Vibrational forces

1. VIBRATIONS
Introduction
A mechanical system is said to be vibrating when its component parts are undergoing periodic
(that is cyclically repeated) oscillations about a central configuration (usually the statical
equilibrium position). It can be shown that any system, by virtue of its inherent mass and
elasticity, can be caused to vibrate by externally applied forces. The duration and severity of the
vibration will depend on the relation between the external forces and the mechanics of the
system.
Although vibration can sometimes be used to advantage- as in cleaning or mixing machines- its
presence is generally undesirable for three main reasons:-
1) Structural damage of a fatigue nature may be caused by the cyclical fluctuation of
loading.
2) Physical discomfort may be experienced by personnel associated with the system (for
example, passengers in a vehicle).
3) Noise, itself a vibration of air molecules, may be generated by a mechanical vibration.
A vibration is characterized and assessed by three parameters:
i. Amplitude
ii. Frequency
iii. Phase
a. Amplitude is the maximum displacement for the central position( measured as a linear or
angular quantity).
b. Frequency is the reciprocal of the period (the time for one complete cycle of vibration)
and is expressed in HZ (cycles/s).
c. Phase is a measure of the instant at which a vibration passes through a central position
and in mechanical systems is usually only of importance when the relation between two
vibrations or between the motions of two parts of a system is being considered.
Free vibrations
If a mechanical system is displaced from its equilibrium position and then released, the restoring
force (arising from either spring elements as in a vehicle suspension, material stiffness as in
tensional or bending systems or gravitational forces as in a pendulum) will cause a return
towards the equilibrium position. There will inevitably be an overshoot on the other side and so
on, resulting in what is called a free vibration.

2. This type of vibration arises from an initial input of energy that is continually changing from the
potential (or strain energy) form to the kinetic form as the system moves between its extreme
positions and its mid- position. In a free vibration the system is said to vibrate at a natural
frequency.
Damping
Due to various causes there will always be some loss( that is, “dissipation”) of mechanical
energy during each cycle of vibration and this effect is called Damping
A free vibration will die away ( that is “decay”) due to damping, though if the damping forces
are small enough they will have little influence on the frequency and are often neglected to
simplify the mathematics.
Degrees of freedom
Single- degree-of- freedom systems can vibrate in only one mode(for example, a pendulum
swinging in a vertical plane ). In practice many systems have more than one degree of- freedom
( a vehicle may pitch, bounce or roll) each having its own natural frequency. Finally there are
the ”continuous” systems such as beams, where mass and elasticity are inextricably linked and in
theory an infinite number of modes of vibration possible.
Fortunately it is rarely necessary to calculate more than one or two of the lower frequencies at
which such systems can vibrate.
Forced vibrations
When a harmonically varying external force or displacement (known as the ”excitation”) is
applied to a single-degree-of-freedom systems at rest it is found that the vibration initiated is a
combination of one motion at the natural frequency together with one at the” forcing” frequency.
However, the natural frequency component will die out after sufficient time has elapsed for
damping to take effect. The system is then said to be performing “steady- state forced
vibrations”. These occur at the excitation (or forcing) frequency and with an amplitude that
depends on the ratio between forcing and natural frequency and the level of damping. When the
forcing frequency is equal to the natural frequency the system is said to be in resonance and the
amplitude may build up to a very high value ( limited only by damping and the physical
constraints of the system). If the excitation consists of a number of harmonic components at n
different frequencies and the system has N degrees of freedom, then there will be “Nn” possible
resonant conditions that is, values of frequency at which resonance will occur).
Self- excited vibrations
There are a number of cases where vibrations can be sustained by energy fed in from steady
external source (as distinct from the fluctuating excitation required for forced vibrations). The

3. necessary cyclical vibrations are produced by the motion of the system itself. Examples are stick-
slip vibrations such as brake squeal) arising from the difference between static and kinetic values
of friction, also aircraft flutter and other wind- excited vibrations (for example, of bridge
structures, power cables and slips rigging) in which “lift” forces fluctuate with the attitude and
lateral velocity of the vibrating member.
The feature of self-excited vibrations is that they take place at a natural frequency of the system
and in this way they can be distinguished from forced vibrations.
From the foregoing it is readily seen that, whatever might be the cause of vibration, the first
requirement is to determine the natural frequencies of the system. From this will follow an
appreciation of the possible resonant conditions that might arise and a means of diagnosing the
source of the excitation.
Undamped free vibration
Fig (a)

