The document discusses forced vibration in mechanical systems. It defines forced vibration as vibration under the influence of external forces. Periodic, harmonic forcing causes steady state vibration with constant amplitude. Common sources of periodic forcing include unbalanced rotating or reciprocating masses. The response of a system to periodic forcing contains components at both the forcing frequency and the natural frequency. Over time, the natural frequency response dies out, leaving only the steady state response at the forcing frequency.

1. UNIT IV FORCED VIBRATION 9
Response of one degree freedom systems to periodic forcing – Harmonic disturbances –Disturbance
caused by unbalance – Support motion –transmissibility – Vibration isolation vibration measurement.
4. Forced Vibration
There is no influence of external forces in the free vibration of a system. Think of a child swinging on
a swing. A child swinging freely on a swing will do so consistently. It consistently takes two to three
seconds to a complete each cycle of swinging. Think of an oscillating pendulum. The vibration of a
pendulum needs no external force to vibrate. A machine left to vibrate freely will tend to vibrate at its
natural oscillation (vibration) rate.
In forced vibration of a system, it is under the influence of external force applied on the system.
The vibration of a machine like a drill is forced vibration. It needs an external force to vibrate.
External forcesare appliedforces. Anappliedforce isaforce that isappliedtoan objectbya personor
anotherobject.If a personispushinga deskacrossthe room, thenthere isan appliedforce actingupon
the object.The appliedforce isthe force exertedonthe deskbythe person. Similarlythe force applied
by the wavesonthe boat.
A boat anchored in a bay is subjected to repeated waves slapping on the sides of the boat. The boat
rocks as long as the waves continue to act on the boat.
Most machine vibrations are also similar. The vibrations are due to repeating forces acting
on its components.
Repeating (Periodic!) forces in machines are mostly due to the rotation of imbalanced (uneven
rotor, bent shaft), misaligned (improper mounting, distortion due to fastening torque), worn
(worn belt, gear teeth), or improperly driven (intermittent bush contact in motors, misfiring
in IC engine cylinder, uneven air supply) machine components.
4.1 Types of forces causing vibration
Types of forces (forcing function!) that cause vibration are:
1. Random forcing
2. Impulsive forcing
3. Periodic forcing
1. Random forcing:

2. These are unpredictable and non-deterministic (difficult to determine the magnitude using
expression)
Examples: Ground motion during earth quake, jet engine noise.
2. Impulsive forcing: They are short duration forces and non-periodic. The vibrations die out soon
and are not significant.
Examples: Rock explosion, gun firing, punching die.
3. Periodic forcing: These are predictable and deterministic (magnitude given by an expression).
Example: Centrifugal (unbalanced) forces caused by an eccentric rotating mass, Inertia
(unbalanced) forces caused by reciprocating mass in an IC engine.
4.1.1 Types of forced vibrations
Classifications based on forces causing vibration
I. Forced vibrations are classified according to the type of forces causing the excitation:
1. Random vibration
2. Transient vibration
3. Steady state vibration
1. Random vibration:
Vibrations caused by random forces are called random vibration.
Examples: Ground motion during earth quake, jet engine noise.
2. Transient vibration:
Vibrations caused by impulsive forces and which are of short duration are called transient
vibration.
Examples: Rock explosion, gun firing, punching die.
3. Steady state vibration:
Periodic vibration with constant amplitude is called steady state vibration.
Example: Centrifugal (unbalanced) forces caused by an eccentric rotating mass under steady
state, Inertia (unbalanced) forces caused by reciprocating mass in an IC engine under steady
state.
II. Classifications based on damping resistance
Forced vibrations are also classified into two types based on the existence of damping
resistance:
1. Undamped forced vibration (c = 0)
2. Damped forced vibration (c ≠ 0)
where ‘c’ is damping coefficient N/(m/s)
1. Undamped forced vibration (c = 0):
When the damping resistance is nil in a forced vibration, it is called undamped forced vibration.
2. Damped forced vibration (c ≠ 0):
When the damping resistance is present in a forced vibration, it is called damped forced
vibration.
4.2 Periodic forcing

3. A force which acts continuously at specific intervals of time is called periodic force. There are two
types of periodic forcing:
1. Harmonic forcing and
2. Non-harmonic forcing.
1. Harmonic forcing: When the applied load (force!) varies as a sine or cosine function, it is called
harmonic forcing.
Example:
Sine loading: f(t) = F sin (ωt)
Cosine loading: f(t) = F cos(ωt)
The response of a system to a harmonic excitation (loading!) is called harmonic response.
2. Non-harmonic forcing: . Any harmonicfunction such as sin or cosine is periodic.But the converse
is nottrue. Non-harmonic forcing is a type of periodic function that is discontinuous. The restoring
force is independent of displacement. It can be represented by either a discrete or combination
of sin and cos functions.
Example: F(t+tp) = F(t) where t is time, tp is time period
4.2.1 Response to periodic forcing
Forces acting on any vibrating machine systemcan be represented in mathematical equation
𝑚𝑥̈ + 𝑘𝑥 = 𝑘 𝑦( 𝜔𝑡)
𝑤ℎ𝑒𝑟𝑒 𝜔𝑡 = 𝑛𝜋, 𝑛 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟.
Consider a system as shown in Fig.14.5.1

