- (3.5)
√(K-mω2)2 + (cω)2
and tan  = OB
BA
tan  =
=
Cωx
KX-mω2x
=
Cω
K-mω2
Cω
K-mω2
 = tan-1
Cω
K-mω2
-------- (3.6)
If X and  are expressed in non-dimensional form it enables a concise graphical presentation
of results. Therefore, divide the numerator and the denominator by K.
F0 /K
X =
2
tan  =
Cω /K
mω2 )
(1K
mω 2
√
cω 2
(1) +(
)
K
K
k
Further, the above equations can be expressed in terms of the following quantities
Fo
K = Xst
- Zero frequency deflection
Deflection of spring mass system under the steady force Fo should not be mistaken as Δst =
mg
m K1
=
K
ωn2
C
K =
2ζ
Kωn
Thus X =
Xst
√(1 - ω2 )2 + (2 ζ. ω)2
ωn2
ωn
ω = r = frequency ratio
ωn
Xst
X=
--------- (3.7)
√ (1 - r2)2 +( 2 ζ. r)2
X
where
is called magnification factor, amplification factor, or amplitude ratio.
Xst
M=
X
:
Xst
It is the term by which Xst is to be multiplied to get the amplitude.
2ζ
ωn ω
tan =
... tan =
(1 -
2
ω
ωn2
)
=
2ζr
1-r2
2ζr
1-r2
-------- (3.8)
Thus the steady state response xp = X Sin (ωt - ), in which
X and  are as given above.
Total solution x = xc + xp
For under damped conditions:
as t
, xc
0 i.e., the transient response dies out. Complete solution consists only
steady state response only.
x = X Sin (ωt - )
--------- (3.9)
As mentioned above, the transient vibrations die out very soon and hence the system vibrates
with steady response amplitudes. The behaviour of the system can be best understood by
plotting frequency response curves as given below, in figure 3.4 and 3.5.
Frequency Response Curves:
Magnification Factor vs Frequency Ratio for Different amounts of Damping
ζ=0
Magnification
ζ=0
ζ = 0.25
Figure 3.4
Phase lag vs frequency ratio for different amounts of damping.
ζ=0
ζ = 0.25
ζ = 0.5
ζ = 0.707
Phase Angle, ,
ζ = 1.0
ζ = 2.0
0.5
1.0
1.5
2.0
2.5
3.0
Figure 3.5
Frequency Ratio r = (ω/ωn)
The following characteristics of the magnification factor (M) can be observed.
1) For damped system (ζ =0); M
as r
1.
2) Any amount of damping (ζ >0) reduces the magnification factor (M) for all values of
forcing frequency.
3) For any specified value of r, a higher value of damping reduces the value of M
4) When the force is constant (r =0), M =1.
5) The amplitude of the forced vibrations becomes smaller with increasing value of forced
frequency. i.e M
0,as r
.
6) For 0< ζ < 1/ √2 (0 < ζ <0.707), the maximum value of M occurs when r=√(1-2ζ 2) or
ω= ωn√(1-2ζ 2), which is lower than the Undamped natural frequency ω n and the
damped natural frequency ωd = ωn√(1-2 ζ 2),
7) The maximum value of X (when r=√ (1 - 2 ζ 2) is given by (X/Xst)= 1/[2 ζ√ (1-ζ2)] and
the value of X at ω= ωn is given by (X/Xst ) = 1/2 ζ
8) For ζ >1/√2, the graphs of M decreases with increasing values of r.
The following characteristics of the phase angle  can be observed from the graph
1) For undamped system the phase angle is 00 for 0<r<1, and 1800 for r>1. This implies
that the excitation and response are in phase for 0<r<1 and out of phase for r>1 when
ζ =0.
2) For ζ >0 and 0<r<1 the phase angle is given by 0 0<<900, implying that the response
lags excitation.
3) For ζ >0 and r>1, the phase angle is given by 90 0<< 1800, implying that the response
leads excitation.
4) For ζ >0 and r=1, the phase angle is =900 implying that the phase difference between
the excitation and response is 900.
5) For ζ >0 and large values of r, the phase angle  ω approaches 1800 implying that the
response and excitation are out of phase.
The damping factor ζ has a large influence on amplitude and phase angle in the region where
r = 1(resonance).The phenomenon represented be frequency response curve can be further
better understood by means of vector diagram as follows. Consider three different cases as
(1) ω/ ωn << 1 (2) ω/ ωn = 1 (3) ω/ ωn >> 1
Case (1): ω/ ωn << 1 for which ω should be very small
At very low frequencies, when ω is very small, the inertia for m ω 2x and the damping force
Cωx are very small.
