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1
Georgios H. Vatistas August 2006
Mass-Spring-Damper Mechanical system and its Electrical
Analogue.
There is large number of mechanical devices where MSD
(Mass-Spring-Damper) system constitutes the fundamental building
block for their mathematical modeling. The basic mechanical
components of this system are the spring and the damper. There are
complete courses in the engineering curriculum that deal
systematically with this topic. Some of these courses are: Modeling
Simulation and Analysis of Physical Systems (MECH 370),
Fundamentals of Control Systems (MECH 373), Mechanical Vibrations
(MECH 373), Fluid Power Control (MECH 448), Basic Circuit Analysis
ELEC 273), Principles of Electrical Engineering (ELEC 275), and
several others. Here we will only give a brief introduction to the
subject, always in the context of the first engineering math course
(ENGR 213). The Spring For a linear spring:
F s = k y The spring constant k is the slope of the Fs - y
curve. It has units:
k ! F sy
= Nm
=
kg m
s 2
m =
kg
s 2
Square bracket [] means units of the parameter(s) inside the
bracket.
If the spring is stiff then the value of k is relatively large,
if it soft k is small.
0
y
F s
unstrechedposition
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2
Fs (
N )
y ( m )
k is the slope
The Damper
A damper is a dissipative device that converts mechanical energy
into heat. This device is mathematically modeled by:
F d = c V = c dy
dt, F d always opposes the motion
The units of c: c !
F dV
= Nm / s
= N sm
= kg m
s 2 s 1
m =
kgs
0
y
c
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3
A relatively large value of c indicates hefty opposition to
motion (and high
dissipation), while a small value indicates the opposite. The
Spring-Mass System
0
W = m g
yo
m
no load (unstreched possition)
loaded(static equilibrium)
y
k
Fs (
N )
y ( m )
k is the slope
Fs o
y o 0
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4
When the solid body with mass m is attached the spring will
stretch until Fs = W. At this point the system is in static
equilibrium, or,
forces acting on the body ! = 0 or W - F s o = W - k y o = 0 ! W
= k y o
Starting from this static position let us begin to pull the mass
down with a varying velocity (an acceleration dV dt).
0
W = m g
yo
m
loaded(static equilibrium)
y
k
Fso
Fso
Fs
Fs
no loadpossition
Newtons 3rd law
0
y
positionwithout
the mass
Equilibrium possition(with the
mass attached
________________________________________________________________________
When a solid material is cut for the sake of analysis then the
internal forces that bind the solid together must be included in
the free-body diagram, see for example the solid rod before and
after the cut.
before after
F i F i
+ ve x - direction Note: that when we put the two parts (on the
right) together to obtain the uncut rod, the total force is from
Newtons 2nd law Fi - Fi = 0.
________________________________________________________________________
From Newtons 2nd law: ddt
m V = forces acting external to the body !
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5
d
dt m V = m d V
dt + V d m
dt = W - F s o -F s = W - k y o - k y
Since we are dealing with a solid body, dm/dt = 0. Also, from
above W = k yo. Then, Newtons second law applied to the mass
(free-body-diagram)gives,
m d Vdt
= - k y or m d 2 y
dt 2 = - k y remember that V =
d y
dt
Rearranging:
d 2 y
dt 2 + k
m y = 0 with characteristic equation r 2 + k
m = 0
Alike to all equations, the differential equation must possess
dimensional homogeneity i.e. all terms must have exactly the same
units:
d 2 y
dt 2 !
y
t 2 = m
s 2 i.e. units of acceleration
d indicates an action; take the infinitesimal difference and as
such it has no units. Similarly d 2 has no units.
km
y !
kg
s 2
kg = m
s 2 i.e. again units of acceleration
Hence, the equation it does possess dimensional homogeneity.
The above equation is a second order ordinary linear equation
with constant coefficients, its general solution is:
y t = A cos km
t + B sin km
t
The argument of transcendental functions such as for example the
trigonometric, exponential, logarithmic etc must be dimensionless
(clear number),
km
t !
kg
s 2
kg s = 1 which is a clear number
Let us now consider the case where the mass is pulled down until
y = yin, it is stopped for a while, and then at t = 0 is released.
The initial conditions are then:
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6
i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 Application
of the second initial condition yields,
dy
dt
t = 0 = - A k
msin k
m 0 + B k
m cos k
m 0 = 0 or B = 0
The first initial condition gives that A = yin. = 0.2 m.