4. A roller is guided in the groove of a face cam generates motion from the point O.
It is a single degree of freedom system
The displacement is expressed by the relation
𝑦( 𝜔𝑡) = 𝜔0(1 − cos 𝜔𝑡)
The far end of the spring of stiffness ‘k’ connected to the sliding mass ‘m’ is assumed as zero.
The roller is assumed to be in its extreme left position when the cam begins zero.
[Note: By appropriately choosing the initial zero position (A), we can also express the
displacement by the relation 𝑦 𝐴( 𝜔𝑡) = 𝑦0 sin 𝜔𝑡 ]
The differential equation of motion is now
𝑚𝑥̈ + 𝑘𝑥 = 𝑘𝜔0(1 − cos 𝜔𝑡)
The solution of this differential equation is
𝑥 = 𝐴 cos 𝜔 𝑛 𝑡 + 𝐵 sin 𝜔 𝑛 𝑡 + 𝑦0 [1 −
cos 𝜔𝑡
1 − (
𝜔
𝜔 𝑛
)
2]
Substituting the boundary conditions,
𝑥(0) = 0, 𝑥̇(0) = 0
𝐴 =
(
𝜔
𝜔 𝑛
)
2
𝑦0
[1 − (
𝜔
𝜔 𝑛
)
2
]
𝑎𝑛𝑑 𝐵 = 0
Substituting the coefficient values,
The complete simplified equation of motion is

5. 𝑥 =
(
𝜔
𝜔 𝑛
)
2
𝑦0 cos 𝜔 𝑛 𝑡
[1 − (
𝜔
𝜔 𝑛
)
2
]
−
𝑦0 cos 𝜔𝑡
[1 − (
𝜔
𝜔 𝑛
)
2
]
+ 𝑦0
The total motion contains a constant plus two vibrations of differing amplitudes and
frequencies.
The first term on the right hand side
(
𝜔
𝜔 𝑛
)
2
𝑦0 cos 𝜔 𝑛 𝑡
[1 − (
𝜔
𝜔 𝑛
)
2
]
is called starting transient. This is not forcing frequency. This is a vibration at natural frequency
ωn. The presence of friction (damping!) would cause this term to die out after a short period of
time.
𝐹𝑜𝑟
𝜔
𝜔 𝑛
= 0, 𝑤𝑒 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑟𝑖𝑔𝑖𝑑 𝑏𝑜𝑑𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 = 𝑦0[1 − cos 𝜔𝑡]
If the spring is replaced by a rigid member then k and ωn become very large. So the mass
exactly follows the cam motion.
4.2.2 Harmonic disturbances
A part of any moving or rotating machinery is often subjected to forces which vary
periodically with respect to time. The parts which are metal elements have both mass and
elasticity. Therefore there is a vibration exists. Since machines normally operate at constant
speeds and constant output, vibratory forces may have a constant amplitude over a period of
time. These varying forces also may vary in magnitude with speed and output. It is general
practice to analyse the vibration problems assuming periodically varying force of constant
amplitude.
4.2.3 Equation of motion
Here we shall consider the application of a single sinusoidal force. The solution contains
components of motion at two frequency levels. One at forcing frequency and the other at
natural frequency of the system.
Since damping is always present in actual systems, component at natural frequency level
becomes insignificant after a certain time.
Therefore the component that contains only forcing frequency remains. This motion is called
steady state motion.
The equation of motion is
𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 𝐹0 cos 𝜔𝑡

6. Complete solution
The complete solution of this equation is
𝑥 = 𝑒−𝜉𝜔 𝑛 𝑡( 𝐶1 sin 𝑞𝑡 + 𝐶2 cos 𝑞𝑡) +
𝐹0 cos( 𝜔𝑡 − 𝜙)
√( 𝑘 − 𝑚𝜔2)2 + 𝑐2 𝜔2
(i) Transient response
The first term on the right hand side (RHS) of the complete solution is a transient term. The
exponential part will cause it to decay in a short period of time.
(ii) Steady state response
When the transient part dies out, only the second part remains. Since this is neither the
beginning nor the end, it is called steady state solution.
The following expressions are introduced to obtain simplified solution.
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝜔2
𝑋
𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑐𝜔𝑋
𝑆𝑝𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑘𝑋
𝐸𝑥𝑐𝑖𝑡𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝐹0
𝜔 𝑛 = √
𝑘
𝑚
𝜉 =
𝑐
𝑐 𝑐
𝑐 𝑐 = 2𝑚𝜔 𝑛
𝑋
(
𝐹0
𝑘⁄ )
=
1
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
(
𝐹0
𝑘⁄ ) = Deflection that a spring of stiffness ‘k’ would experience if acted upon by a force 𝐹0 .
The plot of frequency response namely Amplitude ratio vs Frequency ratio is shown in Fig.15.6

