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

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

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’.

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)