Therefore, the position of the top of the mass is then given
by:
y t = 0.2 cos ! t where ! (circular frequency) = k/m in rads /
s
Since the cosine function appears in the solution, the mass will
oscillate. Let us now plot amplitude y as a function of time for
different values of k/m. The results are given in the following
figure.
-0.4
-0.2
0
0.2
0.4
0 2 4 6 8 10 12
1 4 9
y (
t )
in
m
t in s
k / m = 1 4 9
The period of oscillations T is,
T =
2 !"
= 2 ! m/k in s while the frequency is given by:
f = 1
T = 1
2 ! k/m in cycles / s or hz
Frequency f is known as the natural frequency of the system and
will designated henceforth by fn.
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7
It is clear that this system will undulate indefinitely. The
reason behind this
unrealistic behavior is that in the mathematical modeling of
this system the energy dissipation due to friction inside the
spring and air have been neglected. The last paradox (contrary to
what is expected) will be rectified in a subsequent section. Is
also apparent that as the k/m value goes up, the number of
oscillations within the same time interval (or fn) of the system
increases. On one hand, if m is kept constant, increasing the
stiffness of the spring (k) will result into a higher frequency of
oscillations. On the other hand, increasing the mass m while
keeping k constant will result into a lower frequency of the
system.
A practical way to find fn of the system, by only one simple
measurement, is as follows.
The static deflection of the spring-mass system was given
previously by:
W = k y o ! m g = k y o ! g
y o = k
m Then
f n = 1
2 ! k/m = 1
2 ! g/y o in hz
Therefore, the natural frequency of the system cam be obtained
by loading the
mass to the sprig and measure the static deflection yo. The
Mass-Spring-Damper System
As in the previous section, starting from the static equilibrium
position let us now begin to pull the mass down with a varying
velocity (an acceleration dV/ dt). Remember that the damper opposes
the motion. Application of Newtons second law yields,
m d 2 y
dt 2 = - F s - Fd = - k y - c
d y
dt ! m
d 2 y
dt 2 + c
d y
dt + k y = 0
or
d 2 y
dt 2 + c
m d y
dt + k
m y = 0
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8
position
without
the mass0
W = m g
m
y
k
Fs
Fs
c
Fd
Fd
0
y
Equilibrium
position
(with the mass
attached
position
without the
mass
The above equation is a second order, ordinary, homogeneous,
linear equation with constant coefficients. Its characteristic
equation is:
r 2 + c
m r +
k
m = 0 ! r1, 2 =
- c
m c
m
2 - 4 k
m
2 Depending on the values of k, m, and c the roots of the
characteristic equation could be (A) two real and distinct, (B) two
real double, and (C) two complex conjugates. The three cases will
be examined next. CASE A
Let k = 1 N/m, m = 1 kg, and c = 3 Ns / m. Then
r1, 2 = - 3
2 5
2
The general solution is therefore
y t = A exp - 3
2 + 5
2 t + B exp - 3
2 - 5
2 t
At first the arguments of the exponential function appears to
have the units of s ([t]
s). This off course it is not true. In order to simplify the
algebra we have not included the units of r 1,2 which are:
r1, 2 ! -
c
m c
m
2 - 4 k
m
2 !
