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7/30/2019 Chapter 1 & 2 - Introduction to Vibrations and Single DOF Systems (1)
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There are two general classes of vibrations - free
and forced.
Free vibration
takes place when a system oscillates under theaction of forces inherent in the system itself or
after an initial disturbance.
Example : oscillation of a simple pendulum
Forced vibration
Vibration that takes place under the excitation of
external forces.
Often a repeating type of force, e.g. the oscillation
arises in machines such as diesel engine.
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Natural frequency
Property of the dynamic system established
by its mass(m) and stiffness(k).
Resonance
A condition whereby the frequency of the
external force coincides with one of the
natural frequencies of the system. The system undergoes a dangerously large
oscillation.
Causing failure of structures such as bridges,
buildings, turbines and airplane wing etc.
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Degree of freedom (DOF)
The number of independent coordinates
required to describe the motion of the system.
Free particle undergoing general motion in
space will have 3 DOF.
Rigid body, will have six degrees of freedom,
i.e. 3 components of position and angles
defining its orientation.
Continuous elastic body will require an infinite
number of coordinates to describe motion.
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Period motion the motion is repeated in equal
intervals of time, T.
Periodthe repetition time, tof the oscillation and
its reciprocal. Frequency, f
If the motion is designated by the time function x(t),
1f
txtx
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Harmonic motion is often represented as the
projection on a straight line of a point that is
moving on a circle at constant speed. With the angular speed of the line O-P designated
by , the displacementxcan be written as
)1(....sin tAx
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is generally measured in radians per second
and referred as angular frequency. The
relationship is
Where T= period(s), f= frequency(cycles/s) of theharmonic motion.
Differentiating the x in eq. (1) twice, the velocity
and acceleration of system
)2(....22
f
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Differentiating the x in eq. (1) twice, the velocityand acceleration of the system turned out as
)3(....)sin(cos 2
tAtAx
)4(....)sin(sin22
tAtAx
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Basic vibration model of a simple oscillatory
system consists ofa mass, a spring and a damper
or dashpot.
Spring force-deflection relationship is consideredto be linear following the Hooke's law,
The viscous damping generally representeddashpot, is described by force proportional to the
velocity,
NkxF
NxcF
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Spring-mass system and FBD
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a simple undamped spring-mass system.
One degree of freedom (DOF) as its described by a
single coordinatex.
Applying Newton's second law,
)5(....mgwk
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With x chosen to be positive downward, together with
all quantities, i.e. force, velocity, and acceleration.
Once again applying Newtons second law of motion;
Since , therefore
)( xkwFxm wk
)6(....kxxm
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Defining the circular frequency by the equation;
Eq. (6) can be written as
n
)7(....
m
kn
)8(....02
xx n
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Eq.(8), a homogeneous second order linear differential
equation has the following solution
Where A and B are the necessary constants andevaluated from initial conditions and
And eq.(9) is reduced to
)9(....cossin tBtAxnn
)0(x )0(x
)10(....cos)0(sin)0(
txtx
x nnn
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The naturalperiod of the oscillation is established from
And the natural frequency is
)11(....2 k
m
2n
)12(....211
mkfn
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These quantities can be expressed in terms of the static
deflection, by observing eq.(5), thus eq.(12)
can be expressed as
Note: depend only on the mass and stiffnessof the system.
)13(....2
1
gfn
mgk
nnf and,
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Viscous damping force is expressed by the equation
c is a constant of proportionality.
By adding the damper, the equation of motion
becomes
)14(....xcFd
)15(....)(tFkxxcxm
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Solution of this equation divided into two parts.
If the homogenous solution correspondsphysically to that offree-damped vibration.
If the solution of the system is added with
the particular solution that due to the excitation
irrespective of the homogenous solution.
