mechanical vibration

Eng5932 - Mechanical Vibrations 1

1 Introduction

1.1 De�nitions

� Vibration

�A motion that repeats itself after a time interval or oscillation (e.g. pendulum,plucked guitar string).

� Vibrating system

�Consists of a mass, spring and possibly a damper.

�Mass (or inertia element for rotating systems) is a rigid body that stores kineticenergy

� Spring stores potential energy

�Damper dissipates energy

1.2 Spring Elements

� A spring element is generally assumed to have no mass and no damping. It will storepotential energy due to stretch (or compression) or twist (for torsional springs).

� For a linear spring the spring force is proportional to the change in length of the springelement:

F = kx (1)

where, x is the measured deformation of the spring from its undeformed length, andk is the spring constant.


� The work done deforming a spring from x1 to x2 is:

U12 =

Z x2

x1

F dx =

Z x2

x1

kx dx =k

2(x22 � x21) (2)

and this work is stored as potential energy.

� The spring constant of a helical spring is:

k =Gd4

8nD3(3)

wehere G is the shear modulus of the spring material, d is the diameter of the wire,D is the mean coil diameter, and n is the number of active turns of the coil.

� A rod can act as a spring (or consider an elevator cable when the winding drumsuddenly stops). Consider a rod being stretched by a force F :

� For linear elastic behaviour Hooke�s Law applies:�

�= E

Since � = F=A and � = �x=L, then:

FL

A�x= E


or

F =AE

L�x = k�x

i.e. the linear spring constant for a rod (or cable in tension) is:

k =AE

L(4)

1.3 Damping Elements

� Damping is the mechanism by which vibrational energy is converted into heat orsound.

� A damping element is assumed to have zero mass and elasticity.

� There are three standard models of damping: (1) viscous damping; (2) Coulomb ordry friction damping; and (3) hysteretic damping.

� Viscous Damping:

�Fluid dynamic drag is used to dissipate energy (e.g. air drag on a pendulum,drag as a liquid is forced through an ori�ce (shock absorber)).

�Viscous damping is the most common form of damping used in vibrating systems.

�The damping force is proportional to the rate of change of length of the dampingelement or the relative velocity between the two ends of the damping element.

F = c _x (5)

�Note: Kinematics of a rigid body is used to give the position, velocity andacceleration of various components of a suspension system. The distance betweenthe mounting points of the spring is used to give the spring force. The rate ofchange of the distance between the mounting points of the damper is used togive the damping force. If the spring and damper are coaxial, only one length isof interest.

� Coulomb Damping:

�Coulomb damping is constant in magnitude but opposite in direction to themotion.

�Due to the friction between rubbing surfaces (dry or with insu¢ cient lubrication).

F = �kN (6)

� Hysteretic, Solid or Material Damping:


�As a material is deformed, the work done on the material is stored as strainenergy or heat.

� Since energy is absorbed, a body subjected to material damping shpows a hys-teresis loop on a stress-strain diagram.

1.3.1 A Simple Viscous Damper

� Consider two parallel plates separated by a distance h. The bottom plate is �xed andthe top plate moves to the right at a constant speed v. The space between the platesis occupied by a �uid with dynamic viscosity �.

� Solution of the Navier-Stokes equation show that the velocity pro�le is linear, i.e.Couette �ow.

� The shear stress in the �uid is:

� = �@u

@y= �

v

h


� The drag (i.e. frictional) force exerted by the �uid on the bottom plate is:

F = A� jy=0 =�A

hv

� where c = �A=h is the damping constant and the damping force behaves linearly wrtv.


2 Free Vibration of a One Degree of Freedom System

2.1 Undamped Motion of a Single DOF System

� Consider a mass, m, resting on a �at frictionless surface and connected to a rigidsupport by a linear spring, k:

2.1.1 Equation of Motion

� We will need an equation governing the motion of the mass as it is moved from itsequilibrium position. Such an equation of motion can be derived from Newton�s 2ndLaw using the following steps:

1. Sketch the system and de�ne a suitable positive co-ordinate direction to de�nethe motion of the mass.

2. Determine the static equilibrium position of the system. It is common (andrecommended) to de�ne the positive co-ordinate from the equilibrium position.

3. Draw a free body diagram (FBD) of the mass, assuming positive displacement,velocity and acceleration are given to it. Draw the forces acting on the mass.

4. Use Newton�s 2nd Law to derive an equation of motion.


+!X

Fx = max

m�x = �kx

orm�x+ kx = 0 (7)

� Equation (7) is the equation governing the free vibration of the mass.

� What about a mass, m, suspended from a rigid support by a linear spring, k?

+ #X

Fx = max

m�x = �kx+ (W � k�st)

But, from statics:W = k�st

so the governing equation is:m�x+ kx = 0

� Note: The same governing equation arises when x is de�ned from the equilibriumposition (this is the advantage of de�ning x measured from equilibrium).

2.2 Solution of the Governing Equation

� The governing equation for undamped free vibration of a 1 DOF mass-spring systemis an homogeneous ordinary di¤erential equation (ODE):

m�x+ kx = 0 (8)


� Assume a solution of the form:x(t) = Cest

then:_x(t) = Csest

�x(t) = Cs2est

� Substitution of the assumed solution into Eq. (8) gives:

mCs2est + kCest = 0

orms2 + k = 0 (9)

which is called the characteristic equation.

� The solution of Eq. (9) is:

s = �� k

m

�1=2= �i!n (10)

where

!n =

�k

m

�1=2(11)

is the natural frequency of the undamped system. It is the frequency of oscillation ofthe undamped system.

� The two values of s are the roots or eigenvalues of the characteristic equation. Sinceboth values of s will satisfy the governing equation, the solution for x(t) is written as:

x(t) = C1ei!nt + C2e

�i!nt (12)

where C1 and C2 are constants to be determined from initial conditions. It is di¢ cultto interpret the motion given by Eq. (12), so let�s make it look pretty.