4. W=mg
F=kx
NB
Fig (b)
The simplest type of vibrating motion is undamped free vibration, represented by the model in
fig (a) above.
The block has a mass “m” and is attached to a spring having a stiffness “k”. Vibrating motion
occurs when displacing the block a distance “x” from its equilibrium position and allowing the
spring to restore it to its original position. As the spring pulls on the block, the block will proceed
to move out of equilibrium when x=0. Provided the supporting surface is smooth, oscillation will
continue indefinitely.
The time dependent path of motion of the block may be determined by applying the equation of
motion to the block when it is in the displaced position x. The free body diagram is shown in fig
(b) above.
The elastic restoring force
F=kx
Is always directed toward the equilibrium, position, whereas the acceleration “f: is assumed to
act in the direction of positive displacement

5. Noting that
𝑓 =
𝑑2
𝑥
𝑑𝑡2
= 𝑥̈
We have:
∑ 𝑓
𝑥 = 𝑚𝑓
𝑥
Or, −𝑘𝑥 = 𝑚𝑓
𝑥 = 𝑚𝑥̈
It can be seen in this equation that the acceleration is proportional to the position. Motion
described in this manner is called simple harmonic motion. Rearranging the terms into a
“standard form” gives:
𝑥̈ + 𝑤2
𝑥 = 0 ______ (1)
This constant w is called the “circular frequency”, expressed in rad/s and in this case:
𝑤 = √
𝑘
𝑚
________ (2)
Equation (1) may also be obtained by considering the block to be suspended, as shown in fig ( c )
below and measuring the displacement y from the block’s equilibrium position. The free body
diagram is shown in fig. (d). When the block is in equilibrium, the spring exerts an upward force.
𝑓 = 𝑤 = 𝑚𝑔,

6. Fig ( c)
F=w+ky
W
Fig (d)

7. On the block, hence, when the block is displaced a distance y downwards from this position, the
magnitude of the spring force is
𝑓 = 𝑤 + 𝑘𝑦
Applying the equation of motion gives
∑ 𝑓
𝑦 = 𝑚𝑓
𝑦
or −𝑤 − 𝑘𝑦 + 𝑤 = 𝑚𝑦̈
or −𝑘𝑦 = 𝑚𝑦̈
or 𝑦̈ + 𝑤2
𝑦 = 0
which is the same form as equation (1) where w is defined by equation (2).
Equation (1) is a homogeneous, second order, linear, differential equation with constant
coefficients. It can be shown that, using the methods of differential equation, that the general
solution of this equation is
𝑥 = 𝐴 sin 𝑤𝑡 + 𝐵 cos 𝑤𝑡
Where A and B represent two constants of integration. The block’s velocity and acceleration are
determined by taking successive time derivatives, which yields:
𝜗 = 𝑥̇ =
𝑑
𝑑𝑡
(𝐴 sin 𝑤𝑡 + 𝐵 cos 𝑤𝑡
= 𝐴𝑤 cos 𝑤𝑡 − 𝐵𝑤 sin 𝑤𝑡
____________ (4)
And 𝑓 = 𝑥̈ =
𝑑
𝑑𝑡
(𝐴𝑤 cos 𝑤𝑡 − 𝐵𝑤 sin 𝑤𝑡)
= −𝐴𝑤2
sin 𝑤𝑡 − 𝐵𝑤2
cos 𝑤𝑡
____________ (5)
When equations (3) and (5) are substituted into equation (1), the differential equation is indeed
satisfied and therefore equation (3) represents the true solution to equation (1).