- c
m c
m
2 - 4 k
m
2
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9
= - 3
2
kg
s kg 3
2
kg
s kg
2 - 4
kg
s 2 kg = - 3
2 1s
32
1s
2 - 4 1
s 2 1s
= - 3
2 5
2 1
s Therefore, the arguments of the exponentials functions are
dimensionless,
exp - 3
2 5
2 1s
t
Like before, let the initial conditions be:
i. t = 0, y = yin = 0.2 m ii. t = 0, Vin = dy/dt = 0 From the
first and second initial conditions yield respectively
1
5 = A + B and A -
3
2 +
5
2 + B -
3
2 -
5
2 = 0
Solution of the previous system of equations gives
A = 3 + 5
10 5
and B = - 3 + 5
10 5
Then
y t = 1
10 5 3 + 5 exp - 3
2 + 5
2 t + - 3 + 5 exp - 3
2 - 5
2 t
Because the velocity V (t) is dy/dt, then
V t = - 1
5 5 exp - 3
2 + 5
2 t - exp - 3
2 - 5
2 t
CASE B
Let us now decrease the value of c to 2 Ns / m while keeping the
rest of the parameters the same. In this case the roots of the
characteristic equation are r1 = r2 = -1/2. The general solution is
thus
y t = A t + B exp - t
2
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10
Imposition of the initial conditions (i) and (ii) yields
A = t
10 and B =
t
5
Therefore the position y ( t ) and the velocity V ( t ) are
given by
y t = 1
5 1 + t
2 exp - t
2 and V t = - t
20 exp - t
2
CASE C
Decrease now c further, say to 1 Ns / m. while keeping the rest
of the parameters constant. In this case the roots of the
characteristic equation are complex conjugate:
r1, 2 = - 1
2 i 3
2
Then
y t = exp - 1
2 t A cos 3
2 t + B sin 3
2 t
The initial conditions yield
A = t
5 and B =
t
5 3
Therefore the position and velocity are given respectively
by
y t = 1
5 exp - t
2 cos 3
2 t + 1
3
sin 3
2 t
and
V t = 2
5 3 exp - t
2 sin 3
2 t
The discriminant of the characteristic equation is,
! = cm
2 - 4 k
m
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11
-0.05
0
0.05
0.1
0.15
0.2
0.25
0 2 4 6 8 10
A y
B y
C y
y (
t )
in
m
t in s
A. OVER DAMPED
B. CRITICALLY DAMPED
C. UNDER DAMPED
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-0.1
-0.05
0
0.05
0 2 4 6 8 10
A V
B V
C V
V (
t )
in
m /
s
t in s
C. UNDER DAMPED
B. CRITICALLY DAMPED
A. OVER DAMPED
For CASE A. greater than zero (c is relatively large) the system
is OVER DAMPED
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CASE B. equal to zero the system is CTITICALLY DAMPED
CASE C. less than zero (c is relatively small) the system is
UNDER DAMPED
The position and velocity for the three above case are plotted
in the figures given
above.
It is evident from thee graphs that as t increases ( t ) the
system tends to static equilibrium state i.e. y ( t ) 0 and V ( t )
0. The reason behind this behavior is that in all the cases
considered here we have accounted in the mathematical model for
dissipation of mechanical energy (damper). The system will only
oscillate (with a diminishing amplitude) when it is under damped.
The analysis considered here constitutes the basic methodology in
the design of car, truck, locomotive, and aircraft suspension
systems. In this case however, the spring instead of being
stretched it is compressed. Nevertheless, the analysis is the
same.
k c
m
0
y
positionwithout
the mass
Equilibrium possition(with the
mass attached
In the case of an aircraft, in addition to the damping of any
bumps on the runway
(providing thus comfort to passengers) it must also absorb the
touchdown impact during landing.
When the car mechanic wishes to find out if the car suspension
needs repairs (in addition to other diagnostic techniques) it
pushes the vehicle down and then lets it go. If the car oscillates
excessively, the mechanic then knows that the damping provided is
not enough (under damped system), the hydraulic damper has lost
liquid (due to leaks) and it must be replaced. Resonance of
Spring-Mass-Damper System
Let us now return to the last problem and add an externally
applied force F ( t ). The equation in this case is,
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13
m d 2 y
dt 2 + c
d y
dt + k y = F t
or
d 2 y
dt 2 + c
m d y
dt + k
m y = 1
m F t = f t
position
without
the mass0
W = m g
m
y
k
Fs
Fs
c
Fd
Fd
0
y
Equilibrium
position
(with the mass
attached
position
without the
mass
Suppose ` f t = A sin ! f t Then
d 2 y
dt 2 + c
m d y
dt + k
m y = A sin ! f t
First simplify the system by removing the damper
d 2 y
dt 2 + k
m y = A sin ! f t
The homogeneous solution is
y h t = B cos km
t + D sin km
t Assume a particular solution of the form:
y p t = Q cos ! f t + M sin ! f t
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14
Then
dy p
dt = - Q ! f sin ! f t + M ! f cos ! f t
and
dy p
2
dt 2 = - Q !f
2 cos ! f t - M !f 2 sin ! f t
Inserting yp and yp into the differential equation we have:
Q km
- !f 2 cos ! f t + M
km
- !f 2 sin ! f t = 0 cos ! f t + A sin ! f t
From the above equation
Q km
- !f 2 = 0 and M k
m - !f
2 = A " Q = 0 and M = A
km
- !f 2
The total solution is then
y t = B cos k
m t + D sin k
m t + A
km
- !f 2
sin ! f t
Application of the following initial conditions: i. t = 0, y = 0
ii. t = 0, Vin = dy/dt = 0 gives:
y t = A
1 - !f 2
sin ! f t - ! f sin t
Remember k = 1 N/m and m = 1 kg. Suppose that the amplitude of
the externally applied force F ( t ) divided by the mass i.e. f(t)
is equal to 0.1 (m/s2), then
y t = 1 10 1 - !f
2 sin ! f t - ! f sin t
The natural circular frequency n is
! n = k
m = 1 rads
s
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15
It is not difficult to see that as f n = 1, y ( t ) .