0)( tF
0)( tF
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Particular solutionxp(t) orSteady-state Responsexss.
i. If F(t) = F0 = constant(step input), the we have
The solution is assumed in the same form as the input that
is
From which we obtain,
Applying the initial conditions
Finally, the total solution is
0Fkxxcxm
constant)( Ctxp
0)( KFCtxp
00 )0(and)0( xxxx
textex
tx dt
d
t
d
n nn
sinsin)( 00
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i. If F(t) = F0 sin(t) (sinusoidal input), then we have
We have
The solution is assumed in the form as the input but with
phase angle, that is
)sin(0 tFkxxcxm
)sin()( 0 tXtxp
)sin(0 tFkxxcxm
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Homogenous solution:
Then, assume a solution of the form;
Wheres is a constant. Substituting into the differentialequation eq.(16),
)16(....0 kxxcxm
)17(....stex
0)(2 stekcsms
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Satisfying all values oft;
Eq.(18) is a characteristic equation and has two roots,
solved by using the quadratic formula;
)18(....0
2
m
k
sm
c
s
)19(....22
2
2,1m
k
m
c
m
cs
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Hence the general solution is given by the equation:
WhereA andB are constants to be evaluated from the
initial condition and
Eq.(19) substitute into (20) gives:
)20(....21tsts
BeAex
)21(....))2/(())2/(()2(
22
tmkmctmkmctmc BeAeex
)0(x )0(x
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The first term , is simply an exponentially
decaying function of time.
As the damping term is larger than , the
exponents in the previous equation are real numbersand no oscillations are possible. This condition is
referred as overdamped.
when the damping term is less than , the
exponent becomes an imaginary number,
tmce
2
22mc mk
22mc mk
tmcmki 22
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Thus,
The terms of eq.(21) within the parentheses are
oscillatory. Hence, this is referred as underdamped.
In limiting case between the oscillatory and non-oscillatory motion where , the resultant
of the exponent is zero. Then, the corresponding case
is said as critical damping, cc.
tm
c
m
k
itm
c
m
k
e
tmcmki22
2
2sin2cos
2
mkmc 22
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Any damping can then be expressed in terms of the
critical damping by a non dimensional number zetha,called the damping ratio.
And
)22(....222 kmmm
kmc nc
)23(....
cc
c
)24(....22
nc
m
c
m
c
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i. Oscillatory motion Underdamped Case:
The frequency for damped oscillation is equal to:
)25(....22
11
titi
t nnn BeAeex
)26(....12 2 nd
d
)0.1(
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Damped oscillation )1(
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ii. Non oscillatory Motion Overdamped Case:
The motion is exponentially decreased with time as shown
below:
A periodic motion
)27(....11 22 tt nn
BeAex
)0.1(
)0.1(
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ii. Critically Damped Motion:
Three types of response with initial displacement is
shown in the figure below:
Critically Damped Motion
)28(....)( tneBtAx
)0.1(
)0(x
)0.1(
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Commonly created by the unbalance in rotating
machinery. It is less likely to occur compared to periodic
and other types of excitation.
Basically required in order to understand how the systemwill respond to more general types of excitation.
Considering a single DOF system with viscous damping
excited by a harmonic force . The equation of
motion becomes
)29(....sin tFkxxcxm o
tFo sin
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Viscously Damped System with Harmonic Excitation
Solution to this system, eq.(29) consists of two parts
Complementary function:- solution of thehomogeneous equation, which for this case is a
damped free vibration.
Particular integral particular solution of a steady-
state oscillation of the excitation.
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Particular solution:
WhereX- the amplitude of oscillation
- the phase of the displacement with respect to
the excitation force
By substituting eq.(30) into eq.(29), the amplitude and thephase are solved.
Recall that in the Harmonic Motion the phases of velocity
and acceleration are ahead of the displacement by 90 and
180 respectively.
)30(....)sin( tXx
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Vector Relationship for Forced Vibration with Damping
From the above diagram, it is clearly seen that
and
)31(....
222 cmk
FX o
)32(....tan2
1
mk
c
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Expressing eq.(31) and eq.(32) in non-dimensional term,
divide the numerator and denominator of eq.(31) and
eq.(32) by k, ones obtain:
And
)33(....
1
222
k
c
k
m
kF
X
o
)34(....
1
tan2
k
m
k
c
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Further expressing in terms of the following quantities:
oscilltionundampedoffrequencynaturalm
kn
dampingcriticalmc nc 2
factordampingc
c
c
n
c
c k
c
c
c
k
c
2
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The non-dimensional expression for phase and amplitude
then become:
and
)35(.....