� Euler�s formula states:e�i!nt = cos!nt� i sin!nt

� Substitution of Euler�s formula into Eq. (12) gives:

x(t) = C1[cos!nt+ i sin!nt] + C2[cos!nt� i sin!nt]

orx(t) = (C1 + C2) cos!nt+ i(C1 � C2) sin!nt


� Since x(t) is a displacement, it must be real, therefore, (C1 + C2) must be real and(C1 � C2) must be imaginary. For this to be true C1 and C2 must be complexconjugates:

C1 = a+ ib ; C2 = a� ib

� so:x(t) = 2a cos!nt� 2b sin!nt

� Then the displacement x(t) can be written as:

x(t) = A1 cos!nt+A2 sin!nt (13)

where A1 and A2 are new constants evaluated from the initial conditions of the system.Two conditions are required (i.e. the same number as the order of the governingequation).

� Specifying the initial displacement and velocity at t = 0:

x(t = 0) = xo

_x(t = 0) = _xo

� and solving for A1 and A2:A1 = xo

_x(t) = �A1!n sin!nt+A2!n cos!nt

_xo = !nA2

A2 =_xo!n

� So the solution to Eq. (8) is:

x(t) = xo cos!nt+_xo!nsin!nt (14)

i.e. harmonic motion. Equation (14) is the motion of any system whose governingequation can be written in the form of Eq. (8).

� Note:

1. If xo = 0:

x(t) =_xo!nsin!nt

2. If _xo = 0:x(t) = xo cos!nt


� Equation (13) can be written in a more convenient form by de�ning:

A1 = A cos� (15)

A2 = A sin� (16)

� Square and add Eqs. (15) and (16) to �nd A:

A =�A21 +A

22

�1=2=

x2o +

�_xo!n

�2!1=2

� Divide Eq. (16) by Eq. (15) to �nd �:

� = tan�1�A2A1

�= tan�1

�_xo

xo!n

�� Substitution of Eqs. (15) and (16) into Eq. (13) gives:

x(t) = A cos� cos!nt+A sin� sin!nt

� Using the trigonometric identity:

cos(x� y) = cos x cos y + sinx sin y

� Thenx(t) = A cos(!nt� �) (17)

where the amplitude A and phase angle � are:

A =

x2o +

�_xo!n

�2!1=2(18)

� = tan�1�

_xoxo!n

�(19)

� This is a very convenient form of the solution for x(t) as it makes the motion quiteobvious. The velocity and acceleration of mass m are:

_x(t) = �!nA sin(!nt� �) = !nA cos(!nt� �+�

2) (20)

�x(t) = �!2nA cos(!nt� �) = !2nA cos(!nt� �+ �) (21)

i.e. _x(t) leads x(t) by �=2 (or 90o) and �x(t) leads x(t) by � (or 180o).



� The motion x(t) can also be expressed as a sine wave by de�ning A1 and A2 in Eq.(13) as:

A1 = A sin�o

A2 = A cos�o

Then:

A =�A21 +A

22

�1=2=

x2o +

�_xo!n

�2!1=2

�o = tan�1�A1A2

�= tan�1

�xo!n_xo

�


� Substitution of the de�nitions of A1 and A2 into Eq. (13) gives:

x(t) = A sin�o cos!nt+A cos�o sin!nt

� Using the trigonometric identity:

sin(x+ y) = sinx cos y + cos x sin y

� Thenx(t) = A sin(!nt+ �o) (22)

with the same magnitude as Eq. (17), but a new phase angle, �o:

�o = tan�1�xo!n_xo

�(23)

2.3 Free Vibration with Viscous Damping

� Consider a mass, m, resting on a �at frictionless surface and connected to a rigidsupport by a linear spring, k, and a viscous damper, c:

� We will need an equation governing the motion of the mass as it is moved from itsequilibrium position. Such an equation of motion can be derived using Newton�s 2ndLaw.

2.3.1 Equation of Motion

� The FBD for a 1 DOF mass-spring-damper system is shown below:


� To derive the equation of motion assume a positive displacement, velocity and accel-eration of the mass m.

� Newton�s 2nd Law gives:

+!X

Fx = m�x = �Fs � Fd = �kx� c _x

where, for a linear spring Fs = kx, and a viscous damper Fd = c _x.

� The governing equation can be rearranged to give:

m�x+ c _x+ kx = 0 (24)

This equation governs the motion of all 1 DOF systems in translation with a linearspring and viscous damper.

2.3.2 Solution of the Governing Equation

� The governing equation for viscously damped vibration of a 1 DOF system is anhomogeneous ODE:

m�x+ c _x+ kx = 0

� As for undamped motion, assume a solution of the form:

x(t) = Cest

� Substitution of the assumed solution into the governing equation, Eq. (24), gives thefollowing characteristic equation:

mCs2est + cCsest + kCest = 0

orms2 + cs+ k = 0 (25)


i.e. a quadratic equation in s. The roots of the characteristic equation are:

s1;2 =�c�

pc2 � 4mk2m

= � c

2m�r� c

2m

�2� k

m(26)

� So the solution to Eq. (24) is:

x(t) = C1es1t + C2e

s2t

or

x(t) = C1e

�� c2m+q( c2m)

2� km

�t+ C2e

�� c2m�q( c2m)

2� km

�t

(27)

where, C1 and C2 are constants obtained from initial conditions. This form of thesolution is hard to interpret so other forms are used.

� De�ne:

1. Critical Damping, cc, as the damping constant c which makes the radical in Eq.(26) zero: � cc

2m

�2� k

m= 0

or

cc = 2m

rk

m= 2

pkm = 2m!n (28)

2. Damping ratio, �, as the ratio of the damping constant c to the critical dampingconstant cc:

� = c=cc (29)

� Using Eqs. (28) and (29):c

2m=

c

cc� cc2m

= �!n (30)

� Using Eqs. (30) and the de�nition of !n, Eq. (11), the roots s1 and s2, Eq. (26), canbe written in terms of � and !n:

s1;2 =

��

q�2 � 1

�!n (31)

� The solution of Eq. (24) can be rewritten in the following form:

x(t) = C1e

��+

p�2�1

�!nt + C2e

(��p�2�1)!nt (32)

� Note: if � = 0, the solution reduces to that for undamped free vibration, Eq. (12).