7. 4.2.4 Dynamic magnifier
Dynamic magnifier (Magnification factor!) is the ratio of maximum displacement of the forced
vibration (amplitude!) to the static deflection due to static force.
𝑋 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒𝑑 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛
𝐹0
𝑘
= 𝑆𝑡𝑎𝑡𝑖𝑐 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑋
(
𝐹0
𝑘⁄ )
=
1
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
4.2.5 Frequency response

8. 4.3 Disturbance causedby unbalance
Most of the machines like motors, compressors, engines produce vibrations due to rapid
rotation of a small unbalanced mass.
The differential equation for such systems is
𝑚𝑥̈ + 𝑘𝑥 = 𝑚 𝑢 𝑒 𝜔2
cos 𝜔𝑡
𝑚 = 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑛𝑔 𝑚𝑎𝑠𝑠
𝑚 𝑢 = 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑎𝑠𝑠
𝑒 = 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦
4.3.1 Rotating unbalance

9. The following expressions are introduced to obtain simplified solution.
𝑚 = 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑛𝑔 𝑚𝑎𝑠𝑠
𝑚 𝑢 = 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑎𝑠𝑠
𝑒 = 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦
The simplified solution is
𝑋
( 𝑚 𝑢 𝑒
𝑚⁄ )
=
𝜔2
𝜔 𝑛
2
√(1−
𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
The plot of frequency response namely Amplitude ratio vs Frequency ratio is shown in Fig.16.2
Fig17.34P589
From the plot we conclude that at high speeds, where the frequency ratio (
𝜔
𝜔 𝑛
)is greater than
unity, amplitude X can be reduced only by reducing the mass and eccentricity of the rotating
unbalance.

10. Relative motion:
𝑍
𝑦0
=
𝜔2
𝜔 𝑛
2
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
4.3.2 Reciprocating unbalance
In a reciprocating engine, there is an unbalanced primary inertia force given by 𝑚𝜔2
𝑟 cos 𝜃
This primary unbalanced force always acts along the line of stroke with varying magnitude.

11. This reciprocating unbalance can be considered as the horizontal component of an imaginary
rotating mass m kept at crank radius r (crank pin!).
Now balancing is done in the same way as we have done for rotating unbalance.
Let the balancing be done by balancing mass mB at radius rB.
The reciprocating unbalance force = Horizontal component of rotating balancing mass
𝑚𝜔2
𝑟 cos 𝜃 = 𝑚 𝐵 𝜔2
𝑟𝐵 cos 𝜃
However the balancing mass also has a vertical component of its centrifugal force namely
𝑚 𝐵 𝜔2
𝑟𝐵 sin 𝜃
This remains unbalanced. Therefore when we attempt to balance horizontal component, a
vertical unbalance is introduced!
To minimize the unbalance, a compromise is therefore made. Only a fraction (c ) of the
reciprocating mass is balanced.
𝑐𝑚𝑟 = 𝑚 𝐵 𝑟𝐵
If there exists both reciprocating as well as rotating unbalance in a machine, then
𝑚 𝐵 𝑟𝐵 = ( 𝑚 𝑢 + 𝑐 𝑚) 𝑟
4.3.3 Damping factor (ξ)
Damping factor (damping ratio!) is the ratio of actual damping coefficient, c to critical damping
coefficient, cc.
𝜉 =
𝑐
𝑐 𝑐
𝐼𝑓 𝜉 > 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑜𝑣𝑒𝑟 𝑑𝑎𝑚𝑝𝑒𝑑.
𝐼𝑓 𝜉 = 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑
𝐼𝑓 𝜉 > 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑢𝑛𝑑𝑒𝑟 𝑑𝑎𝑚𝑝𝑒𝑑

12. 4.3.4 Steady state response
The simplified solution for steady state response X is
𝑋
( 𝑚 𝑢 𝑒
𝑚⁄ )
=
𝜔2
𝜔 𝑛
2
√(1−
𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
4.3.5 Absolute amplitude

13. The simplified solution for absolute amplitude X is
𝑋
( 𝑚 𝑢 𝑒
𝑚⁄ )
=
𝜔2
𝜔 𝑛
2
√(1−
𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
4.3.6 Relative amplitude
The relative amplitude Z is

14. From the plot we conclude that at high speeds, where the frequency ratio (
𝜔
𝜔 𝑛
)is greater than
unity, amplitude X can be reduced only by reducing the mass and eccentricity of the rotating
unbalance.
Relative motion:
𝑍
𝑦0
=
𝜔2
𝜔 𝑛
2
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
4.3.7 Phase angle
The direction of 𝐹0 cos( 𝜔𝑡) will always be ahead of 𝑘𝑋 cos( 𝜔𝑡 − 𝜙) by angle 𝜙.
The phase angle 𝜙 is
𝝓 = 𝐭𝐚𝐧−𝟏
2𝜉 𝜔
𝜔 𝑛
⁄
(1 −
𝜔2
𝜔 𝑛
2)
4.3.8 Resultant force on motor
Resultant force on motor is the resultant of the forces exerted by the spring and dashpot
4.3.9 Resonance speed
Let N be the speed of the driving shaft of the motor at which resonance occurs.
The angular speed at which resonance occurs is given by
𝜔 = 𝜔 𝑛 = √
𝑘
𝑚
Resonance speed N is obtained from the relation
𝜔 =
2𝜋𝑁
60
4.3.10 Amplitude of resonance
Amplitude of resonance X is obtained by substituting 𝜔 = 𝜔 𝑛 in the relation
𝑋
( 𝑚 𝑢 𝑒
𝑚⁄ )
=
𝜔2
𝜔 𝑛
2
√(1−
𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
𝑋
(
𝑚 𝑢 𝑒
𝑚⁄ )
=
1
√(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
=
1
2𝜉 𝜔
𝜔 𝑛
⁄