8. The integration constants A and B in equation (3) are generally determined from the initial
conditions of the problem. For example, suppose that the block in Fig (a) has been displaced a
distance x1 to the right from its equilibrium position and given an initial (positive) velocity,
𝜑, directed to the right. Substituting
𝑥 = 𝑥1 𝑎𝑡 𝑡 = 0
Into equation (3), we get:
𝑥1 = 𝐴 sin 0 + 𝐵 cos 0
= 0 + 𝐵
Or 𝐵 = 𝑥1
Since 𝜑 = 𝜑1 𝑎𝑡 𝑡 = 0, we can substitute in equation (4) thus:
𝜑1 = 𝐴𝑤 cos 0 − 𝐵𝑤 sin 0
=Aw-0
Or 𝐴 =
𝜑1
𝑤
Of these values are now substituted into equation (3), we get:
𝑥 =
𝜑1
𝑤
sin 𝑤𝑡 + 𝑥1 cos 𝑤𝑡
_____________ (6)
Equation (3) may also be expressed in terms of simple sinusoidal motion.
Let, 𝐴 = ∁ cos ∅ _________ (7)
And, 𝐵 = ∁ sin ∅ _________ (8)
Where ∁ 𝑎𝑛𝑑 ∅ are new constants to be determined in place of A and B. Substituting in equation
(3) yields:
𝑥 = ∁ cos ∅ sin 𝑤𝑡 + ∁ sin ∅ cos 𝑤𝑡
Since,
sin(𝜃 + ∅) = sin 𝜃 cos ∅ + cos 𝜃 sin ∅
Then: 𝑥 = ∁ sin(𝑤𝑡 + ∅)
__________ (9)
If this equation is plotted on an x versus wt axis, the graph shown in fig (e) below is obtained.
The maximum displacement of the block from its equilibrium position is defined as the

9. amplitude of vibration. From either the figure or equation (9) the amplitude is ∁. The angle∅ is
called the phase angle since it represents the amount by which the curve is displaced from the
origin when t=0. The constants ∁ and ∅ are related to A and B by equations (7) and (8). Squaring
and adding these two equations,𝐴2
+ 𝐵2
= 𝐶2
cos2
∅ + ∁2
sin2
∅
Fig (e) : Graph of x and wt.
: 𝐴2
+ 𝐵2
= 𝐶2
(cos−1
∅ + sin−1
∅
= 𝑐2
(1)
: 𝐶2
= 𝐴2
+ 𝐵2
Hence The amplitude, c, is given by
𝐶 = √𝐴2 + 𝐵2
____________ (10)
If equation (8) is divided by equation (7), we get :
𝐵
𝐴
=
𝐶
𝐶
cos ∅
sin ∅
=
1
tan ∅

10. Hence the phase angle, ∅, is given by
tan ∅ =
𝐵
𝐴
Or, ∅ = tan−1
(
𝐵
𝐴
) ________ (11)
Note that the sine curve of equation (9) completes one cycle in time.
T=T
When wt= 2𝜋
Or, T=
2𝜋
𝑊
_________ (12)
This length of time is called a period, as shown in fig (e) above. Usin equation (2), the period
may also be represented by:
𝑇 = 2𝜋√
𝑚
𝑘
______ (13)
The frequency, f, is defined as the number of cycles completed per unit of time, which is the
reciprocal of the period, i.e.
𝑓 =
1
𝑇
=
𝑤
2𝜋
_________ (14)
Or, 𝑓 =
1
2𝜋
√
𝑘
𝑚
__________ (15)
The frequency is expressed in cycles/s. This ratio of units is called a hertz (HZ), where :
1 HZ = 1 cycle/s
= 2𝜋 rad/s
When a body or system of connected bodies is given an initial displacement from its equilibrium
position and released, it will vibrate with a definite frequency known as the natural frequency.
This type of vibration is called free vibration, provided no external forces except gravitational or
elastic forces act on the body during the motion. Also, if the amplitude of vibration remains
constant, the motion is said to be undamped.
The undamped free vibration of a body having a single degree of freedom has the same
characteristics as simple harmonic motion of the block and spring discussed above.
Consequently, the body’s motion is described by a differential equation of the name fom as
equation (1), i.e.
𝑥̈ + 𝑤2
𝑥 = 0 _________ (16)

