simple harmonic motion (shm) and electrical oscillations · 2016. 11. 8. · simple harmonic motion...
TRANSCRIPT
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Simple Harmonic Motion (SHM) and Electrical Oscillations Dr Andrew French
22
0 022 i t
d x dxx A e
dt dt
-
SHM equation from driven mechanical oscillations
x
m
-29.81msg l
l
0cosmg F t
2dx
m k x ldt
Consider a particle of mass m suspended from a light elastic string from a fixed surface. The string has natural length l. Assume a Hookean law of elasticity i.e. restoring force is proportional to extension. The elastic constant in this case is k. Also assume mass is subject to air resistance which is proportional to velocity and mass m. The mass is also pulled ‘driven’ via an oscillatory force of magnitude F0 and frequency f = /2p
In the absence of any driving force, the mass rests at string extension l. It is assumed at time t = 0 that extension from this equilibrium point, (x), is zero and the mass is at instantaneous rest.
By Newton’s Second Law: 2
02cos 2
d x dxm mg F t m k x l
dt dt
2 f p oscillation frequency
-
SHM equation from driven mechanical oscillations
x
m
-29.81msg l
l
0sinmg F t
2dx
k x ldt
By Newton’s Second Law:
2
02sin 2
d x dxm mg F t m k x l
dt dt
At equilibrium x = 0, F0 = 0
0 mg k l
mgl
k
Hence:
2
02
2
0
2
2 sin
2 sin
d x dxm m kx F t
dt dt
Fd x dx kx t
dt dt m m
This equation has a very similar form to the generic equation of Simple Harmonic Motion (SHM)
22
0 022 sin
d x dxx A t
dt dt
-
2 20( ) cos sintx t Ae t B t
0 01
2 2
0 0
0
sin costan
sin
sin
cos
x x B B
x B
x BA
0
22 2 2 2
0
1
2 2
0
4
2tan
AB
2 2
max 0when 2B
22
0 022 sin
d x dxx A t
dt dt can have oscillatory solutions if 0
The general solution is:
‘Transient solution’ (which exponentially decays) Steady state oscillation at same frequency as driving force
The steady state amplitude exhibits resonance phenomenon i.e. it has a maximum at a particular ‘resonance frequency’
Transient amplitude and phase
-
( ) sinx t B t
0
22 2 2 2
0 4
AB
2 2
max 0when 2B
22
0 022 sin
d x dxx A t
dt dt can have oscillatory solutions if 0
Steady state oscillation at same frequency as driving force
The steady state amplitude exhibits resonance phenomenon i.e. it has a maximum at a particular ‘resonance frequency’
22 2 2 2
max 0
2 2 2
0
2 2 2
0
2 2 2
0
2 2
0
when 4 0
2 ( 2 ) 8 0
2 0
2
2
dB
d
-
2 2
max 0
22 2 2 2
0
22 2 2 2 2 2
0 0 0
4 2 2 4
0
2 2 4
0
0max 2
2 2 2 2
0
0max 2 2 4
0
0max 2 2
0
2
4
2 4 2
4 4 8
4 4
4
4 4
2
AB
AB
AB
22
0 02
0
22 2 2 2
0
1
2 2
0
2 sin
sin
4
2tan
d x dxx A t
dt dt
Ax t
-
SHM parameters for the driven mechanical oscillator
x
m
-29.81msg l
l
0sinmg F t
2dx
k x ldt
22
0 022 sin
d x dxx A t
dt dt
2
0
22 sin
Fd x dx kx t
dt dt m m
10 2
00
kf
m
FA
m
p
2 2
res 0 2 Resonance frequency
Mechanical oscillator
SHM
Natural oscillation frequency
0 02 f p
-
( )
2 2 ( )
0 0
2 2
0 0
0
2 2
0
0
22 2 2 2
0
2 2
0
1
2 2
0
2
2
2
4
arg 2
2tan
i t
i t i t
i
x Be
i Be A e
i Be A
AB
i
AB
i
( )
cos sin
i tx Be
x B t iB t
22
0 02
0
22 2 2 2
0
1
2 2
0
2 cos
cos
4
2tan
d x dxx A t
dt dt
Ax t
22
0 02
0
22 2 2 2
0
1
2 2
0
2 sin
sin
4
2tan
d x dxx A t
dt dt
Ax t
Solving the SHM equation (steady state) using complex variables
22
0 022 i t
d x dxx A e
dt dt
-
Inductor Capacitor
Resistor ‘driving’ AC input V RV
I
0
2
02
2
012
1
1cos
sin
sin
R C L
R
C
L
RL LC
V V V V
V IR
V IdtC
dIV L
dt
dIIR Idt V t L
C dt
dI I d IR L V t
dt C dt
d I dI VI t
dt dt L
Q CV
dQI CV Idt
dt
LV CV
22
0 02
0
22 2 2 2
0
1
2 2
0
2 sin
sin
4
2tan
d x dxx A t
dt dt
Ax t
2 00 0
1, ,
2
VRA
L LC L
Steady state solution to the LCR circuit
Steady state solution to SHM equation
0 cosV V t
Sum of potential differences
EMF – ‘back EMF’ due to induction in the coil
Let current I flow through the circuit. The net EMF V - VL must equal the sum of the potential drops across each electrical component.