21
1
222
nn
oF
Xk
2
1
2
tan
n
n
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Eq.(35) and eq.(36) indicate that the non-dimensional
amplitude , and the phase angle, are functions of
the frequency ratio, and the damping factor, can be
plotted as below:
oFXk
n
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For mechanical system as shown, derive its mathematical
model and determine its frequency of vibration in hertz if
k=100N/m and m=2kg. Obtain the displacementx if the initial
conditions are x(0) =0.05m and (0)=0 m/s. What is themaximum velocity and acceleration attained by the mass?
mk
x
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FBD:
Applying newtons second law, yields
or (mathematical model or equation of motion)
The frequency of the vibration is given by
which is the undamped natural frequency of the system and in the unit of hertz,
mkx
x
m m
x
: xmF xmkx )1(....0 kxxm
sradm
kn 07.7
2
100
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For the undamped free vibration, the displacement is given by
which is the complimentary orhomogeneous solution of the forced vibration.
Differentiating eq.(1) yields,
Substituting the initial conditions into eq.(1) and (2), we
obtain
Hzf nn 125.12
07.7
2
)1(....)sin( tCtx n
)2(....)cos( tCtx nn00 )0(and)0( xxxx
0sin0 xCx 0cos0 xCx n
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Solving for
Substituting the given initial conditions ,we have
and
Differentiating eq.(2), yields
2
0
2
0 )( nxxC
yieldsandC
001tan xx n smxmx /0)0(and05.0)0(
m05.0)07.70(05.0 22 A rad571.1007.705.0tan1
(2)....m)rad571.107.7sin(05.0)( ttx
(4)....m/s)rad571.107.7sin()07.7(05.0)()( 22 ttxta
(3)....m/s)rad571.107.7cos()07.7(05.0)()( ttxtv
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Notice that sine and cosine function oscillates between 1 to -1, therefore, the
maximum values occur when this function is equal 1 or -1. Hence, we
m/s3535.0)1)(07.7(05.0max v
22
max m/s5.2)1()07.7(05.0
a
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The logarithmic decrement , is introduced to determine the reduction trend of
two successive amplitudes. It is applied for underdamped system. An example
of decrement curve is as shown in the figure below:
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The ratio of two successive amplitudes ofa full cycle distant such as A1andA2are measured and computed as
The logarithmic decrement is defined as the natural logarithm of this ratio
For many systems, the first amplitude A0
is somehow not very clear and
because of that the general form is written as
)37(....)ln(2
1
A
A
)36(....)()(
2
1
1
1
2
1
dn
dn
n
ee
e
Ce
Ce
A
ATt
t
t
t
)38(....)ln(...)ln()ln(14
3
2
1
k
k
A
A
A
A
A
A
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From eq.(36), we get
which then rearrange into
if the value of is very small, then and eq.(40) simplifies to
)40(....4
22
)39(....1
2
1
22
22
n
n
d
ndn
)41(...2
112
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When value of is too small, the decrement of two successive amplitude is not
so significant and the amplitude ratio close to 1 where .
thus, in order to overcome this condition, we have to consider n number of
cycles for measuring the logarithmic decrement, the modified formula is given
as follow;
Where
)42(....1
2ln
2
n
A
An
nk
k
01ln
knk
k
AnAA
aftercyclesamplitudeanisamplitudeknownanyis
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When value of is too small, the decrement of two successive amplitude is not
so significant and the amplitude ratio close to 1 where .
thus, in order to overcome this condition, we have to consider n number of
cycles for measuring the logarithmic decrement, the modified formula is given
as follow;
Where
)42(....1
2ln
2
n
A
An
nk
k
01ln
knk
k
AnAA
aftercyclesamplitudeanisamplitudeknownanyis
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Given that the amplitude of free vibration is reduced by 65%
from its maximum value after 4 complete cycles, determine
for this system
a) The damping ratio, and
b) The corresponding value of the viscous damping
coefficient form = 5 kgand k = 0.5 kN/m.
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Let the maximum amplitude be A0, then after 4 complete
cycles (n = 4),A4 = 0.35A0 (since it reduces by 65%).
By using eq.(42);
With n = 4 and k = 0
nk
k
A
A
nln
1
262.0
35.0
1ln4
1
35.0ln4
1ln4
1
0
0
4
0
A
A
A
A
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Using eq.(40)
Since the value of is quite small, eq.(41) could also be
applied, from which we get
(b) Applying
0417.0262.04
262.0
22
0417.0)142.3(2
262.0
msNkmc /.17.4))5(5002)(0417.0(2
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Consider the mechanical system as shown in figure. If the
10-kg block is moved 0.16m to the right of the equilibrium
position released from rest ort = 0 s, determine
(a) The frequency of the resulting motion, and
(b) The displacement of the block at t = 0.3 s.