11. Hence, if the circular frequency is of the body is known, the period of vibration T, the natural
frequency, f, and the other vibrating characteristics of the body can be established using the
above equations.
PROCEDURE FOR ANALYSIS
As in the case of the block and spring, the circular frequency, w, of a rigid body or system of
connected rigid bodies having a single degree of freedom can be determined using the following
procedure:
1) Free- body diagram
Draw the free- body diagram of the body when the body is displaced by a small amount from its
equilibrium position. Locate the body with respect to its equilibrium position by using an
appropriate inertial coordinate q. The acceleration of the body’s mass centre (or centre of
gravity), G. or the body’s angular acceleration, ∝, should have a sense which is in the positive
direction of the position coordinate. If it is decided that the rotational equation of motion
∑ 𝑚𝑝 = ∑(𝑚𝑘𝑝)
Is to be used, then it may be beneficial to also draw the kinetic diagram since it graphically
accounts for the components.
𝑚(𝐺)𝑥
𝑚(𝐺)𝑦
And 𝐼𝐺 ∝
And thereby makes it convenient for visualizing the terms needed in the moment sum.
∑(𝑚𝑘)𝑝
2) Equation of motion
Apply the equation of motion to relate the elastic or gravitational restoring forces and
couples acting on the body to the body’s accelerated motion.
3) Kinematics
Using kinematics, express the body’s accelerated motion in terms of the second time
derivative of the position coordinate, 𝑞.
̈ Substitute this result into the equation of motion
and determine w by rearranging the terms so that the resulting equation is of the form:
𝑞̈ + 𝑤2
𝑞 = 0

12. EXAMPLE 1:
𝑙
𝜃
s
Determine the period of vibration for the simple pendulum shown above, if the bob has a mass
“m” and is attached to a cord of length “l”. Neglect the size of the bob

13. SOLUTION
n t
T
𝜃
W=mg
Free- body diagram
Motion of the system will be related to the position coordinate (𝑞 =)𝜃, as sketched above. When
the bob is displaced by an angle, the restoring force acting on the bob is created by the weight
component.
𝑚𝑔 sin 𝜃.
Furthermore, Gt acts in the direction of increasing s (or 𝜃 ).
Equation of motion:
Applying the equation of motion in the tangential direction, since it involves the restoring force,
yields
∑ 𝑓𝑡 = 𝑚𝑓𝑡
Or −𝑚𝑔 sin 𝜃 = 𝑚𝑓𝑡 _________ (17)

14. Kinematics:
𝑓𝑡 =
𝑑2
𝑠
𝑑𝑡2
= 𝑠̈
Since s may be related to 𝜃 by the equation
𝑠 = 𝑙𝜃
Or 𝑓 = 𝑙𝜃̈
Hence in equation (1)
𝑚𝑔 sin 𝜃 = 𝑚. 𝑙𝜃̈
Or 𝑚𝑙𝜃 + 𝑚𝑔 sin 𝜃 = 0
̈
Or 𝑙𝜃 + 𝑔 sin 𝜃 = 0
̈
Or 𝜃̈ +
𝑔
𝑙
sin 𝜃 = 0
_____________ (18)
The solution of this equation involves the use of an elliptical integral.
For small displacements, however,
sin 𝜃 ≈ 𝜃
Hence in equation (2) above,
𝜃̈ +
𝑔
𝑙
𝜃 = 0 ________ (19)
Comparing this equation with
𝑥̈ + 𝑤2
𝑥 = 0
Which we had seen previously, and which is the “standard form” for simple harmonic motion,
then it can be seen that:
𝑤2
=
𝑔
𝑙
Or 𝑤 = √
𝑔
𝑙
From equation (12), the period of time required for the bob to make one complete swin is
therefore:
𝑇 =
2𝜋
𝑊

15. = 2𝜋√
𝑙
𝑔
___________ (20)
EXAMPLE 2:
b=0.3m
a=0.2m

16. The 10- kg mass rectangular plate shown above is suspended at its centre from a rod having a
torsion stiffness k=1.5 Nm / rad. Determine the natural period of vibration of the plate when it is
given a small angular displacement 𝜃 in the plane of the plate.
Solution :
T=w
𝑚 = 𝑘𝜃
W
The above sketch shows the free-body diagram of the plate in motion. Since the plate is
displaced in its own plane, the torsional restoring moment created by the rod is
𝑚 = 𝑘𝜃
This moment acts in the direction opposite to the angular displacement 𝜃. The angular
acceleration 𝜃̈ acts in the direction of positive 𝜃.