15. 𝑋
(
𝑚 𝑢 𝑒
𝑚⁄ )
=
1
2𝜉 𝜔
𝜔 𝑛
⁄
4.4 Support motion
Support motion is also called base excitation. In many situations, the excitation is created
(applied!) by the base or support configuration. Example: sinusoidal profile of a road.
4.4.1 Absolute amplitude
Absolute harmonic displacement of support is y = Y sin ωt
Absolute displacement of mass m is x.
Absolute differential equation of motion is
𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 𝑌√ 𝑘2 + 𝑐2 𝜔2 sin( 𝜔𝑡 + 𝛼)

16. 𝑭 𝟎 = 𝒀√ 𝒌 𝟐 + 𝒄 𝟐 𝝎 𝟐
4.4.2 Relative amplitude
Relative amplitude is z = y-x, where x is absolute displacement of mass m.
Relative differential equation of motion is
𝑚𝑧̈ + 𝑐𝑧̇ + 𝑘𝑧 = 𝑚𝜔2
𝑌(sin 𝜔𝑡)
𝐹0 = 𝑚𝜔2
𝑌
4.4.3 Circular frequency of vibration
Natural circular frequency of motion 𝜔 𝑛 = √
𝑘
𝑚
𝑇𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 =
𝑊𝑎𝑣𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
→ 𝑡 𝑝 =
𝜆
𝑣
4.4.4 Vertical amplitude of vibration
Steady state amplitude due to excitation of support,

17. 𝑿 =
𝒀√ 𝟏+( 𝟐𝝃
𝝎
𝝎 𝒏
)
𝟐
√[ 𝟏−(
𝝎
𝝎 𝒏
)
𝟐
]
𝟐
+( 𝟐𝝃
𝝎
𝝎 𝒏
)
𝟐
4.5 Vibrationisolation
Vibrations are produced in all machines having motion of unbalanced masses. These vibrations
will be transferred to the foundation or base supports upon which the machines are installed.
This will result in noise, wear, and failure of machine as well as the structure. Hence this is not
desirable and needs to be eliminated (isolated!). If not at least diminish to acceptable limits.
4.5.1 Transmissibility
Transmissibility is defined as the ratio of force or displacement transmitted to the foundation to
the vibrating force or displacement applied by the unbalance. Transmissibility is a measure of
effectiveness of the vibration isolating material.
1. Force transmissibility
In order to reduce the transmitting forces to the foundation, machines are mounted on springs
and dampers or some other isolation materials like cork, rubber which have these properties.
Definition: Force Transmissibility is defined as the ratio of force transmitted to the foundation
to the vibrating force applied by the unbalance. Transmissibility is a measure of effectiveness of
the vibration isolating material.
Force transmissibility is also called as isolation factor.
TransmissibilityvsFrequencyratio
The transmitted force to the foundation is a vector sum of spring force (kX) and damping force
(cωX). The spring force and damping force act perpendicular to each other. Therefor the
resultant transmitted force Ft is
𝐹𝑡 = √( 𝑘𝑋)2 + ( 𝑐𝜔𝑋)2 = 𝑋√𝑘2 + ( 𝑐𝜔)2.
𝑋
(
𝐹0
𝑘⁄ )
=
1
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
Multiplying the above two equations
𝐹𝑡 𝑘𝑋
𝐹0
=
𝑋√𝑘2 + ( 𝑐𝜔)2
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2

18. →
𝐹𝑡
𝐹0
=
1
𝑘
√𝑘2 + ( 𝑐𝜔)2
√(1 −
𝜔2
𝜔 𝑛
2)
2
+ (2𝜉 𝜔
𝜔 𝑛
⁄ )
2
→
𝐹𝑡
𝐹0
=
√1+(
𝑐𝜔
𝑘
)
2
√(1−
𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛⁄ )
2
(1)
We know that 𝜔 𝑛 = √
𝑘
𝑚
; 𝜉 =
𝑐
𝑐 𝑐
; 𝑐 𝑐 = 2𝑚𝜔 𝑛; → 𝑚 =
𝑐 𝑐
2𝜔 𝑛
=
(
𝑐
𝜉
)
2𝜔 𝑛
=
𝑐
2𝜉 𝜔 𝑛
→ 𝜔 𝑛 = √
𝑘(2𝜉𝜔 𝑛)
𝑐
→ 𝜔 𝑛
2
=
𝑘(2𝜉𝜔 𝑛)
𝑐
→ 𝜔 𝑛 =
𝑘2𝜉
𝑐
→
𝑐
𝑘
=
2𝜉
𝜔 𝑛
Substituting the value for (
𝑐
𝑘
) in equation (1)
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦, 𝜀 =
𝐹𝑡
𝐹0
=
√1 +(2𝜉 𝜔
𝜔 𝑛
)
2
√(1 − 𝜔2
𝜔 𝑛
2)
2
+(2𝜉 𝜔
𝜔 𝑛
⁄ )
2
At resonance,
𝜔
𝜔 𝑛
= 1
Hence
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦, 𝜀 =
𝐹𝑡
𝐹0
=
√1 + (2𝜉)2
2𝜉
With no damper, c=0 and 𝜉 =
𝑐
𝑐 𝑐
= 0
Hence
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦, 𝜀 =
𝐹𝑡
𝐹0
= ±
1
√(1 − 𝜔2
𝜔 𝑛
2)
2
Transmissibilityregions