� It should be obvious that due to the �2 � 1 term in the radical di¤erent types ofsolution will arise if � < 1, � = 1 and � > 1. These three cases correspond tounderdamped, critically damped, and overdamped cases, respectively. The solutionsfor these di¤erent cases will now be considered separately.


2.3.3 Case 1: Underdamped Motion (� < 1, c < cc)

� If � < 1, then �2� 1 < 0, andp�2 � 1 = i

p1� �2, therefore, the roots s1 and s2 can

be written as:

s1;2 =

�� i

q1� �2

�!n

= ��!n � i!d (33)

where !d is the damped frequency of oscillation:

!d =

q1� �2 !n (34)

� The solution to Eq. (24) can be written in the following form:

x(t) = C1e(��!n+i!d)t + C2e

(��!n�i!d)t

orx(t) = e��!nt

�C1e

i!dt + C2e�i!dt

�(35)

� Compare Eq. (35) with Eq. (12) for undamped motion. The solution will be harmonicwith an exponentially decaying amplitude. The frequency of oscillation will be !d =p1� �2!n, i.e. less than !n for undamped motion. Due to the similarities of Eqs.

(12) and (35) the same techniques may be used to rewrite Eq. (35) in more easilyinterpreted forms.

x(t) = e��!nt [(C1 + C2) cos!dt+ i(C1 � C2) sin!dt]

orx(t) = e��!nt

�C 01 cos!dt+ C

02 sin!dt

�Then:

x(t) = Xe��!nt sin (!dt+ �o) (36)

orx(t) = Xe��!nt cos (!dt� �) (37)

� Where the �nal two forms of the solution clearly show that the motion is harmonicwith frequency !d and exponential decay in the amplitude.

� The constants C 01 and C 02 can be derived from the initial conditions:

x(t = 0) = xo

_x(t = 0) = _xo


� and solving for C 01 and C 02:C 01 = xo (38)

_x(t) = ��!ne��!nt�C 01 cos!dt+ C

02 sin!dt

�+ e��!nt

��!dC 01 sin!dt+ !dC 02 cos!dt

�_xo = ��!nC 01 + !dC 02

C 02 =_xo + �!nxo

!d(39)

� The amplitude X and phase angles � and �o are de�ned as follows:

X =q(C 01)

2 + (C 02)2 (40)

�o = tan�1(C 01=C02) (41)

� = tan�1(C 02=C01) (42)

The amplitude X is the maximum amplitude that would exist at t = 0, for zero phaseangle.

� A typical underdamped system response is shown below.


� The motion is harmonic, with exponential decay of amplitude and damped frequencyof oscillation, !d, which is always less than !n since � < 1.

� Vehicle suspension systems are examples of underdamped systems.

2.3.4 Case 2: Critically Damped Systems (� = 1, c = cc)

� If � = 1 the roots s1 and s2 are not distinct:

s1 = s2 = �cc2m

= ��!n = �!n (43)

� To determine the motion take the limit of Eq. (36) as � ! 0.

lim�!1

x(t) = lim�!1

e��!nt�C 01 cos!dt+ C

02 sin!dt

�= e�!nt

�(C 01 + C

02!dt

�Remember:

!d =

q1� �2!n

lim !0

(cos ) = 1

lim !0

(sin ) =

� So the solution can be written as:

x(t) = (C 01 + C02!dt)e

�!nt

� Using Eqs. (38) and (39), the motion of a critically damped system is given by:

x(t) = (xo + ( _xo + !nxo) t) e�!nt (44)


� The motion is not harmonic, it is aperiodic and decreases to zero as t increases dueto the e�!nt term.

� Critical damping, cc, is the lowest value of c that will not produce oscillations.

� A typical example of critically damped motion is shown below with an underdampedcase of the same system for comparison.

2.3.5 Case 3: Overdamped System (� > 1, c > cc)

� For this case, the roots s1 and s2 are both real, distinct and less than 1.

s1 =

�� +

q�2 � 1

�!n (45)

s2 =

��

q�2 � 1

�!n (46)

and s2 < s1.


� The solution is:

x(t) = C1e(��+

p�2�1)!nt + C2e

(��p�2�1)!nt (47)

� For initial conditions:

x(t = 0) = xo

_x(t = 0) = _xo

the constants are:

C1 =xo!n(� +

p�2 � 1) + _xo

2!np�2 � 1

(48)

C2 =�xo!n(� �

p�2 � 1)� _xo

2!np�2 � 1

(49)

� The motion is aperiodic. It has a similar form to the critically damped solution,however, it overshoots the maximum amplitude of the critically damped system andtakes longer to return to equilibrium.


� A good example of an overdamped system is the closer on a screen door which takesforever and an age to close.

2.3.6 Summary

� Underdamped motion:

� is harmonic with exponential decay of amplitude.

�The frequency of oscillation of underdamped motion (!d =p1� �2!n) is always

less than the frequency of oscillation of undamped motion (!n).

�Underdamped motion will not reach the same peak amplitude as undampedmotion, unless the phase angle is zero.

�The underdamped case is the only solution that results in oscillatory motion.

�Underdamped systems return to equilibrium quickly, but continue to oscillate.In the context of a suspension system the wheels would be in position to respondto another bump more quickly than a critically damped system.

� Critically damped motion:


� is aperiodic with exponential decay to equilibrium position.

�Critically damped systems are quickest to return to the equilibrium postion.

�A critically damped system would transmit too much force to car passengers.Vehicle suspensions are underdamped.