17. Equation of motion
∑ 𝑚𝑜 = 𝐼𝑜𝜑
Or – 𝑘𝜃 = 𝑙0𝜃̈
Or 𝑙0𝜃̈ + 𝑘𝜃 = 0
Or 𝜃̈ +
𝑘
𝐼𝑜
𝜃 = 0
Since this equation is in the “standard form” the circular frequency is given by:
𝑤2
=
𝑘
𝑙0
Or 𝑤 = √
𝑘
𝑙0
The moment of inertia Io, of the plate about an axis coincident with the rod is given by
𝑙0 =
1
12
𝑚(𝑎2
+ 𝑏2)
Hence, 𝑙0 =
1
12
∗ 10 ∗ [(0.22) + (0.32)]
= 0.108𝑘𝑔𝑚2
The natural period of vibration is, therefore,
𝑇 =
2𝜋
𝑊
= 2𝜋√
𝐼𝑜
𝑘
= 2𝜋√
0.108
1.5
=1.695

18. EXAMPLE 3
Fig. Q3
Fig. Q3 above shows a uniform bar of mass m and pivoted at o. The bar is kept in equilibrium by
springs of constants k and 2k at A and B. A concentration mass Mc is at the point C.
Determine the undamped natural frequency.
Solution
The sketch below shows the forces which are acting on the system
Let, IG= moment of inertia of the bar about the pivot

19. 2k (2a𝜃)
IG 𝜃̈
𝜃 0
𝑘𝑎𝜃 mca 𝜃̈
𝑚.
𝑎
2
𝜃̈
Forces in the system
Taking moments about the pivot we have:
𝐼𝐺𝜃̈ + 𝑚(
𝑎
2
)2
𝜃̈ + 𝑚𝑐𝑎2
𝜃̈
= −(8𝑘𝑎2
+ 𝑘𝑎2)𝜃
But 𝐼𝐺of bar of length l about the centre
𝑚𝑙2
12
Or 𝐼𝐺 =
𝑚
12
(3𝑎)2
=
3
4
𝑚𝑎2
[(
3
4
𝑚𝑎2
+
1
4
𝑚𝑎2
) + 𝑚𝑐𝑎2
] 𝜃̈ = −9𝑘𝑎2
𝜃
Or (𝑚 + 𝑚𝑐)𝜃 + 9𝑘𝜃 = 0
̈
Or 𝜃 = −
9𝑘
𝑚+𝑚𝑐
𝜃
̈
= −𝑤𝑛
2
𝜃

20. 𝑤𝑛
2
=
9𝑘
𝑚+𝑚𝑐
Or 𝑤
𝑛=√
9𝑘
𝑚+𝑚𝑐
___________ (21)
TORSIONAL VIBRATIONS
Torsional single- degree- of- freedom vibration system.
The above figure shows a system of two inertias of magnitude 𝐼1 𝑎𝑛𝑑 𝐼2 supported in
“frictionless” bearings and joined by a “light” shaft of torsional stiffness k equal to torque per
radian twist, GJ/l). If 𝜃1𝑎𝑛𝑑𝜃2are the angular displacements of 𝐼1 𝑎𝑛𝑑 𝐼2
respectively from the
untwisted position then the torque in the shaft is 𝑘(𝜃1 − 𝜃2) in the pause tending to decrease
𝜃1 𝑎𝑛𝑑 𝑡𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝜃2. Note that this torque will change sign during a vibration, since 𝜃1 𝑎𝑛𝑑𝜃2
can take positive and negative values.

21. In practice torsional vibrations are frequently superimposed on a steady running speed but this
does not affect the equations of motion, which are as follow:
𝐼1𝜃̈ = −𝑘(𝜃1 − 𝜃2)
𝐼2𝜃2
̈ = 𝑘(𝜃1 − 𝜃2)
Dividing by 𝐼1 𝑎𝑛𝑑 𝐼2 respectively and subtracting we get:
𝜃̈1 − 𝜃̈2 = −
𝐾
𝐼1
(𝜃1 − 𝜃2)
−
𝑘
𝐼2
(𝜃1 − 𝜃2)
= −𝑘(𝜃1 − 𝜃2) [
1
𝐼1
+
1
𝐼2
] ___________ (22)
Writing 𝜃 = 𝜃1 − 𝜃2, the twist in the shaft, the above equation becomes:
∅̈ = −𝑘∅ [
𝐼1 + 𝐼2
𝐼1𝐼2
]
Or ∅̈ + 𝑘 (
𝐼1+𝐼2
𝐼1𝐼2
) ∅ = 0 ___________(23)
From this equation,
𝑤𝑛
2
=
𝑘(𝐼1 + 𝐼2)
𝐼1𝐼2
Or 𝑤𝑛 = √
𝑘(𝐼1+𝐼2)
𝐼1𝐼2
______________ (24)
The natural frequency, 𝑓
𝑛, is given by:
𝑓
𝑛 =
𝑤𝑛
2𝜋
=
𝐼
2𝜋
√
𝑘(𝐼1+𝐼2)
𝐼1𝐼2
_________ (25)
If one end of the shaft is fixed (equivalent to an infinite value of one inertia, say 𝐼2), and the
inertia at the free end is I (for 𝐼1) then equation (25) becomes:
𝑓
𝑛 =
1
2𝜋
√
𝑘
𝐼
_________ (26)
In the two- inertia case, because during vibration the shaft is being wound up first in one
direction then in the other, the inertias at the ends must be moving in opposite directions (they