19. A plot of 𝜀 𝑣𝑠 (
𝜔
𝜔 𝑛
) can be drawn for different values of ξ.
The observations from the plot are
1. Critical value of (
𝜔
𝜔 𝑛
) = √2
2. When (
𝜔
𝜔 𝑛
) > √2, 𝑡ℎ𝑒𝑛 𝜀 < 1
3. When (
𝜔
𝜔 𝑛
) < √2, 𝑡ℎ𝑒𝑛 𝜀 > 1
4. When (
𝜔
𝜔 𝑛
) = √2, 𝑡ℎ𝑒𝑛 𝜀 = 1
5. When (
𝜔
𝜔 𝑛
) = 1, 𝑡ℎ𝑒𝑛 𝜀 = ∞
6. Up to (
𝜔
𝜔 𝑛
) = 1, increasing (
𝜔
𝜔 𝑛
) increases ε
7. Beyond (
𝜔
𝜔 𝑛
) = 1, increasing (
𝜔
𝜔 𝑛
) decreases ε
8. Note that as damping 𝑐 𝑜𝑟 𝜉 is increased, (
𝜔
𝜔 𝑛
) decreases. Therefore for various
regions, the effect of damping varies as above.
Include examples from excel file Q15and
Example: A machine supportedsymmetricallyonfourspringshasa mass of 80 kg.
The mass of the reciprocatingpartis2.2 kg whichmove througha vertical stroke of 100 mm withsimple harmonic
motion.Neglectingdamping,determinethe combinedstiffnessof the springssothatthe force transmittedtothe
foundationis1/20th
of the impressedforce.
The machine crankshaft rotatesat 800 rpm.
If underactual workingconditions,the dampingreducesthe amplitudesof successive vibrationsby30%,find:
(i) the force transmittedtothe foundationat800 rpm,
(ii) the force transmittedtothe foundationatresonance,and
(iii) the amplitude of the vibrationsatresonance.

20. Unit– 4 November/December2006; Unit – 4 November/December2008;
Unit– 4 May/June 2009; Unit– 4 November/December2009
Unit– 4 NOVEMBER/DECEMBER 2010
KJ10.38.22
Given: Mass of machine m = 80 kg; No. of support springs n = 4;
Mass of reciprocating parts mR = 2.2 kg; Vertical stroke of SHM L = 0.1 m;
Neglect damping; Force transmitted to foundation = (1/20) impressed force;
Speed of machine crank shaft = 800 rpm;
Damping reduces amplitudes of successive vibrations by 30%;
Find: (a) Combined stiffness of spring;
(b) Force transmitted to the foundation at 800 rpm;
(c ) Force transmitted to the foundation at resonance; and
(d) Amplitude of vibration at resonance.
Solution:
mass,m= 80 kg
Stroke length, L = 0.1 m; Hence, eccentricity e=crank radius r = L/2= 0.05 m
Assuming no damping, i.e. c = 0 and ξ=0 we find ωn
Speed, N = 800 rpm
Circular frequency, 𝜔 = 2𝜋𝑓 = 2𝜋𝑥13.3333
=
Logarithmic decrement, 𝜕 =
1
𝑛
ln (
𝑥0
𝑥 𝑛
) = 𝜉𝜔 𝑛 𝑡 𝑑 =
2𝜋𝜉
√1 − 𝜉2
=
1
1
𝑙𝑛 (
1
0.7
) =
Damping Factor, 𝜉 =
𝑐
𝑐 𝑐𝑟
= √
𝜕2
4𝜋2 + 𝜕2 =
Impressed force, 𝐹0 = 𝑚0 𝜔2 𝑒 = 2.2𝑥83.80952 𝑥0.05 =
Force transmitted, 𝐹𝑇 = ɛ 𝑥 𝐹0 = 0.05𝑥702.404 =
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜, ɛ =
𝐹𝑇
𝐹0
=
√1 + [2𝜉 (
𝜔
𝜔 𝑛
)]
2
√[1− (
𝜔
𝜔 𝑛
)
2
]
2
+ [2𝜉 (
𝜔
𝜔 𝑛
)]
2
= 0.05