� Overdamped systems:

� take a long time to return to the equilibrium position. This is not practicalfor suspension systems. These systems are useful for controlled, slow return toequilibrium without oscillation. Screen door closers are overdamped.

2.3.7 Logarithmic Decrement

� Consider the motion of an underdamped system.

� The ratio of displacements at two times separated by one period of damped oscillationis:

x1x2=Xe��!nt1 cos(!dt1 � �)Xe��!nt2 cos(!dt2 � �)

� But t2 = t1 + �d where �d = 2�=!d.

� Since the angle in the cosine terms would be di¤erent by 360o (2� radians or one fulloscillation) the cosines would be equivalent.

x1x2=

e��!nt1

e��!n(t1+�d)

� De�ne the logarithmic decrement as the natural log of the displacement ratios:

� = lnx1x2= �!n�d = �!n

2�

!np1� �2

=2��p1� �2

=2�

!d� c

2m(50)


� Note: in the log domain the amplitude decay is constant. In the �gure above thedashed line is Xe��!nt.

� The damping ratio, �, can be de�ned as a function of �:

� =�q

(2�)2 + �2(51)

� How can logarithmic decrement be used?

1. Given a system with a known !n and both amplitude ratio and �d are speci�ed! �, i.e. the damping ratio and thus the damping constant. Suspension design.

2. Given a system with known !n and � ! �, i.e. the amplitude ratio. Suspensionanalysis.

� Experimental measurement of time and displacement can be used to determine thedamping constant. It is not necessary to use only one period of oscillation, however,as any numnber of periods may be used:

x1xm+1

=x1x2� x2x3� x3x4� � � xm

xm+1

where m is the number of cycles. Note: m may be a partial cycle.

� Sincex1x2= e�!n�d

thenx1xm+1

= em�!n�d

or

ln

�x1xm+1

�= m�

and

� =1

mln

�x1xm+1

�(52)


3 Forced Vibration - Harmonic Excitation

� Forced vibration is caused by an external applied force or displacement.

� Harmonic excitation implies the applied force is of the form F (t) = Foei(!t+�) of

F (t) = Fo cos(!t + �) or F (t) = Fo sin(!t + �). The phase angle, �, is often chosento be zero as only the steady state response of the system is of interest.

3.1 Equation of Motion

� Consider a 1 DOF spring-mass-damper system subjected to an external appled force.

� The governing equation is:m�x+ c _x+ kx = F (t) (53)

� Equation (53) is a nonhomogeneous ODE that has solution x(t) that is a sum of anhomogeneous solution xh(t) and a particular solution xp(t).

� The homogeneous solution satis�es:

m�x+ c _x+ kx = 0

i.e. the equation for free vibration. The solution xh(t) will, therefore, decay to zerofor any under, over or critically damped system.

� The solution to Eq. (53) will, after some transient during which the homogenoussolution decays to zero, reduce to the particular solution, xp(t).


� Harmonic response will be considered separately for undamped and damped motion.

3.2 Undamped Motion

� Consider Eq. (53) for the undamped case when F (t) = Fo cos(!t):

m�x+ kx = Fo cos!t (54)

� The homogeneous solution is:

xh(t) = C1 cos!nt+ C2 sin!nt (55)

where !n =pk=m.

� Since F (t) is harmonic, the particular solution is also harmonic. Propose:

xp(t) = X cos!t (56)

� Substitue Eq. (56) into Eq. (54):

�m!2X cos!t+ kX cos!t = Fo cos!t

then

X =Fo

k �m!2 =�st

1��!!n

�2 (57)

where�st = Fo=k (58)

is the static de�ection due to force Fo.


� The total solution is:

x(t) = xh(t) + xp(t) = C1 cos!nt+ C2 sin!nt+Fo

k �m!2 cos!t

� Using the initial conditions: x(t = 0) = xo, _x(t = 0) = _xo:

xo = C1 +Fo

k �m!2

So:

C1 = xo �Fo

k �m!2 (59)

_x(t) = �C1!n sin!nt+ C2!n cos!nt�Fo

k �m!2! sin!t

then_xo = C2!n

or

C2 =_xo!n

(60)

� And the total solution for the motion is:

x(t) =

�xo �

Fok �m!2

�cos!nt+

_xo!nsin!nt+

�Fo

k �m!2

�cos!t (61)

� Interesting stu¤ can be seen by writing Eq.(57) in a non-dimensional form:

X

�st=

1

1��!!n

�2 (62)

which is callled the amplitude ratio (or amplitude factor or magni�cation factor). Itis the ratio of the dynamic amplitude to the static amplitude of the motion.

� A plot of X=�st versus !=!n yields:


� As shown by the �gure there are three distinct behaviours.

� Case 1: 0 < !=!n < 1

�The frequency of the input force is less than the natural frequency of the systemand the denominator in Eq. (62) is positive.

�The steady state motion is given by Eq. (61) and the harmonic response of thesystem, xp(t), is in phase with the forcing function.


� Case 2: 1 < !=!n <1

�The frequency of the input force is greater than the natural frequency of thesystem and the denominator in Eq. (62) is negative, therefore, the amplitiuderatio is negative. The motion of the system, xp(t), Eq. (56), will be out of phasewith the external force.

�As !=!n ! 1, X ! 0, i.e. zero response to a high frequency input. In otherwords there is insu¢ cient time for the system to respond to the input. Later thisconcept will be used to show why you should always drive faster.

� Case 3: !=!n = 1 ! Resonance

�The frequency of the input force equals the natural frequency of the system.


�X = 1, i.e. resonance and if energy is continuously fed into the system it willfail. Let�s prove it.