Given k = 40 N/m and that the force exerted by the dashpot
is 1.8 N when the speed of the block is 0.12 m/s
m
kx
c
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FBD: Kinetic Diagram:
Applying newtons second law, yields
or (mathematical model or equation of motion)
The frequency of the vibration is given by
mkx
x
m m
: xmF xmxckx )1(....0 kxxcxm
srad210
40
m
kn
c
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And the viscous damping coefficient is obtained from the relation
For the underdamped free vibration, the displacement is given by
where
ed)(underdamp375.0)10(402
15
2
km
c
N.s/m1512.0/8.1/ xFc
)1(....)sin()( tCetx d
tn
xcF
rad/s854.1375.0121 22 nd
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By differentiating eq.(1), we get
Substituting the initial condition into eq.(1) and
(2), we obtain
Solving for
)3(...sin)0( 0xCx
sin0xC
)2(....)cos()sin()( tCetCetx d
t
dd
t
nnn
00 )0()0( xxandxx
)4(...cossin)0( 0xCCx dn
yields,and C
nd xxx 0001tan
n
d
dn
dn
xx
x
xx
x
xxx
00
0
00
0
000
tan
tan
cossinsinsin)4(eq.intoCSubs.
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Substituting the given initial conditions
And the displacement in meters is
Hence, at time t = 0.3 s gives
m1358.0)186.1)3.0(854.1sin(1726.0)3.0()3.0(75.0 ex
1726.0)186.1sin(16.0 C
m/s0)0(m16.0)0( xandx
m)186.1854.1sin(1726.0)( 75.0 tetx t
rad186.1)2375.016.00(854.116.0tan 1
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Determine the natural frequency of the spring-mass system in bothrad/s and Hz. Given k = 2.5 kN/m and m = 20 kg.
m
k x
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For the spring-mass-damper system shown in figure below, determine
(a) The natural frequency in rad/s
(b) Damping ratio
Classify the system as underdamped, critically damped, oroverdamped.
Given k = 2.5 kN/m, m = 20 kg and c = 120 N/m/s
m
k x
c
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For the spring-mass-damper system shown in figure below, determine
(a) The natural frequency in rad/s
(b) The Damping ratio
Classify the system as underdamped, critically damped, oroverdamped.
Given k = 3 kN/m, m = 10 kg and c = 0.5 N/m/s
c
m
k
y
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An industrial press is mounted on a rubber pad to isolate it from itsfoundation. If the rubber pad is compressed 5 mm by the self-weight
of the press, find the natural frequency of the system.
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Consider a rod of mass m, length l, pivoted at O, and is connected toa spring of stiffness kat a distance a from O and also carries a loadM
at a free endB which actuated by the forcef(t).
M
kO
a
B
f(t)
l
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Firstly, equate FDB and kinetic diagram of the system as in figurebelow,
Knowing that, the spring displacement is ,
the acceleration of the mass centre of the rod is ,
acceleration of end B is ,
And the moment of inertia of the rod is
Mg
O
B
f(t)k(y+st)
mg
A
G
FBD
MaB
O
B
maG
A
G
KD
I
ay
2laG
IaB 2
12
1 mlI
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Then, applying Newtons 2nd law, yields
At static equilibrium, we have
From eq.(1) we obtain
effectiveOexternalO MM+
)1(...)(22))((2)()( lMl
ll
mIaak
l
mglMgltf st
)2(...0)(2
)(
ak
lmglMg st
)3(...)(2212
1))(()(
2lMl
llmmlaakltf
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Alternatively, eq.(4) can be derived by considering only moment ofexternal forces without the effect of the weight since its moment will
cancel out with the moment of static equilibrium.
where simplifying eq.(5)
becomes
which is identical to eq.(4).
)5(...)()()( OIaakltf
)6(...)(3
1 222 ltfkaMlml
:)( OO IM+
2
,
2
,,,3
1MlIandmlIwithIII loadOrodOloadOrodOO
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Some may consider the vertical moment arm when writing eq.(5) as
However, since we assumed that the angular displacement is too
small, . Therefore eq.(7) becomes
Which eventually yields eq.(4) or (6).
1cosandsin
)7(....)cos)(sin(cos)( OIaakltf
)8(....))(()( OIaakltf
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A rod OA hinged at O and connected to spring at end A which wasinitially in equilibrium is released abruptly. If the stiffness of the spring is
given as k = 100N/m, as well as the mass and length of the rod are
given as m = 2.0kg and l = 250mm respectively. Determine the
mathematical model of the system and its natural frequency.