21. Hence , Resonant frequency,
The four springs in parallel,
Example:Findthe stiffnessof eachspringwhenarefrigeratorunithavingamassof 30 kg isto be
supportedbythree springs.The force transmittedtothe supportingstructure isonly10% of the
impressedforce.The refrigeratorunitoperatesat420 r.p.m.(16) Unit – 4 November/December
2005
JK10.36.20
Given: Mass of refrigerator unit m = 30 kg; No. of support springs n =3;
Force transmitted to support structure = 0.1 impressed force
𝜔 𝑛 = √
𝑘 𝑒𝑞
𝑚
= √
𝑘 𝑒𝑞
80
= 12.2887
∴ 𝑘 𝑒𝑞 = 𝑛 𝑠 𝑘; 𝑘 =
𝑘 𝑒𝑞
𝑛 𝑠
=
26758.2
4
=
Impressed force, 𝐹0 = 𝑚0 𝜔2 𝑒 = 2.2𝑥18.28872 𝑥0.05 =
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜, ɛ =
𝐹𝑇
𝐹0
=
√1 + [2𝜉 (
𝜔
𝜔 𝑛
)]
2
√[1 − (
𝜔
𝜔 𝑛
)
2
]
2
+ [2𝜉 (
𝜔
𝜔 𝑛
)]
2
=
√1 + 2𝑥0.05665
√[2𝑥0.05665]2 + 0
=
Force transmitted at resonance , 𝐹𝑇 = ɛ 𝑥 𝐹0 = 8.88219𝑥36.7926 =
𝑀. 𝐹.=
𝐴
𝑋0
==
1
√[1 − (
𝜔
𝜔 𝑛
)
2
]
2
+ [2𝜉 (
𝜔
𝜔 𝑛
)]
2
=
1
√[1 − 1]2 + [2𝑥0.05665(1)]2
=
Amplitude at resonance, A = M.F. x 𝑋0 = 𝑀. 𝐹.
𝐹0
𝑘 𝑒𝑞
=
36.7926
26758.2
=
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠
𝜔
𝜔 𝑛
= 1 𝑎𝑛𝑑 𝜉 = 0.05665,

22. Speed of the unit = 420 rpm
Find: (a) Stiffness k of each spring;
(b) Deduce the expression for transmissibility.
Solution:
mass,m= 30 kg;
420 rpm
Hence, f=420/60= 7 Hz
44 rad/s
0.1
No damping; Hence, damping coefficient, c = 0 Ns/m
0
Substituting in the equation,
11
13.2665 rad/s
5280 N/m
Since the three springsare inparallel,
Therefore stiffness of each spring, 1760 N/m
2. Motion transmissibility
Definition: Motion transmissibility is the ratio of motion transmitted to the foundation to the
vibrating motion applied by the unbalance. Transmissibility is a measure of effectiveness of the
vibration isolating material.
Motion transmissibility is also called as Amplitude transmissibility.
In the case of forced excitation due to support excitation, Motion Transmissibility is the ratio of
absolute amplitude of mass of the body to the amplitude of base excitation.
Transmissibility is the same whether it is force transmissibility or motion transmissibility.
Speed, N =
Circular frequency, 𝜔 = 2𝜋𝑓 = 2𝜋𝑥7 =
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦, ɛ =
𝐹𝑇
𝐹0
=
Damping Factor, 𝜉 =
𝑐
𝑐 𝑐𝑟
=
0
𝑐 𝑐𝑟
=
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜, ɛ =
√1 + [2𝜉 (
𝜔
𝜔 𝑛
)]
2
√[1 − (
𝜔
𝜔 𝑛
)
2
]
2
+ [2𝜉 (
𝜔
𝜔 𝑛
)]
2
=
√1 + 0
√[1 − (
44
𝜔 𝑛
)
2
]
2
+ 0
= 0.10
(
44
𝜔 𝑛
)
2
=
1
0.10
+ 1 =
Natural frequency, 𝜔 𝑛 =
44
√11
=
𝜔 𝑛 = √
𝑘 𝑒𝑞
𝑚
=; 𝐻𝑒𝑛𝑐𝑒, 𝑘 𝑒𝑞 = 𝑚 𝜔 𝑛
2 = 30𝑥13.26652 =
𝑘 𝑒𝑞 = 3 𝑘
𝑘 =
𝑘 𝑒𝑞
3
=
58080
3
=