�Equation(61) can be rewritten as:

x(t) = xo cos!nt+_xo!nsin!nt+

Fok �m!2 (cos!t� cos!nt)

or

x(t) = xo cos!nt+_xo!nsin!nt+ �st

0B@cos!t� cos!nt1�

�!!n

�21CA (63)

�Consider the response at ! = !n. The last term in Eq. (63) is unde�ned when! = !n, so use l�Hôpital�s Rule:

lim!!!n

0B@cos!t� cos!nt1�

�!!n

�21CA = lim

!!!n

0BB@ dd! (cos!t� cos!nt)

dd!

�1�

�!!n

�2�1CCA

= lim!!!n

"�t sin!t� 2!!2n

#

=!nt

2sin!nt

� So, the system response at resonance is:

x(t) = xo cos!nt+_xo!nsin!nt+

�st!nt

2sin!nt (64)

which is harmonic with linear increase in amplitude.


3.2.1 Total Response

� Similar to free vibration, x(t), can be written as follows:

x(t) = A cos(!nt� �) +�st

1��!!n

�2 cos!t (65)

� Note: Eq. (3.17) in the text is incorrect. The negative sign in front of the secondterm should be a positive, and then Eq. (3.170 is redundant.

3.2.2 Beat Phenomena

� Beating occurs when ! is close to, but not equal to, !n.

� The amplitude will build up and diminish in a regular pattern.


3.3 Damped Motion

� Consider a 1 DOF system with a linear spring and viscous damper subjected to aharmonic applied force.

� The governing equation is:

m�x+ c _x+ kx = Fo cos!t (66)

� Assume a particular solution of form:

xp(t) = X cos(!t� �) (67)

with derivatives:_xp(t) = �!X sin(!t� �)�xp(t) = �!2X cos(!t� �)

� Substitution of the assumed solution into Eq. (66) gives:

X�(k �m!2) cos(!t� �)� c! sin(!t� �)

�= Fo cos(!t) (68)

� The trigonometric relations:

cos(!t� �) = cos(!t) cos(�) + sin(!t) sin(�)

sin(!t� �) = sin(!t) cos(�)� cos(!t) sin(�)can be used in Eq. (68) to give:

X�(k �m!2)(cos(!t) cos(�) + sin(!t) sin(�))� c!(sin(!t) cos(�)� cos(!t) sin(�))

�= Fo cos!t

or

X�cos!t((k �m!2) cos�+ c! sin�) + sin!t((k �m!2) sin�� c! cos�)

�= Fo cos!t

(69)


� Equating LHS and RHS coe¢ cients of cos!t and sin!t in Eq. (69):

X�(k �m!2) cos�+ c! sin�

�= Fo (70)

X�(k �m!2) sin�� c! cos�

�= 0 (71)

� Equations (70) and (71) can be solved for X and �.

� To solve for X, square each equation and then add the resultants to eliminate �:

X2�(k �m!2)2 cos2 �+ 2c!(k �m!2) cos� sin�+ (c!)2 sin2 �+ (k �m!2)2 sin2 �� 2c!(k �m!2) cos� sin�+ (c!)2 cos2 �

�= F 2o

orX2�(k �m!2)2 + (c!)2

�= F 2o

� So:X =

Fo

[(k �m!2)2 + (c!)2]1=2(72)

� and from Eq. (71):

� = tan�1�

c!

k �m!2

�(73)

� A typical plot of a forcing function and steady-state system response, xp(t), is shownbelow. Here, the system motion lags the forcing function.

� Using the de�nitions of undamped natural frequency, !n, damping ratio, �, staticde�ection, �st, and frequency ratio, r:

!n =pk=m

� =c

cc=

c

2m!n! c = 2�m!n

�st =Fok


r =!

!n(74)

the amplitude ratio (or magni�cation factor, M), X=�st, can be obtained from Eq.(72):

X =Fo=kh

(1� m!2

k )2 + ( c!k )2i1=2

=�sth

(1� ( !!n )2)2 + (2� !

!n)2i1=2

� So:

M =X

�st=

24 1� � !

!n

�2!2+

�2�

!

!n

�235�1=2 = 1p(1� r2)2 + (2�r)2

(75)

� Similarly, �, Eq. (73) can be written in terms of r and �:

� = tan�1

264 2� !!n

1��!!n

�2375 = tan�1� 2�r

1� r2

�(76)

� Plots of M = X=�st and phase angle, �, versus frequency ratio are shown below forvarious values of �.

� Magni�cation factor comments:


1. Any amount of damping reduces M at all frequency ratios. Damping slowssystem response and leads to the phase angle, see Eq. (76).

2. As � ", M # at any r, i.e. more damping ! less motion.

3. As r ", M #, i.e. the system does not respond to high frequency inputs.

4. Damping produces a signi�cant reduction in M near r = 1, i.e. damping is veryimportant near resonance as it controls the motion.

5. For what value of r would M be a maximum, and what is Mmax?

dM

dr= 0 =

�(1� r2)2 + (2�r)2

��3=2��12

��2(1� r2)(�2r) + (2�)2(2r)

�0 = �2(1� r2) + (2�)2

� So:r =

q1� 2�2 (77)

then

Mmax =

�(1� (1� 2�2))2 + (2�

q1� 2�2)2

��1=2= (4�2(1� �2))�1=2

=1

2�p1� �2

(78)

� If � is known, Eq. (77) gives r and Eq. (78) givesMmax and Xmax =Mmax�stgives an estimate of the maximum displacement.

�Measurement of Xmax will give an estimate of � (if Fo and k are known).

� Phase angle comments:

1. For � > 0, 0 < r < 1, then 0o < � < 90o, i.e. the response lags the input.

2. For � > 0 r > 1 then 90o < � < 180o, i.e. response is out of phase with input(similar to undamped system).

3.3.1 Total Response (Damped System)

� The complete expression for the response of the damped 1 DOF system to a harmonicinput is:

x(t) = Xoe��!nt cos(!dt� �o) +X cos(!t� �) (79)

� Note: the �rst term in Eq. (79) is the homogeneous solution and it will decay to zero,and the solution will reduce to the steady-state solution, i.e. the second term whichis the particular solution.