22. are said to be in anti- phase, that is, a phase difference of 180’ ). If their amplitudes are
𝐴1 𝑎𝑛𝑑 𝐴2 then the figure below shows that there is a point of no twist (called a node) in the
shaft such that:
𝑙1
𝑙2
=
𝐴1
𝐴2
_____________ (27)
Node
A2 𝐼2
𝑙1
𝐴2
The mode can be treated as a fixed end, so that if 𝑘1 𝑎𝑛𝑑 𝑘2 are the stiffness of the shaft on
either side of the mode, equation (26) can be written
𝑓
𝑛 =
1
2𝜋
√
𝑘1
𝐼1
=
1
2𝜋
√
𝑘2
𝐼2
Giving:
𝑘1
𝑘2
=
𝐼1
𝐼2
_________ (28)
But shaft stiffness is inversely proportional to length, that is
𝑘1
𝑘2
=
𝑙2
𝑙1
_________ (29)
Eliminating
𝑘1
𝑘2
between equations (27), (28) and (29),
𝑙1
𝑙2
=
𝐴1
𝐴2
=
𝐼2
𝐼1
________ (30)

23. DAMPED FREE VIBRATIONS
Damping Mechanisms
As pointed out earlier, a free vibration of a real- life mechanical system dies away (decays) after
some time.
The decay is caused by some loss (dissipation) of mechanical energy during each cycle of
vibration. Damping forces may be encountered as a result of dry friction (coulomb friction)
between rigid bodies, by fluid friction when a rigid body moves in a fluid or by the internal
friction between the molecules of a seemingly elastic body such as a rubber support.
In practical situations, engineers occasionally have to try to eliminate it. However, when
damping is particularly desirable, it may be introduced artificially as with the vibration dampers
(shock- absorbers) on automobiles. The dashpot represented schematically in the figure below is
an example of a viscous damper. It comprises a piston in
Fig. 13

24. Schematic representation of a viscous damper
A cylinder filled with oil. The piston is pierced by small holes, so that laminas flow of oil can
occur as the piston moves. As in all cases of viscous damping the damping force on the piston is
directly proportional to the instantaneous velocity of the piston and always opposes the direction
of motion of the piston. The viscous damping force, 𝑓𝑑, is given by
𝑓𝑑 = ∁𝜑
= ∁𝑥̇
Where 𝜑 is the instantaneous velocity and c is the damping coefficient (constant) which is equal
to the damping force at unit velocity. The unit of c is NS/m.
Analysis of Damped Free Vibration
Fig. 14
A mass spring- dashpot system.
Let us consider a damped mechanical system comprising a body mass m supported on a spring of
stiffness k and a dashpot of damping coefficient c, as shown in the figure above. The free- body
diagram of the system is shown below at displacement x from the equilibrium position. In the
free-