Fo
Kx

CωX
2
mω X
Figure 3.6

(ωt-)
x
This results in small values of  as shown in fig .The impressed force F0 is almost equal and
opposite to spring force KX. Thus for very low frequencies, the phase angle tends to zero and
the impressed force wholly balance the spring force
Case (2): when ω/ ωn = 1
Kx
Fo
x
CωX
mω2X
Figure 3.7
With increased value of ω, the damping force Cωx and inertia force m ω 2 x increase. The
phase angle also increases. If ω is increased to such an extent that phase angle =900, the
force polygon becomes a rectangle as shown. The spring force and inertia vectors become
equal and opposite.
KX = m ω2 x
ω = √(K/m) = ωn
ω = ωn
ω / ωn =1
This is the response condition of the system during which the forcing frequency is equal to
natural frequency of the system. Also the impressed force is completely balanced by the
damping force.
CωX= F0
X= F0 /Cω = F0/K/ Cω/K
X=Xst / 2 ζ (ω/ ωn)
X=Xst / 2 ζ
(ω/ ωn) = 1
Xr /Xst = 1/2 ζ
Xr = Amplitude at resonance
Case (3): when ω / ωn >>1
x
Fo
KX
mω2X
CωX
Figure 3.8
At very large values of ω > approaches 1800, the inertia force becomes very large, where as
the spring force and damping force vectors becomes negligibly small. The improved force is
wholly utilized in balancing the inertia force.

1800 i.e., Fo = m ω2x
X = Fo / m ω 2
NUMERICAL EXAMPLES:
3.1) A machine part of mass 2.5 Kgs vibrates in a viscous medium. A harmonic exciting
force of 30 N acts on the part and causes resonant amplitude of 14mm with a period of
0.22sec. Find the damping coefficient. If the frequency of the exciting force is changed to
4Hz, determine the increase in the amplitude of forced vibration upon removal of the damper.
Data: m = 2.5Kg, F0 = 30N, X = 14mm, τ = 0.225sec
Part 1: At Resonance
ωn = forcing frequency = 2/ τ = 28.56 rad/sec
At resonance: ω = ωn = 28.56 rad/sec
ωn = √(K/m) = 28.56 rad/sec
K = 2039 N/m
Amplitude at resonance
Fo/K
X=
√ [1 - r2] 2 + [2ζr] 2
As ω/ ωn = 1,
X = (F0/K)/2ζ = 0.014
... ζ = 0.526
Damping coefficient = C = Cc ζ = 2m ωn ζ
= 2*2.5*28.56*0.526
= 75.04 N/m/s
C = 0.07504 Ns/m
Part (2): When f = 4 Hz
Forcing
ω = 2*fn = 25.13 rad/sec
Frequency
ωn = 28.56 rad/sec, unchanged
Amplitude of vibration with damper
Xa =
Fo/K
√ [1 - r2] 2 + [2ζr] 2
= 0.01544m
Amplitude of vibration without damper
Xb = (30/2039)/(0.2258)
= 0.0652m
Increase in Amplitude = 0.0652 – 0.0155 = 0.0497m
Amplitude = 49.7mm
3.2) A body having a mass of 15 kgs, is suspended from a spring which deflects 12mm due to
the weight of the mass. Determine the frequency of free vibrations. What viscous damping
force is needed to make the motion a periodic at a speed of 1mm/sec.
If when, damped to this extent, a disturbing force having a maximum value of 100N
and vibrating at 6Hz is made to act on the body. Determine the amplitude of ultimate motion.
Solution:
Data: m = 15Kg; F0 = 100 N; f = 6Hz; Δst = 12mm;
(a) fn = (1/2)√(g/ Δ st) = 4.55Hz
(b) The motion becomes aperiodic, when the damped frequency is zero or when it is critically
damped (ζ = 1).
ω = ωn = √(g/ Δ) = 28.59 rad/sec
C = Cc = 2m ωn = 2*15*28.59 = 857 N/m/s
= 0.857 N/mm/s
Thus a force of 0.857 N is required at a rate of 1mm/s to make the motion a periodic.
(c)
X=
F0
√(K-mω2)2 + (cω)2
ω = 2f = 2*6 = 37.7 rad/sec, f0 = 100 N
fn = (1/2)(√(K/m)
... K = 12,260 N/m
X = 0.00298m
= 2.98mm.
Condition for peak amplitude of vibration
(Expression for peak amplitude)
The frequency at which the maximum amplitude occurs can be obtained as follows.
M=
X
Xst
Xst
 X = √ [1 - r2] 2 + [2ζr] 2
i.e., for a system acted upon by a known harmonic force, the amplitude depends only on (ω/
ωn). Hence for X to be maximum √ [1 – r2] 2 + [2ζr] 2 should be minimum.

dx
dx
d(ω/ ωn)  d(r) ([1 - r2] 2 + [2ζr] 2) = 0
2(1 - r2) 2 (-2r) + 4ζ2r = 0
2(1 - r2) + 4ζ2r = 0
42ζ2r = 0 = 2(1 - r2)
2ζ2 = 1-r2
r2 = 1 -2ζ2
r = √1 -2 ζ2
(ω/ ωn) peak = √1 -2 ζ2
(
ω
ωn
)
(
ωp
ωn
)
peak
peak
= √1 -2 ζ2
= √1 -2 ζ2
---------- (3.10)
ωp = frequency at which peak amplitude occurs.