� Xo and �o are obtained from initial conditions: x(t = 0) = xo and _x(t = 0) = _xo

xo = Xo cos(��o) +X cos(��)= Xo cos(�o) +X cos(�) (80)

_x = Xo(��!ne��!nt cos(!dt� �o)� !de��!nt sin(!dt� �o))�X! sin(!t� �)

_xo = Xo(��!n cos(��o)� !d sin(��o))�X! sin(��)= Xo(��!n cos(�o) + !d sin(�o)) +X! sin(�) (81)

� It is easier to solve for Xo and �o with numbers from a particular problem, ratherthan derive a messy general solution.


4 Harmonic Motion of the Base

� Consider a 1 DOF viscously damped system subjected to a harmonic motion of thebase.

� The motion of the base is de�ned by:

y(t) = Y sin!t (82)

� The spring force and damping force will be functions of both x and y:

Fs = k(x� y)Fd = c( _x� _y)

� the governing equation for the motion is:

m�x+ c( _x� _y) + k(x� y) = 0 (83)

Note: there is no forcing function.

� Since y(t) = Y sin!t, the governing equation can be written as:

m�x+ c _x+ kx = c _y + ky

= c!Y cos!t+ kY sin!t

= A sin(!t� �) (84)

where

A = Ypk2 + (c!)2 (85)

� = tan�1��c!k

�(86)


� Comparison of Eq. (84) with Eq. (53) shows that they are of the same form (i.e. adamped 1 DOF system with a harmonic forcing function). The forcing motion is nottransmitted directly to the mass, however, as it is transmitted through the spring anddamper (e.g. a suspension system). The particular solution is then:

xp(t) =Ypk2 + (c!)2

((k �m!2)2 + (c!)2)1=2sin(!t� �1 � �) (87)

where

�1 = tan�1�

c!

k �m!2

�(88)

i.e. similar to Eqs. (??, (??) and (??).

� Using the trigonometric identities for angle addition, Eq. (87) can be written as:

xp(t) = X sin(!t� �) (89)

whereX

Y=

�k2 + (c!)2

(k �m!2)2 + (c!)2

�1=2=

�1 + (2�r)2

(1� r2)2 + (2�r)2

�1=2(90)

� = tan�1�

mc!3

k(k �m!2) + (c!)2

�= tan�1

�2�r3

1 + (4�2 � 1)r2

�(91)

� The ratio of the amplitudes of the response to base input, X=Y , is called the displace-ment transmissibility, Td.

� The variation of Td and � with frequency ratio are shown below for various values of�.


� Td has a maximum value at:

rm =1

2�

�q1 + 8�2 � 1

�1=2� The force, F , acting on the base is

F = c( _x� _y) + k(x� y) = �m�x (92)

� Using Eq. (89) for x:

F = m!2X sin(!t� �) = FT sin(!t� �)

where FT = m!2X is the amplitude of the force transmitted to the base. Using Eq.(90):

FTkY

= r2X

Y(93)

where FT =kY is the force transmissibility.

� The behaviour of force transmissibilty with frequency ratio is plotted below for variousvalues of damping ratio.


� De�ne z = x� y as the motion of the mass relative to the base, then Eq. (84) can bewritten as:

m�z + c _z + kz = �m�y = m!2Y sin!t (94)

� Comparison of Eq. (??) with Eq. (84) gives the solution:

z(t) = Z sin(!t� �1) (95)

where

Z =m!2Yp

(k �m!2)2 + (c!)2= Y

r2p(1� r2)2 + (2�r)2

(96)

�1 = tan�1�

c!

k �m!2

�= tan�1

�2�r

1� r2

�(97)

� The variation of Z=Y with frequency ratio for several values of � is shown below.



5 System Modelling

5.1 Combinations of Springs

� In practical systems several linear springs may be used in combination (e.g. foursprings used in a car).

� An e¤ective spring constant can be developed for spring combinations in parallel andseries.

5.1.1 Springs in Parallel

� Consider the case where two linear springs support a platform.

� If a force ~W is applied to the platform such that it does not rotate, the springs willobtain the same static de�ection, �st.

� Using the FBD of the platform when it is in static equilibrium:

+ "X

Fy = 0 = F1 + F2 �W

� But

F1 = k1�st

F2 = k2�st

� thereforeW = k1�st + k2�st

� We want an equivalent spring keq that will support forceW and attain the same staticde�ection, �st.

W = keq�st


� Sokeq = k1 + k2

� For any combination of parallel springs:

keq =

nXi=1

ki (98)

5.1.2 Springs in Series

� Consider two linear springs k1and k2 attached in series:

� When load ~W is applied to spring 2 the springs will de�ect a di¤erent amount, unlessk1 = k2.

� Consider an FBD for each spring:

� Each spring will carry the same load, W :

W = k1�1 = k2�2 = keq�st = keq(�1 + �2)

� De�ne an equivalent linear spring keq that would carry load W , but de�ect �st =�1 + �2.


� Note:

�1 =W

k1=keqk1�st

�2 =W

k2=keqk2�st

� Then:

�st = �1 + �2

=keqk1�st +

keqk2�st

� or1

keq=1

k1+1

k2

� For any combination of springs in series:

1

keq=

nXi=1

1

ki(99)

5.1.3 General Method for Combinations of Springs

� Since springs are used to store potential energy it is possible to equate the potentialenergy stored due to the deformation of a new equivalent spring to that stored in theoriginal spring system.

� e.g. springs in parallel

1

2k1�

2st +

1

2k2�

2st =

1

2keq�

2st

�! keq = k1 + k2


� e.g. spings in series:

1

2k1�

21 +

1

2k2�

22 =

1

2keq�

2st

� But

�1 =W

k1=keqk1�st

�2 =W

k2=keqk2�st

� Then1

2keq�

2st =

1

2k1

�keqk1�st

�2+1

2k2

�keqk2�st

�2�! 1

keq=1

k1+1

k2

5.2 Combinations of Damping Elements

� To obtain equivalent damping elements use the method of equivalence of force betweenthe original combination and the equivalent damper.