23. 𝑴𝒐𝒕𝒊𝒐𝒏 𝑻𝒓𝒂𝒏𝒔𝒎𝒊𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒕𝒚,
𝑨
𝒀
=
√ 𝟏 + [𝟐𝝃 (
𝝎
𝝎 𝒏
)]
𝟐
√[ 𝟏 − (
𝝎
𝝎 𝒏
)
𝟐
]
𝟐
+ [𝟐𝝃 (
𝝎
𝝎 𝒏
)]
𝟐
𝑻𝒓𝒂𝒏𝒔𝒎𝒊𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒕𝒚 𝒓𝒂𝒕𝒊𝒐, 𝜺 =
𝑭 𝑻
𝑭 𝟎
=
√ 𝟏 + [𝟐𝝃 (
𝝎
𝝎 𝒏
)]
𝟐
√[ 𝟏 − (
𝝎
𝝎 𝒏
)
𝟐
]
𝟐
+ [𝟐𝝃 (
𝝎
𝝎 𝒏
)]
𝟐
Phase lag
Phase lagbetween motionof mass (A) andmotion ofsupport (Y) (or) FT and F0 isgiven by the relation:
𝑷𝒉𝒂𝒔𝒆 𝒍𝒂𝒈, 𝝋 = 𝝓 − 𝜶 = 𝐭𝐚𝐧−𝟏 {
𝟐𝝃(
𝝎
𝝎 𝒏
)
[ 𝟏− (
𝝎
𝝎 𝒏
)
𝟐
]
} − 𝐭𝐚𝐧−𝟏 𝟐𝝃 (
𝝎
𝝎 𝒏
)
4.6 Measurement of vibration
The experimental determination of natural frequencies, mode shapes, and damping ratios is called
experimental modal analysis. It is based on vibration measurements that fall within general
designation of model testing. The objective of modal testing is to acquire frequency response
functions (FRFs) that are accurate and extensive, in both frequency and spatial domains. Prior
knowledge of vibration analysis, instrumentation, signal processing, and modal identification are
required to understand the modal testing.
The basic aim of modal testing is to obtain (FRF’s) relating to output vibration responses at a
number of coordinates of interest. They are in the form of accelerations (velocities, displacements)
to input vibration excitations, in the form of driving forces, applied at a given coordinate.

24. Excitation mechanism; Sensing mechanism; Data acquisition and processing mechanism.
4.6.1 Instruments for vibration measurement
Vibration instruments are used to measure frequency, displacement, velocity and acceleration of
vibration. The measured values are displayed for monitoring. The vibration parameters are also
analysed for the purpose of appropriate corrective action.
Vibration instruments comprise of transducer, transmitters, data acquisition, display indicators,
interface to computers and control devices.
Basic measurement system
A typical measurement set up consists of Signal generator, power amplifier,exciter,force transducer,
response transducer, conditioning amplifier and analyser.
Excitationmechanism –It providesinputmotioninthe formof drivingforce appliedata coordinate.The
excitationsignalscanbe in manyforms (impulse,random, steppedsine etc.) Itcan be controlledbothin
frequency and amplitude.
Sensingmechanism –These are sensingdevicesknownastransducers.Piezoelectrictransducersare used
for measuringforce excitation.Transducersgenerateelectricsignalsproportionaltophysical parameters
one want to measure.
Contactlessmotiontransducersimprove the accuracyof dynamicresponse.Laservibrometerisa velocity
transducer which works on the principle of Doppler frequency shift of a laser beam light scatteredfrom
moving surface.
Data acquisitionandprocessingmechanism –The basic objective of the data acquisitionandprocessing
mechanismistomeasure the signalsdevelopedbysensingmechanismsandtoascertainthe magnitudes
and phasesof the excitationforcesandresponses.These are calledanalysers.Theyincorporate functions
based on fast Fourier transform algorithm and provide direct measurement of FRFs
Understanding the principles behind signal acquisition and processing is very important for anyone
involved with doing vibration measurement and analysis. The validity and accuracy of the experimental
results may strongly depend on the knowledge and experience of the equipment user.
Vibration Transducers:
Acceleration sensor:
An accelerometer is a device that measures the vibration, or accelerationof motion of a structure. The
force caused by vibration or a change in motion (acceleration) causes the mass to "squeeze" the
piezoelectricmaterial whichproducesan electrical charge thatis proportional tothe force exertedupon
it. Since the charge is proportional to the force, and the mass is a constant, then the charge is also
proportional to the acceleration.
Piezoelectricaccelerometersrelyonthe piezoelectriceffectof quartz or ceramic crystals to generate an
electrical outputthatisproportional toappliedacceleration. The piezoelectriceffectproducesanopposed
accumulationof charged particleson the crystal.This charge is proportional toappliedforce or stress.A
force applied to a quartz crystal lattice structure alters alignment of positive and negative ions, which

25. resultsinanaccumulationof these chargedionsonopposedsurfaces.These chargedionsaccumulate on
an electrode that is ultimately conditioned by transistor microelectronics.
PiezoelectricMaterial
There are two types of piezoelectric material that are used in PCB accelerometers: quartz and
polycrystalline ceramics. Quartz is a natural crystal, while ceramics are man-made. Each material offers
certain benefits, and material choice depends on the particular performance features desired of the
accelerometer. Quartz is widely known for its ability to perform accurate measurement tasks and
contributes heavily in everyday applications for time and frequency measurements.
Velocity Sensor:
The velocity probe consists of a coil of wire and a magnet so arranged that if the housing is moved,the
magnettendstoremainstationarydue toitsinertia.The relative motionbetweenthe magneticfieldand
the coil induces a current that is proportional to the velocity of motion. The unit thus produces a signal
directly proportional to vibration velocity. It is self-generating and needs no conditioning electronics in
order to operate, and it has a relatively low electrical output impedance making it fairly insensitive to
noise induction.
Applications:
Measurements on Structures or Machinery Casings: Accelerometers and Velocity Sensors are