O
lA
k
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FBD of rod OA,
Equate to KD,
Therefore, its equation of motion is,
O A
kl
mg
:IM
O
2
3
1where)( mlIIlkl OO
)1(....02 klIO
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Take a look at eq.(1), the weight of the rod is not considered in thissolution as it will cancel out with the static equilibrium equation of the
system.
From eq.(1), the undamped natural frequency of the system is
rad/s25.122
)100(33
m
kn
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For the torsional mechanical system as shown, derive itsmathematical model and determine its frequency of vibration if kt =
100Nm/rad, r = 120mm and m = 3kg. Obtain the displacement if
the initial conditions are .rad/s0)0(andrad05.0)0(
Jkt
r
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FBD:
Applying Newton's 2nd law,
Mathematical model is yielded as
The frequency is represented as
)1(....0 tkJor
J
tk
JT22
kg.m0144.03
1where mrJJkt
rad/s3.49
0144.0
35
J
ktn
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The maximum velocity attained by the mass of a simple harmonicoscillator is 10 cm/s and the period of oscillation is 2 s. If the mass is
released with an initial displacement of 2 cm, find
(a) The amplitude
(b) The initial velocity
(c) The maximum acceleration and
(d) The phase angle
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Three springs and a mass are attached to rigid, weightless barPQ asshown in figure below. Find the natural frequency of vibration of the
system.
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Find the natural frequency of vibration of a spring-mass systemarranged on an inclined plane as shown below.
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For plane motion, we have for
(a) A particle, the potential energy and kinetic energy given,
respectively as
(elastic potential energy, where x=0 in the
equilibrium position)
(gravitational potential energy, where x is
measured from the equilibrium position)
(b) A rigid body, the kinetic energy given as
2
2
1kxVe
mgxVg
2
2
1xmT
22
2
1
2
1ImvT G
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Using the principle of conservation of energy, we write the totalenergy as
and that
constanVT ge VVV where
0VTdt
d
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If the given system is a conservative one, the total kinetic energy ofthe system is zero at the maximum displacement, but is maximum at
the static equilibrium point.
On the other hand, the potential energy is vice versa of the above
statement.Hence, the relationship is given as
This is known as Rayleighs Method. Result of the above equation
will yield the natural frequency of the system.
systemtheofenergytotal).().( maxmax EPEK
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Using the energy method, determine the undamped natural frequencyof the block having m shown in the figure below and also its natural
period of vibration. Assume the block does not slip on the surface of
contact as it oscillates.
m
kx
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Applying energy method, we can write
The equation of motion is
From which
constantVT 222
1
2
1kxxmVT
:0
VTdt
d0
2
2
2
2 xkxxxm
0kxxm
thatand
m
kn s284.6
2
k
mT
n
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Find the natural frequency of the pulley system show in figure belowby neglecting the friction and masses of the pulleys. Solve by using
Rayleighs method.
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Use energy method to find the natural frequency of the
system shown in the figure below.
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The system shown in figure below has a natural frequency
of 5 Hz for the following data: m = 10kg, J0=5kg-m2, r1=10cm,
r2=25cm. When the system is disturbed by giving it an initial
displacement, the amplitude of free vibration is reduced by
80% in 10 cycles. Determine the values ofkand c.
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A boy riding a bicycle can be modeled as a spring-mass-damper
system with an equivalent weight, stiffness, and damping constant of
800N, 50,000N/m and 1,000N-s/m, respectively. The differential setting of
the concrete blocks on the road caused the level surface to decrease
suddenly as indicated in the figure. If the speed of the bicycle is 5m/s(18km/hr), determine the displacement of the boy in the vertical
direction. Assume that the bicycle is free of vertical vibration before
encountering the step change in the vertical displacement.
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A weight of 50N is suspended from a spring of stiffness 4000N/m and is
subjected to a harmonic force of amplitude 60N and frequency of 6Hz.
Find
(a) The extension of the spring due to the suspended weight
(b) The static displacement of the spring due to the maximum applied
force, and
(c) The amplitude of the forced motion of the weight.
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Consider a spring-mass-damper system with k = 4000N/m, m = 10kg,
and c = 40N-s/m.
Find the steady-state and total responses of the system under the
harmonic force F(t) = 200 cos10t N and the initial condition
.0andm1.0 00 xx
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