-
Inductor Capacitor
Resistor ‘driving’ AC input R
V
I
2
0
2
2 00 0
sin
1, ,
2
Vd I R dI It
dt L dt LC L
VRA
L LC L
0 cosV V t
LV CV
Steady state solution to the LCR circuit .... cont
02 2
2 2
1
2
sin
1
tan1
V LI t
RC
LC LC
RC
LC
2 2
max 0
2
max
2
max 0
22
max 0 0
when 2
1 1
2
1
1 4
I
RCf
LC LC
RCf f
LC
f f f
p
p
10 2
2 2
0
2
1
1
4
f
fLC
RC
LCf
p
p
p
22
0
10 2
4 1f
fp
p
Steady state solution to SHM equation
22
0 02
0
22 2 2 2
0
1
2 2
0
2 sin
sin
4
2tan
d x dxx A t
dt dt
Ax t
-
2
0
2
2 00 0
sin
1, ,
2
Vd I R dI It
dt L dt LC L
VRA
L LC L
22
0 02
00 2
0 0 0
0
22 2 2
1
2
2 sin
2
sin
1
tan1
d x dxx A t
dt dt
Ax z k
xx t
z k z
kz
z
10 2
0
0 00 0 0 02
0
0
0
0 0
22 2 2
1
2
1
22 2
12
2sin
1
tan1
fLC
fz
f
V fV LCI fCV zf CV
L L
R R LC Ck R RC f RC
L L L LC
zf CVI t
z k z
kz
z
p
pp p
p
p
Using dimensionless variables ...
Note 0 0f CV
is the average current when the maximum amount of charge stored in the capacitor is discharged over one complete period at frequency f0
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0 0
22 2 2
2sin
1
zf CVI t
z k z
p
1
2tan
1
kz
z
10 2
0
0
1
2
ff z
LC f
k f RC
p
p
-
Inductor Capacitor
Resistor ‘driving’ AC input V RV
I
Let current I flow through the circuit. The net EMF V - VL must equal the sum of the potential drops across each electrical component.
0
1
1
R C L
R
C
L
i t
V V V V
V IR
V IdtC
dIV L
dt
dIIR Idt V e L
C dt
Q CV
dQI CV Idt
dt
LV CV
Complex impedance
0
i tV V e
0
0 0
arg( )
1
1e
1
arg( )
i t
i t i t
i Z
dIIR Idt L V e
C dt
I R i L V ei C
Z R i Li C
V I Z e
Z
V IZ
0
i tI I e
Inductor Capacitor
Resistor ‘driving’ AC input
LZ i L1
CZi C
RZ R0
i tV V e
VI
Z
L C RZ Z Z Z
Ohm’s Law, generalized for AC
Impedances for L,C,R components
Sum of potential differences
EMF – ‘back EMF’ due to induction in the coil
-
Inductor Capacitor
Resistor ‘driving’ AC input inV outV
LZ i L1
CZi C
RZ R
I
0
i tV V e
out
in
out
in
out
2in
1
1
R
L C R
V Z
V Z Z Z
V R
VR i L
C
V RC
V RC i LC
out
2 2in
2 2
0
out 0
2in 0
out
2in
2
42 1
4
2
2 1
2
2 1
V f
V ff i
f
V xf
V xf i x
V x
V x i x
p
pp
p
p
p
p
p
10 2
0
0
1fx f
f LC
RC f
p
-
Inductor
Capacitor
Resistor
‘driving’ AC input inV outV
LZ i L
1CZ
i C
RZ R
I
0
i tV V e
out
in
out
in
out
2in
1
1
1
1
C
L C R
V Z
V Z Z Z
V i C
VR i L
C
V
V iRC LC
out
2 2in
2 2
0
out
2in 0
out
2in
1
42 1
4
1
2 1
1
2 1
V
V ff i
f
V
V xf i x
V
V x i x
pp
p
p
p
10 2
0
0
1fx f
f LC
RC f
p
-
Inductor
Capacitor Resistor
‘driving’ AC input inV outVLZ i L
1CZ
i C
RZ R
I
0
i tV V e
out
in
out
in
2
out
2in
1
1
L
L C R
V Z
V Z Z Z
V i L
VR i L
C
V i LC
V RC i LC
2 2
2 2
out 0
2 2in
2 2
0
out
2in
4
4
42 1
4
2 1
fi
V f
V ff i
f
V ix
V x i x
p
p
pp
p
p
10 2
0
0
1fx f
f LC
RC f
p
-
No voltage gain here! better to take output voltage across the capacitor or inductor
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LCR circuits can be used as electrical filters. If a signal consists of a superposition of oscillations at different frequencies, an LCR circuit can be tuned to preferentially boost signal components whose frequencies are near the resonance peak of the circuit. This has enormous application in communications or Radar technology, whereby a weak signal of a known frequency (e.g. a local radio station, or indeed the broadcasts from Voyager!) may be buried in electrical noise ‘across the waveband’.