y t lim! f " ! n = 1
= sin ! n t - ! n sin t 1
10 1 - !f 2
lim! f " ! n = 1
= #
This condition is known as resonance. In reality however the
amplitude of y ( t ) will increase, as f n, but because of friction
(damper) it will never (even theoretically) go to infinity. The
presence of singularities (like in the present situation) is an
indication that an important physical feature (friction in this
case) has not been incorporated into the mathematical
representation of reality.
-1.5
-1
-0.5
0
0.5
1
1.5
0 10 20 30 40 50 60
0.9
0.6
0.3
y (
t )
in
m
t in s
! f
We are now ready to include the damper. The equation is
then,
d 2 y
dt 2 + c
m d y
dt + k
m y = A sin ! f t
Let, like before, k = 1 N/m, m = 1 kg, and c = 1 Ns / m, or
d 2 y
dt 2 +
d y
dt + y = A sin ! f t
The homogeneous solution is
y h t = exp - 1
2 t B cos 3
2 t + D sin 3
2 t
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16
Assume a particular solution of the form:
y p t = Q cos ! f t + M sin ! f t Then
dy p
dt = - Q ! f sin ! f t + M ! f cos ! f t
and
dy p
2
dt 2 = - Q !f
2 cos ! f t - M !f 2 sin ! f t
Inserting yp , yp, and yp into the differential equation we
have: - Q !f
2 cos ! f t - M !f 2 sin ! f t - Q ! f sin ! f t + M ! f cos ! f
t
Q cos ! f t + M sin ! f t = 0 cos ! f t + A sin ! f t From the
above equation
M = A 1 - !f
2
1 - !f 2 2 + ! f
2 and Q = -
A ! f
1 - !f 2 2 + ! f
2
The particular solution is then
y p t =
A
1 - !f 2 2 + ! f
2 1 - !f
2 sin ! f t - ! f cos ! f t
The total solution is then
y t = exp - 1
2 t B cos 3
2 t + D sin 3
2 t
+ A
1 - !f 2 2 + ! f
2 1 - !f
2 sin ! f t - ! f cos ! f t
Application of the initial conditions: i. t = 0, y = 0 ii. t =
0, Vin = dy/dt = 0 yields y t = y h t + y p t
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17
where
y h t = A = 0.1 ! f exp -
1
2 t
1 - !f 2 2 + ! f
2 cos 3
2 t -
1 - 2 !f 2
3 sin 3
2 t
and
y p t =
A = 0.1
1 - !f 2 2 + ! f
2 1 - !f
2 sin ! f t - ! f cos ! f t
Making use of the identity
C 1 cos x + C 2 sin x = C 1 2
+ C 2 2
sin x + arctan C 1
C 2
Since in our case
C 1 = - ! f and C 2 = 1 - !f 2
then the particular solution transforms into:
y p t = ! p sin " f t - arctan " f
1 - "f 2
Similarly
y h t = ! h sin 3
2 t - arctan 3
1 - 2 "f 2
where
! p = A = 0.1
1 - "f 2 2 + " f 2
and ! h = A = 0.1 " f
1
3 1 - 2 "f
2 2+ 1 exp - 1
2 t
1 - "f 2 2 + " f
2
In the following figure the homogeneous and particular solutions
are plotted separately.
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18
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0 5 10 15 20 25 30
B
B
yh (
t )
, y
p (
t )
in (
m )
t in s
yh ( t )
yp ( t )
From the above graphs we conclude the following: (a) Due to the
presence of exp( - 0.5 t ) term, the homogeneous solution will
rapidly
diminish with time to zero*. In this problem, after 20 s the
percentage of the homogeneous solution amplitude of y is less than
0.004 % of the total ( h / x100, = h + p ). Therefore, yh is only
important at the start up phase.