25. Equilibrium position
𝑘𝑥 𝑐𝑥̇ x
𝑚𝑥̈
Fig. 15
Free body diagram for the mass spring- dashpot system
Body diagram, we have indicated only the dynamic forces since, as pointed out previously, the
weight w(=mg) balances the spring force kx and are therefore ignored in drawing the free- body
diagram. Consequently, in drawing the diagram for the analysis of dynamic forces, the static
equilibrium forces w(=mg) and kx are met shown. At the position shown in the free- body
diagram, the mass is subjected to a spring force kx and a damping force 𝑐𝑥̇. The equation of
motion of the mass is given by
𝑚𝑥̈ = −𝑘𝑥 − 𝑐𝑥̇
Or 𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 0 ____________ (55)
Equation (55) is the basic equation for the damped free motion of a single degree- of- freedom
system and may be written in the form:
𝑥̈ +
𝑐
𝑚
𝑥̇ +
𝑘
𝑚
𝑥 = 0
Or 𝑥̈ + 2𝛾𝑥̇ + 𝑤𝑛
2
𝑥 = 0 _________ (56)
Where, 𝛾 =
𝑐
2𝑚
____________ (57)
𝛾is a damping parameter while 𝑤𝑛 is the natural frequency of vibration of the system as defined
earlier.

26. It is worthy to note that for a shaft undergoing damped free torsional vibration the equation of
motion is given by
𝜃̈ + 𝑐𝜃̇ + 𝑘𝜃 = 0 _________ (58)
Which is similar to equation (55) for the spring- mass- dashpot system.
In equation (58), k is as before, the torsional stiffness, c is the damping coefficient which is such
that 𝑐𝜃 gives the resisting torque due to damping. Thus, the analysis that follows applies to
torsional vibration as well.
Solution of the Differential Equation of Motion
The differential equation for damped free vibration expressed by equation (56) is a second-
order differential equation with constant coefficients. The traditional approach for solving such
equations is to assume a solution of the form:
𝑥 = 𝑒∝𝑡
_________ (59)
Where ∝ is a constant.
Substituting (59) into equation (56), we obtain
(∝2
+ 2𝛾 ∝ +𝑤𝑛
2) = 0 __________ (60)
Equation (60) is called the characteristic or auxiliary equation. It has two roots, namely:
∝1,2= −𝛾 ± √𝛾2 − 𝑤𝑛
2
Or ∝1,2= 𝑤𝑛 [−
𝛾
𝑤𝑛
± √(
𝛾
𝑤𝑛
)
2
− 1]
= 𝑤𝑛 [−𝜀 ± √𝜀2 − 1]
_____________ (61)
Where, 𝜀 =
𝛾
𝑤𝑛
=
𝑐
2𝑚𝑤𝑛
____________ (62)
After substituting for 𝛾 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (57), 𝜀 is called the damping factor or damping ratio.
The general solution of equation (56) is given by
𝑥(𝑡) = 𝐴1𝑒∝1𝑡
+ 𝐴2𝑒∝2𝑡
_________ (63)
Where 𝐴1 𝑎𝑛𝑑 𝐴2 are constants evaluated from the initial conditions 𝑥(0)𝑎𝑛𝑑 𝑥̇(0).
Three distinct cases of damping are examined below. These are referred to as :

27. i. Critical damping
𝜉 = 1
ii. Heavy damping
𝜉 > 1
iii. Light damping
𝜉 < 1
i. Critical damping
This damping occurs when the damping coefficient c is such that the damping ratio 𝜉 = 1. From
equation (62), this means that
𝜉 =
𝑐
2𝑚𝑤𝑛
= 1
Or 𝑐 = 𝑐0 = 2𝑚𝑤𝑛 __________ (64)
Where 𝑐0 is the critical damping. From equation (61)
∝1,2= −𝑤𝑛
These are two repeated roots. From the calculus of differential equations the general solution is
given by:
𝑥(𝑡) = 𝐴1𝑒−𝑤𝑛𝑡
+ 𝐴2𝑡𝑒−𝑤𝑛𝑡
= (𝐴1 + 𝐴2𝑡)𝑒−𝑤𝑛𝑡
_______ (65)
The motion is non-oscillatory and the system is said to be critically- damped. Such systems are
of special interest in some engineering applications and instruments such as electrical meters,
where the systems are expected to regain their equilibrium position in the shortest possible time
without oscillation and overshooting.
ii. Heavy damping
This damping corresponds to the case when 𝜀 > 1. In this case the roots
∝1 𝑎𝑛𝑑 ∝2 are distinct and real, culminating in the general solution from
equation (63) becoming:
𝑥(𝑡) = 𝐴1𝑒𝑥𝑝[(−𝜉 + √𝜉2 − 1)𝑤𝑛𝑡]+𝐴2𝑒𝑥𝑝[(−𝜉 − √𝜉2 − 1)𝑤𝑛𝑡] ____ (66)
The motion is non- oscillatory exponentially decreasing function of time. The
motion is called aperiodic, while the system is said to be over-damped.