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Designing a mains noise filter In many applications we would like to filter out electrical signals associated with mains AC at 50Hz
10 2
2 2
0
1
1
4
fLC
LCf
p
p
If L = 0.05, therefore C = 2.03 x 10-4 F
-
inV outV
LZ i L1
CZi C
RZ R
I
0
i tV V e
out
in
out
in
out
in
1
1 1
1
1
1
11
R
R
C L
V Z
VZ
Z Z
V R
VR
i Ci L
V
VRC
i RC iLC
0
10 2
0
1
fx
f
fLC
RC
f
p
‘Knotch filter’
RZ RCZ
out
in
out
in
1
12
11 2
RCi RC i
V LCRCV
i RC iLC
i xV x
Vi x
x
p
p
-
A 50Hz knotch filter which could be used to remove ‘mains hum’ from a signal
-
Low pass filter
Band stop filter Band stop filter
High pass filter Band pass filter
Band pass filter
Other types of filter using just R,L,C configurations
-
More complicated L,C,R filter circuits
LZ i L
1CZ
i C
RZ R
Write down equations for the currents in each loop ... Apply V = IZ ... Solve simultaneously!
-
Impedance matching
Note maximum power is transferred from input to output if impedances
if source impedance is fixed. (If it is adjustable, then setting ZS = 0 will maximise the power dissipated in the load). A change in impedance will cause a fraction of the signal to be reflected, which results in a loss of power conveyed to the output
*
L SZ Z
21
2
2 2 2
2 2
22
2
12 2 2
L L L S S S
L rms L L
S S L S S L
S L L L S S
S L L S L S
S L
L
S L S L
Z R iX Z R iX
P I R I R
V I Z Z V I Z Z
Z Z R iX R iX
Z Z R R i X X
V RP
R R X X
PL is maximized when
*
L S
L S
L L S S
L S
R R
X X
R iX R iX
Z Z
See next page!
-
Maximizing the load power in the ‘power dissipation theorem’ is equivalent to maximizing y given constant a in the equation
2
xy
a x
1a
0.5a
0.3a
2, 0
xy x
a x
2
4
4
3
(1) 2
2
0x a
a x x a xdy
dx a x
a x a x xdy
dx a x
dy a x
dx a x
dy
dx
Hence maxima at
2
1
4
x a
ay
aa a
-
Radiated electromagnetic waves
Guglielmo Marconi Wireless transmission pioneer 1874-1937
James Clerk Maxwell 1831-1879
Henrich Hertz 1857-1894
Maxwell developed the experimental work of Faraday and others into a mathematical theory of electric and magnetic fields. It is a vector theory, encapsulated in four equations which are now know as Maxwell’s Equations. One can combine them to form a wave equation in both the electric and magnetic fields. In each case, the velocity of waves (in ‘free space’ i.e. a perfect vacuum) is
0 0
1c
i.e. the speed of light c = 2.998 x 108 ms-1. This is independent of the frame of reference! ..... A big clue that Albert Einstein used to help him develop the theory of Relativity.
-
1 2
2
0
1
4
Q QF
rp
Permittivity & permeability
Coulomb’s Law of force (F) between two charges (Q1,Q2) separated by distance r
Charles-Augustin de Coulomb 1736-1806
12 -1 2
0
-3 48.85418782 10 kg m s A
0
2
0
NIB
l
N AL
l
7 2 2
0
- 10 s A4 kgm p
Magnetic field strength inside a solenoid of N turns and cross section A
Inductance of a coil of N turns and cross section A
Joseph Henry 1797-1878
-
2
2
2 2
2
2
2 2
0 0
1
1
1
c t
c t
c
EE
BB
Wave equations for electric fields E and magnetic fields B
c = 2.998 x 108 ms-1 independent of any coordinate system! So no matter how fast you are moving, electromagnetic waves always propagate at the same speed .....
-
The Electromagnetic Spectrum
c f
f
f
-
Maxwell’s Equations (from Woan, The Cambridge Handbook of Physics Formulas)