26. usedin gas turbines,axial compressors,small andmid-size pumps. These sensorsdetecthighfrequency
vibration signals related to bearing supports, casing and foundation resonances, vibration in
turbine/compressor vanes, defective roller or ball bearings, noise in gears, etc.
Displacement measurements relative to rotating shafts: Proximity Probes (capacitance or eddy-current)
are used in turbo machinery supported on fluid film bearings, centrifugal compressors, gears and
transmissions,electricmotors,largepumps(>300HP),some turbinesandfans. Thesesensorsdetectshaft
static displacements, unbalance response, misalignment, shaft bending, excessive loads in bearings,
dynamic instabilities, etc.
Accelerometers
Advantages
Simple toinstall
Good response athighfrequencies
StandhighTemperature
Small size
Disadvantages
Sensitivetohighfrequencynoise
Require external power
Require electronicintegrationforvelocityanddisplacement.
Velocity Sensors
Advantages
Simple toinstall
Good response inmiddlerange frequencies
Standhightemperature
Do not require external power
Lowestcost
Disadvantages
Low resonantfrequency&phase shift
Crossnoise
Big andheavy
Require electronicintegrationfordisplacement
Proximity Sensors
Advantages
Measure static anddynamicdisplacements
Exact response atlowfrequencies
No wear
Small andlowcost
Disadvantages
Electrical andmechanical noise
Boundedbyhighfrequencies

27. Notcalibratedforunknownmetal materials
Require external power
Difficulttoinstall
Types of Velocity Sensors:
Electromagneticlinearvelocitytransducers:Typicallyusedtomeasure oscillatoryvelocity.A permanent
magnet moving back and forth within a coil winding induces an emf in the winding. This emf is
proportional tothe velocityof oscillationof the magnet.Thispermanentmagnetmay be attachedtothe
vibrating object to measure its velocity.
Electromagnetictachometergenerators:Usedto measure the angularvelocityof vibrating objects.They
provide an output voltage/frequency that is proportional to the angular velocity. DC tachometersuse a
permanent magnet or magneto, while the AC tachometers operate as a variable coupling transformer,
with the coupling coefficient proportional to the rotary speed.
Types of Acceleration Sensors
Capacitive accelerometers : Used generally in those that have diaphragm supported seismic mass as a
moving electrode and one/two fixed electrodes. The signal generated due to change in capacitance is
post-processed using LC circuits, to output a measurable entity.
Piezoelectric accelerometers : Acceleration acting on a seismic mass exerts a force on the piezoelectric
crystals, which then produce a proportional electric charge. The piezoelectric crystals are usually
preloadedsothat eitheranincrease or decrease inaccelerationcausesachange in the charge produced
by them. But they are not reliable at very low frequencies.
Potentiometric accelerometers : Relatively cheap and used where slowly varying acceleration is to be
measured with a fair amount of accuracy. In these, the displacement of a spring mass system is
mechanicallylinkedtoaviperarm,whichmovesalongapotentiometricresistive element.Variousdesigns
may have either viscous, magnetic or gas damping.
Reluctive accelerometers : They compose accelerometers of the differential transformer type or the
inductance bridge type.The AC outputsof these vary in phase as well as amplitude.Theyare converted
into DC by means of a phase-sensitive demodulator.
Servoaccelerometers:These use the closedloopservosystemsof force-balance,torque-balance ornull-
balance to provide close accuracy. Acceleration causes a seismic mass to move. The motion is detected
by one of the motion-detectiondevices,whichgenerate asignal that acts as an error signal inthe servo-
loop.The demodulatedandamplifiedsignalisthenpassedthroughapassive dampingnetworkandthen
appliedtothe torquingcoil locatedat the axisof rotationof the mass. The torque is proportional tothe
coil current, which is in turn proportional to the acceleration.
StrainGage accelerators:these canbe made verysmall insize andmass.The displacementof the spring-
mass systemisconvertedintoachange in resistance,due to strain,infourarms of a Wheatstone bridge.
The signal is then post-processed to read the acceleration.

28. FLG are NOT REQUIRED
4.1 Applied force (FI):
4.2 Inertia force (FI):
Inertiaforce isa propertyof matterby virtue of whicha bodyresistsany change inlinearvelocity(v).FI =
– m R a G where mR ismass of the reciprocatingbodyinkgand a G islinearaccelerationof the center
of mass of the body in m/s 2
The negative sign indicates that the inertia force acts in opposite direction to that of the acceleration.
Inertia force acts through the centre of mass of the body.
4.3 Inertia Torque (I):
Inertiatorque isa propertyof matterby virtue of whicha body resistsanychange in the angularvelocity
( ω). Inertia torque, TI = – IG α where I G is mass moment of inertia of the body about an axis passing
through the centre of mass in kg m 2
, α is angular acceleration of the body in rad / s 2
The negative sign indicates that the inertia torque acts in opposite direction to that of the angular
acceleration.