28. iii. Light damping
This damping happens or occurs when the damping ratio 𝜉 < 1. For this case,
∝1,2= 𝑤𝑛(−𝜉 ± 𝑖√1 − 𝜉2) ___________ (67)
Where I is a complex number to take care of the fact that
√𝜉2 − 1 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟
𝜉 < 1. 𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡
√𝜉2 − 1 = √−1√1 − 𝜉2 = 𝑖√1 − 𝜉2
Hence equation (67)
Let , 𝑤 = (√1 − 𝜉2)𝑤𝑛 __________ (68)
∴ ∝1,2= (−𝑤𝑛𝜉 ± 𝑖𝑤)
Leading thereby to the general equation
𝑥(𝑡) = 𝐴1𝑒𝑥𝑝(−𝑤𝑛𝜉𝑡 + 𝑖𝑤𝑡) + 𝐴2𝑒𝑥𝑝(−𝑤𝑛𝜉𝑡 − 𝑖𝑤𝑡)
= 𝑒𝑥𝑝[−𝑤𝑛𝜉𝑡(𝐴1𝑒𝑖𝑤𝑡
+ 𝐴2𝑒−𝑖𝑤𝑡
)] ____________ (69)
From the study of complex numbers, we recall
𝑒𝑖𝜃
= cos 𝜃 + 𝑖 sin 𝜃
It can then be shown that equation (69) leads to the general equation
𝑥(𝑡) = 𝐴𝑒𝑤𝑛
𝜉𝑡
sin(𝑤𝑡 + ∅) ________ (70)
Where A and ∅ are constants to be evaluated from initial conditions 𝑥(0)𝑎𝑛𝑑 𝑥̇(0). w, given by
equation (68) is the damped circular frequency of vibration of the body. Equation (70) represents
an oscillatory motion.
The period of the damped free vibration is given by
𝑇𝑑 =
2𝜋
𝑤
=
2𝜋
𝑤𝑛√1 − 𝜀2
=
𝑇𝑛
√1−𝜉2
________ (71)
Where, 𝑇𝑛 is the period of natural vibration.
Also, the frequency of the damped free vibration is

29. 𝑓 =
𝑤
2𝜋
=
𝑤𝑛
2𝜋
√1 − 𝜉2
= 𝑓𝑛√1 − 𝜉2 ____________ (72)
The amplitude of vibration ,x, of the mass is
𝑥 = 𝐴𝑒
−𝑤
𝑛𝜉𝑡
______________ (73)
It is a decaying exponential function of time. Thus, the mass vibrates from an initial amplitude
𝑥0 which decays with time as shown in the figure (16). The system is said to be under damped.
Also because the vibration dies away with time the under damped vibration is often referred to
as a transient.
From the amplitude equation we note that the ratio of the amplitude at time 𝑡1 and the amplitude
at a later time 𝑡2 is given by
𝑥1
𝑥2
=
𝐴𝑒
−𝑤
𝑛𝜉𝑡1
𝐴𝑒
−𝑤
𝑛
𝜉±𝑡2
= 𝑒
𝑤
𝑛𝜉(𝑡2−𝑡1)
____________ (74)
Fig 16
Oscillatory motion of an under damped system.
Let us assume that 𝑡2 − 𝑡1 = 𝑡𝑑.

30. That is, between time 𝑡2 and time 𝑡1 the body has one through one complete cycle. Therefore
𝑥1
𝑥2
= 𝑒
𝜀𝑤
𝑛𝑡𝑑 ____________ (75)
Therefore, 𝛿 = 𝑙𝑛
𝑥1
𝑥2
= 𝜀𝑤𝑛𝑡𝑑 ________________ (76)
Thus the ratio of successive decrement of amplitude is given by equation (75). The factor 𝜀𝑤𝑛𝑇𝑑
is called the logarithmic decrement. It is worth noting that practical values for 𝜀 vary between 0
and 1. Most systems will therefore undergo some transient vibration with the rate of decay
determined by the actual value of 𝜀.