(b) For relatively long time levels, the solution tends to the
particular solution.
Therefore, when the transient effects have died down the system
vibrates harmonically following closely the particular solution. At
this level, the system oscillates with a frequency equal to the
frequency of the forcing function (f).
We have seen previously that when the damping was neglected as f
n the
amplitude of oscillations of the system was tending to infinity.
This however is not true in real situations (when friction is
present). The maximum amplitude at steady state (when the transient
effects are over) for different values of c are given in the
following figure. NOTE: Only the cases c = 0 Ns / m and c = 1 Ns /
m (k = 1 N/m, m = 1 kg) were treated thoroughly here. One however
can produce the rest of the curves following exactly the same
procedure as before but with different c values.
The maximum amplitude of the particular case is obtained by:
dd ! f
A = 0.1
1 - !f 2 2 + ! f 2
= ! f 2 ! f
2 - 1
1 - !f 2 2 + ! f 2
3/2
= 0
or
* Theoretically when t .In reality however, infinity may be
closer than we think. If a property cannot be sensed (detected,
measured) its existence is uncertain. In engineering, t
representing infinity may be taken as value of time, at which the
magnitude of y reaches the limit of certainty of the best available
measuring instrument.
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19
! f 2 ! f 2 - 1 = 0
The root f = 1/2 gives the frequency where the amplitude is
maximum (at 0.115 m).
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 0.5 1 1.5 2 2.5 3
!f ( in rads / s )
" (
in m
)
c = 1
0.3
0.5
0.7
00
" = A = 0.1
1 - !f 2 2 + c 2 ! f
2
Electrical Analog the RLC (resistance-inductance-capacitance)
System (Electrical filter)
The study of physical phenomena by analogy is a common and
useful method of scientific inquiry and technical application that
has been used in many physical areas. Two systems, that may not
necessarily resemble physically each other, are considered to be
analogous, if and only if, both are described by the same set of
equations (including the boundary and initial conditions). Such
systems can be found in dynamics of bodies, electricity, magnetism,
thermodynamics, hydraulics, etc. Recently advances in optics were
triggered by accumulated knowledge in fluid mechanics. Therefore,
this powerful technique can be used to cross-fertilize different
classifications of scientific and technical knowledge.
In order to demonstrate the method we consider here the
electrical circuit, shown bellow, which consists from an electrical
source Vs, and inductor L, a capacitor C, and a resistor R.
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20
L
C
R V RV S i or Vout
Application of Kirchhofs second law: the sum of voltage drop in
each of the
components is equal to the electrical voltage source, we have:
Voltage drop across Governing Equation components
____________________
L: L d i t
dt
R: i t R L
d i t
dt + i t R + 1
C q t = V s t
C: i t R ____________________ Where: C the capacitance (in F (s
A / V , in fundamental units: s4 A2 m-2 Kg -1) i is the current (in
A) L the inductance (in H = V s A-1 , in fundamental units: m2 kg
s-2 A-2) q the electrical charge in the capacitor (Q = A s) R the
resistance (in = V / , in fundamental units: m2 kg s3 A2) t the
time (in s) VS the voltage source (V, in fundamental units: m2 kg
s3 A1) VR The voltage drop across the resistor (V, in fundamental
units: m2 kg s3 A1 )
The current i in the circuit and voltage Vout are given by
i t = d q t
dt and V out = i t R = R
d q t
dt
respectively. Due to the first of the above two identities, the
governing equation transforms into:
L d 2 q t
dt 2 + R
dq t
dt + 1
C q t = V s t in V
Dimensional homogeneity of the equation:
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21
Dividing through the equation by L yields
d 2 q t
dt 2 + R
L
dq t
dt + 1
C L q t = 1
L V
s t
Assuming a harmonic variation for the source we have:
d 2 q t
dt 2 + R
L dq t
dt + 1
C L q t = Aampl sin ! f t in
Vs
" Aampl = Vs
Comparing the mechanical and electrical systems Mechanical
system
d 2 y
dt 2 + c
m d y
dt + k
m y = A sin ! f t
initial conditions:
i. t = 0, y = 0 ii. t = 0, Vin = dy/dt = 0 Electrical system
d 2 q t
dt 2 + R
L dq t
dt + 1
C L q t = A ampl sin ! f t
initial conditions:
i. t = 0, q = 0 ii. t = 0, iin = dq/dt = 0 We see that the two
differ by scaling factors (constants). If now we let:
y ! q
c
m ! R
L
km
! 1 L C
(R/L = 1, 1/LC = 1, and Aampl = 0.1 as in the last example),
then the expression for q(t) is:
q t = ! h sin 3
2 t - arctan 3
1 - 2 "f 2
+ ! p sin " f t - arctan " f
1 - "f 2
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22
where
! p = Aampl = 0.1
1 - "f 2 2 + " f 2
and ! h = Aampl = 0.1 " f
1
3 1 - 2 "f
2 2+ 1 exp - 1
2 t
1 - "f 2 2 + " f
2
Because i(t) = dq(t)/dt, the current through the circuit can be
obtained by
differentiating the above equation with respect to time. i t
=
d q t
dt = ! h
3
2 cos 3
2 t - arctan 3
1 - 2"f 2
+! p " f cos " f t - arctan " f
1 - "f 2
The voltage across the resistor (VR or Vout), can then be
calculated multiplying the resulting equation by R. Vout t = R i t
= ! h
3
2 R cos 3
2 t - arctan 3
1 - 2"f 2
+! p " f R cos " f t - arctan " f
1 - "f 2
The capacitors charge and current through the circuit versus
time are plotted in the subsequent figure. Note that this graph
represents also the velocity of the mechanical system. It is
evident that the magnitude of the current through the circuit
depends on the frequency of the source. As the circular frequency
approaches the value of 1/2, the current amplitude increases. For
high frequencies the amplitude is relatively insignificantly small.
Therefore this circuit is used to filter out frequencies other than
the frequency that will resonate the system. This is the reason
that the RLC circuit is also known as a filter. Filters can be
found in many electrical and electronic systems.
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0 5 10 15 20 25 30 35 40
q i
t , s
q ( t ) i ( t )
q (
t )
, i (
t )
Q
, A
Because of the similarity between the two systems, any of the
developments given before for the mechanical system is also
applicable to the electrical analog.
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23
For example consider the LC circuit shown schematically
below.
L
C
V s
A
B
The governing equation is:
d 2 q t
dt 2 + 1
C L q t = 0
Initially, switch A is closed and B is open. When the capacitor
is charged with 0.2 Q, switch A is opened and then B is closed.
Initial conditions i. t = 0, y = qin = 0.2 Q ii. t = 0, iin = dq/dt
= 0 A Using the similarity transformation relationships
y ! q
c
m ! R
L
km
! 1 L C
The solution is: q t = 0.2 cos ! t In the equivalent mechanical
system is given by:
! (circular frequency) = k/m In the electrical analogue (k/m
1/(LC)
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24
! (circular frequency) = 1/LC in rads / s The natural frequency
for the mechanical system is:
f n = 1
2 ! k/m
The natural frequency for the electrical LC circuit analogue
is,
f n = 1
2 ! 1
L C 1
L C = 1
V s A
s AV
= 1s
Other equivalents systems can be constructed. If the solution is
known in one then the solution in the analogous system can be
obtained through a simple variable transformation.
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25
The Power of Analogies (optional)
Two systems, that may not necessarily resemble physically each
other, are analogous if both are described by similar set of
equations, boundary, and initial conditions. A well-known paradigm
of such systems is the shallow water hydraulics and the
2-dimensional compressible gas flows. This renowned likeness
between the two problems suggests that the free surface profile in
hydraulics corresponds to the density variation in gas dynamics.
The refraction angle depends on the free surface rise and the gas
density variation in liquid and compressible vortices respectively.
Since the radial profiles of liquid elevation and density are
similar, when illuminated, they must project analogous refracted
patterns on the image plane.
Hydraulic Vortices
Kiehn, R. M. THE FALACO SOLITON Cosmic Strings in a Swimming
Pool , Physics Department, University of Houston
Berry & Hajanal 1983,Optica Acta, 30 Sterling et al. Phys.
Fluids, 30 (11), November 1987
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26
Gas Dynamics Vortices
Theoretical
Vatistas, 2006, Transactions of CSME, 30 (1)
Experimental helicopter blade tip vortices
Bagai & Leishman 1993, Exp. Fluids, Vol. 15,, pp. 431 -
442.
Mathematical modeling of these two problems shows that the
differential equations and boundary conditions describing the two
phenomena are similar. Therefore, the physical problems must be
analogous. The argument is confirmed by the experimental results
shown above.