5.2.1 Viscous Dampers in Parallel

� Assume the velocity at the end of each damper is equivalent.


� Since the ends of each damper will have the same velocities:F1 = c1 ( _x2 � _x1)

F2 = c2 ( _x2 � _x1)

Feq = ceq ( _x2 � _x1)

� But the equivalent damper will exert the force of both original dampers:Feq = F1 + F2

� So:ceq = c1 + c2

� For any combination of dampers in parallel:

ceq =

nXi=1

ci (100)

5.2.2 Viscous Dampers in Series

� Consider three viscous dampers in series to be replaced by and equivalent viscousdamper ceq that will have the same velocites at its end points.


� Each damper will carry the same force:

F1 = F2 = F3 = Feq

� where

F1 = c1 ( _x2 � _x1)

F2 = c2 ( _x3 � _x2)

F3 = c3 ( _x4 � _x3)

Feq = ceq ( _x4 � _x1)

� From kinematics:

( _x4 � _x1) = ( _x2 � _x1) + ( _x3 � _x2) + ( _x4 � _x3)

� ThereforeFeqceq

=F1c1+F2c2+F3c3

� Then1

ceq=1

c1+1

c2+1

c3

� For any combination of viscous dampers in series:

1

ceq=

nXi=1

1

ci(101)


6 Two Degree of Freedom Systems

6.1 Governing Equations

� Consider the two degree of freedom (2 DOF) system shown below:

� Each mass can move in one dimension, therefore the system has two degrees of free-dom. This system is the most general case where a forcing function is applied to eachmass.

� Use the FBD for each mass to derive the governing equations:

� To de�ne directions of forces assume relative magnitudes for the motion of each mass,e.g. x2 > x1 and _x2 > _x1.

m1�x1 = c2( _x2 � _x1) + k2(x2 � x1)� k1x1 � c1 _x1 + F1(t)m2�x2 = �c2( _x2 � _x1)� k2(x2 � x1)� k3x2 � c3 _x2 + F2(t)

� or


m1�x1 + (c1 + c2) _x1 � c2 _x2 + (k1 + k2)x1 � k2x2 = F1(t)

m2�x2 � c2 _x1 + (c2 + c3) _x2 � k2x1 + (k2 + k3)x2 = F2(t) (102)

� or �m1 00 m2

� ��x1�x2

�+

�(c1 + c2) �c2�c2 (c2 + c3)

� �_x1_x2

�+

�(k1 + k2) �k2�k2 (k2 + k3)

� �x1x2

�=

�F1(t)F2(t)

�(103)

� Note: The governing equation for any 2 DOF system can be written in the followinggeneral form:�

m11 m12

m21 m22

� ��x1�x2

�+

�c11 c12c21 c22

� �_x1_x2

�+

�k11 k12k21 k22

� �x1x2

�=

�F1(t)F2(t)

�(104)

� Here: �m11 m12

m21 m22

�=

�m1 00 m2

��c11 c12c21 c22

�=

�(c1 + c2) �c2�c2 (c2 + c3)

��k11 k12k21 k22

�=

�(k1 + k2) �k2�k2 (k2 + k3)

�� For di¤erent systems, the entries in the matrices on the RHS of the equations wouldchange.

6.2 Free Vibration of an Undamped 2 DOF System

� Simplify the system above for the case of undamped free vibration:


� From the general form of the governing equation for a 2 DOF system Eq. (104) theequations of motion for this system are:�

m11 m12

m21 m22

� ��x1�x2

�+

�k11 k12k21 k22

� �x1x2

�=

�00

�(105)

� For this system, m12 = m21 = 0. This will have signi�cance later.

� So, the governing equations are:

m11�x1 + k11x1 + k12x2 = 0

m22�x2 + k21x1 + k22x2 = 0 (106)

� Can m1 and m2 oscillate at the same frequency and phase angle, but with di¤erentamplitude?

x1(t) = X1 cos(!t+ �)

x2(t) = X2 cos(!t+ �) (107)

� The same phase angle would imply each mass passes through its equilibrium positionat the same time.

� From Eq. (107):

�x1 = �!2X1 cos(!t+ �)�x2 = �!2X2 cos(!t+ �) (108)

� Substitution of Eqs. (107) and (108) into Eq. (105):��m11!

2 + k11�X1 + k12X2

�cos(!t+ �) = 0��

�m22!2 + k22

�X2 + k21X1

�cos(!t+ �) = 0 (109)

� or �(�m11!

2 + k11) k12k21 (�m22!

2 + k22)

� �X1X2

�=

�00

�(110)

� For a nontrivial solution:�� (�m11!2 + k11) k12

k21 (�m22!2 + k22)

�� = 0 (111)

� So:(m11m22)!

4 � (m11k22 +m22k11)!2 + (k11k22 � k12k21) = 0 (112)


� This equation is quadratic in !2. For convenience, rewrite the equation in the fol-lowing form:

a!4 + b!2 + d = 0 (113)

� where:

a = m11m22

b = �(m11k22 +m22k11)

d = k11k22 � k12k21 (114)

� Solving for the two roots gives:

!1 =

�b�

pb2 � 4ad2a

!1=2(115)

!2 =

�b+

pb2 � 4ad2a

!1=2(116)

� These are the two natural frequencies of the system!

� Substitution of Eqs. (115) and (116) into Eq. (110) allows determination of amplituderatios at the two natural frequencies.

(�m11!21 + k11)X1 + k12X2 = 0

(�m22!21 + k22)X2 + k21X1 = 0

� and

(�m11!22 + k11)X1 + k12X2 = 0

(�m22!22 + k22)X2 + k21X1 = 0

� Then the amplitude ratios:

r1 =

�X2X1

�(1)=(�m11!