Where ωp refers to the forcing frequency corresponding to the peak amplitude. No maximum
or peak will1occur when the expression within the radical sign becomes negative i.e., for ζ >
or for ζ > 0.707.
√2
(
ωp
ωn
) = √1 -2 ζ
2
and peak amplitude is given by
(X/Xst)max = 1/[2 ζ(√1- ζ)]
--------- (3.11)
3.3) A machine of mass 25 kgs, is placed on an elastic foundation. A sinusoidal force of
magnitude 25N is applied to the machine. A frequency sweep reveals that the maximum
steady state amplitude of 1.3mm occurs when the period of response is 0.22sec. Determine
the equivalent stiffness and damping ratio of the foundation.
Solution:
Data: F0 = 25N; m = 25 Kgs; Xmax = 1.3mm; τ = 0.22sec
For a linear system, the frequency of response is same as frequency of excitation.
... Excitation frequency = ω = 2f = 2/ τ = 28.6 rad/sec
thus Xmax = occurs, when ω = 28.6 rad/s
Condition for maximum amplitude to occur:
r = √1 -2 ζ2 = ω/ωn
... ωn = ω /(√1 -2 ζ2 ) = 28.6/(√1 -2 ζ2 ) --------------(1)
also we have,
X/Xst =
for Xmax = r =√1 -2 ζ2
1
√ [1 - r2] 2 + [2ζr] 2
Xmax/Xst =
1
√ [1 – (1 -2 ζ2 )] 2 + [4ζ2(1 -2 ζ2 )]
=
1
2 ζ √ (1 -ζ2 )
Xmax/(F0/K) =
1
2 ζ √ (1 -ζ2 )
Xmaxmωn2/F0 =
1
2 ζ √ (1 -ζ2 )
25*0.013* ωn2/25 =
1
2 ζ √ (1 -ζ2 )
Now substitute for ωn2 from eq.(1);
0.013*28.6/(√1 -2 ζ2) =
1
2 ζ √ (1 -ζ2)
1.0633/(√1 -2 ζ2) =
1
2 ζ √ (1 -ζ2)
Squaring and rearranging,
ζ4 - ζ2 +0.117 = 0
Z2 – Z + 0.117 = 0
where ζ2 = Z.
Solving the quadratic equation
ζ = 0.368, 0.93
The larger value of ζ is to be discarded because the amplitude would be maximum only
for ζ < 0.707 ... take ζ = 0.368
... natural frequency ωn =
ω
√ (1 – 2(0.368)2 )
ωn = 33.5 rad/sec
stiffness of the foundation,
K = mωn2 = 25(33.5)2
= 28.05*103 N/m
3.4) A weight attached to a spring of stiffness 525 N/m has a viscous damping device. When
the weight is displaced and released, without damper the period of vibration is found to be
1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude and
phase when the force F=2Cos3t acts on the system.
Solution:
Data: K = 525 N/m; τ = 1.8secs: x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t
... F0 = 2N, ω = 3 rad/sec
X=
Fo/K
√ [1 - r2] 2 + [2ζr] 2
ωn = 2/ τ = 3.49rad/sec
δ = ln(4.2/1.0) = 1.435
ζ=
δ
= 0.22
2
2
√ (4 + δ )
r = ω/ωn = 2/3.49 = 0.573
r2 = 0.328
X=
2/525
√ [1 – 0.328] 2 + [4*0.484*0.328]
X = 5.3mm
 = tan-1(2ζr)
(1-r2)
-1
 = tan (2*0.22*0.573)
(1-0.328)
-1
 = tan (0.375)
 = 20.560
3.5) The damped natural frequency of a system as obtained from a free vibration test is 9.8
cps. During a forced vibration test with a harmonic excitation on the same system, the
frequency of vibration corresponding to peak amplitude was found to be 9.6 cps.
Determine the damping factor for the system and natural frequency.
ωd = 9.8 cps,
ωp = 9.6 cps.
(ωp / ωn) = √1 -2ζ2
ωn = ωd/√1 -2ζ2
 ωp√1 -2ζ2 /ωd = √1 -2ζ2
Solving for ζ: ζ = 0.196
ωn = ωd/ √1 -2ζ2 = 10 cps.
3.6) A reciprocating pump of mass 300 Kgs is mounted at the middle of a steel plate of
thickness 12 mm and width 500 mm and length 2.5 m damped along two edges as shown.
During the operation of the pump, the plate is subjected to a harmonic excitation of F(t) = 50
cos 60 t N. Determine the amplitude of vibration of the plate.
.
12
m = 300 Kgs
F0 = 50 N
ω = 60
K = 192EI/l3 = 176.94*103 N/m
ζ=0
X = F0 /(K-m ω2)2
X = 6.13*10-8mm
2.5 m
500
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