21 + k11)

�k12=

�k21��m22!21 + k22

� (117)

r2 =

�X2X1

�(2)=(�m11!

22 + k11)

�k12=

�k21��m22!22 + k22

� (118)

� !1, !2, r1 and r2 de�ne the natural modes of the system. The bracketed superscriptsindicate the �rst and second natural modes of the system.


� Use of the amplitude ratios allows for a general solution for the motion of the systemto be written as:

~x(1)(t) =

"x(1)1 (t)

x(1)2 (t)

#=

"X(1)1 cos(!1t+ �1)

r1X(1)1 cos(!1t+ �1)

#= �rst mode (119)

~x(2)(t) =

"x(2)1 (t)

x(2)2 (t)

#=

"X(2)1 cos(!2t+ �2)

r2X(2)1 cos(!2t+ �2)

#= second mode (120)

� Using superposition, the motion of each mass can be written as:

x1(t) = X(1)1 cos(!1t+ �1) +X

(2)1 cos(!2t+ �2)

x2(t) = r1X(1)1 cos(!1t+ �1) + r2X

(2)1 cos(!2t+ �2) (121)

where the general solution for X(1)1 , X(2)

1 , �1 and �2 is given by:

X(1)1 =

1

r2 � r1

"fr2x1(0)� x2(0)g2 +

f�r2 _x1(0) + _x2(0)g2

!21

#1=2

X(2)1 =

1

r2 � r1

"f�r1x1(0)� x2(0)g2 +

fr1 _x1(0) + _x2(0)g2

!22

#1=2�1 = tan�1

��r2 _x1(0) + _x2(0)

!1 [r2x1(0)� x2(0)]

��2 = tan�1

�r1 _x1(0) + _x2(0)

!2 [�r1x1(0)� x2(0)]

�(122)

See the derivation for Eq. (5.18) in the text.

� The i-th mode can be excited by setting initial conditions as follows:

x1(t = 0) = X(i)1 _x1(t = 0) = 0

x2(t = 0) = riX(i)1 _x2(t = 0) = 0

(123)


� So what�s that about? Consider the following simple system:

�From Eq. (103) the governing equation is:�m 00 m

� ��x1�x2

�+

�2k �k�k 2k

� �x1x2

�=

�00

�(124)

� So, for reference: �m11 m12

m21 m22

�=

�m 00 m

��k11 k12k21 k22

�=

�2k �k�k 2k

��From Eq. (114):

a = m2

b = �(2km+ 2km) = �4kmd = 4k2 � k2 = 3k2

�From Eqs. (115) and (116):

!1 =

4km�

p(4km)2 � 4(m23k2)

2m2

!1=2=

rk

m(125)

!2 =

4km+

p(4km)2 � 4(m23k2)

2m2

!1=2=

r3k

m(126)

�From Eqs. (117) and (118)

r1 =

�X2X1

�(1)=

��m

�km

�+ 2k

�k

=k�

�m�km

�+ 2k

� = 1 (127)

r2 =

�X2X1

�(2)=

��m

�3km

�+ 2k

�k

=k�

�m�3km

�+ 2k

� = �1 (128)


�The bracketed superscripts indicate the 1st and 2nd natural modes.� 1st mode - amplitudes of both masses are equivalent.� 2nd mode - amplitudes of both masses are the same, but the motion is outof phase (due to the negative sign).

�The modes are often shown on a mode diagram.

�As shown in the mode diagram for the second mode, the point in the middle ofthe second spring remains �xed for all time. This point is called a node.

�The general solution for the motion of the system is (using Eq. (121)).

x1(t) = X(1)1 cos

rk

mt+ �1

!+X

(2)1 cos

r3k

mt+ �2

!

x2(t) = X(1)1 cos

rk

mt+ �1

!�X(2)

1 cos

r3k

mt+ �2

!(129)

where �1, �2, X(1)1 and X(2)

1 are dependent on the intial conditions: x1(0), x2(0),_x1(0) and _x2(0). See Eq. (122) or Eq. (5.18) in the text.

� Let�s use some numbers. Consider the case where m = 100 kg, k = 10 kN=m.Then:

!1 =

rk

m= 10 rad= s

!2 =

r3k

m= 17:32 rad= s

�Consider the following three initial states:1. First mode:

x1(0) = 0:02m

x2(0) = r1x1(0) = 0:02m

_x1(0) = _x2(0) = 0


From Eq. (122) or Eq. (5.18) in the text:

�1 = �2 = 0

X(1)1 =

1

�1� 1 (�1 (0:02)� 0:02) = 0:02m

X(2)1 =

1

�1� 1 (�1 (0:02) + 0:02) = 0m

Then from Eq. (121) or Eq. (129):

x1(t) = 0:02 cos(10t)m

x2(t) = 0:02 cos(10t)m

2. Second mode:

x1(0) = 0:02m

x2(0) = r2x1(0) = �0:02m_x1(0) = _x2(0) = 0

From Eq. (122) or Eq. 5.18 in the text:

�1 = �2 = 0

X(1)1 =

1

�1� 1 (�1 (0:02) + 0:02) = 0m

X(2)1 =

1

�1� 1 (�1 (0:02)� 0:02) = 0:02m


x1(t) = 0:02 cos(17:32t)m

x2(t) = �0:02 cos(17:32t)m3. Arbitrary initial conditions:

x1(0) = 0:02m

x2(0) = �0:01m_x1(0) = _x2(0) = 0

From Eq. (122) or Eq. (5.18) in the text:

�1 = �2 = 0

X(1)1 =

1

�1� 1 (�1 (0:02)� (�0:01)) = 0:005m

X(2)1 =

1

�1� 1 (�1 (0:02)� 0:01) = 0:015m


x1(t) = 0:005 cos(10t) + 0:015 cos(17:32t)m

x2(t) = 0:005 cos(10t)� 0:015 cos(17:32t)m

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