SF2003 1

Oscillations

PY2P10 Professor John McGilp 12 lectures

-damping, forced oscillations, resonance for systems with 1 degree of freedom

(DOF)

-coupled oscillations, modes, normal co-ordinates

-oscillations in systems with many DOF

-transition to a continuous system

-non-linear behaviour

•Syllabus - see https://www.tcd.ie/Physics/undergraduate/freshman_physics/SF_syllabus.php

•Recommended books

Vibrations and Waves by French, Nelson (531.32 L12)

The Physics of Vibration and Waves by Pain, Wiley (531.1 L61)

Vibrations and Waves in Physics by Main, Cambridge (531.1 L8)

•Summary of course – handout, and at https://www.tcd.ie/Physics/local/undergraduate/SF/

SF2003 2

Aims and objectives:

•see similarities in many different physical systems - unitye.g. mechanical, electrical, acoustic, electromagnetic oscillators

•to become familiar with the basic framework and basic

behaviour

•to predict the behaviour quantitatively

SF2003 3

m

k

x

Figure 1.

Mass, m, has acceleration: determined2

2

dt

xdx

by Newton's Second Law: Fs = m where Fs is the spring force.x

If Fs = -kx (negative because acts against x)

Then m = -kxx

m +kx = 0 (1)x

Re-write as + x = 0 where = k/m (2)x 02

02

(angular frequency)

I. Revision: Simple harmonic oscillator (SHO) and simple harmonic motion (SHM)

(Main ch.1,2)

SF2003 4

General real solution:

x Acos( 0t ) (3)

where A is the amplitude and is the initial phase, both determined

by the boundary conditions (often the initial conditions for SHO):

e.g. if the mass is pulled to A1 and released at t = 0, then

0,

0sin)0(

cos)0(

1

0

1

AA

Ax

AAx

Note that ( 0t + ) is called the phase (angle) at time t, the period, , is 2 / 0,

and the frequency, = 1/

Phase:

Discussed in terms of phase lead or phase lag. A phase lead of is equivalent to

a phase lag of . A phase lead of > is called a phase lag of (2 ).

| | = antiphase | | = quadrature

SF2003 5

The phase of the velocity leads the displacement by . The phase of the acceleration leads the

displacement by for SHM [show this!].

Representations:

(a) rotating vector - SHM represented as projection of rotating vector of length A

x Acos( 0t )x

A

x

( 0t+ )

Figure 2.

(b) complex (Argand) plane representation

i axis

real axis, x

A

( 0t+ )Figure 3.SHM )cos(]Re[

diagram vector thereproduces }){exp(

0

0

tAzx

ztiAz

(see Appendix A for maths)

(8)

SF2003 6

Energy

For the mass and spring, the kinetic energy T, the potential energy V and the total energy,

E, are given by:

(9)

constant

2

212

212

21

VTE

kxFdxVxmmvT

During oscillation, the energy is continually exchanged between the two components.

For oscillatory motion, we require an inertial component, capable of storing

kinetic energy, and an elastic component, capable of storing potential energy.

SF2003 7

Analogies (see handout)

LC circuit

L C

charge q, current I

Kirchoff's Law: VL + VC = 0

1200

111

)( and )cos(

)( 0)(00

LCtAq

qIqLCqqCqLqCIL

SF2003 8

SF2003 9

SF2003 13

Figure 5. Damped oscillations

Q=50 Q=16.7

Q=5

SF2003 15

As the time increases, the first term decays away (|p1|>|p2|), leaving

)exp()(

212

11 tppp

Apx

Heavily damped motion is aperiodic, but note that the system may be forced into

periodic motion - see later.

Figure 6 shows damped, aperiodic motion.

Q=0.25

Figure 6.

SF2003 16

(c) critical damping = 2 0

021

21 ppp

- from (17), but tricky. A system returns to zero fastest if critically damped.

Critical damping is very important in the design of

equipment containing parts capable of vibrating.

Q=0.5

Figure 7. Critical damping

(18) )exp()1( 001 ttAx

SF2003 18

Summary (1)

(8) })]{exp(Re[ 0tiAx

•Motion represented by a rotating vector in the complex (Argand) plane: (8) is often written

with only implicit understanding that the real part is taken:

•SHM: (2) )( ere wh0 20

20 mkxx ( is the angular frequency, the

natural frequency of the system)0

(10) the,/ where020 widthmbxxx •Damped SHM:

-heavy damping > 2 0 -critical damping = 2 0

-light damping < 2 0 (13) )}{exp()exp()}{exp(21

21 titAtitAx ff

amplitude_________

(14) 1)2/(for ])2/(1[ 02

021

0f

•Heavily damped motion is aperiodic, but note that the system may be forced into periodic motion.

(20) / where(21) 2 0QQ•The fractional energy decay per cycle is

SF2003 23

Figure 10. Displacment amplitude: light damping

SF2003 24

Figure 11. Displacement

amplitude: heavy damping

Figure 12. Phase lag: light

and heavy damping

Resonance:

phase lag 90o

SF2003 27

Figure 14. Dispersive and absorptive

components

Figure 13.

SF2003 29

Summary (2)

(22) )]Re[exp()/(cos)/( 0020 timFtmFxxx •Forced oscillations:

(24a) ])()[(

)/()(

22220

0 mFA (24b)

)()(tan

220

•Displacement amplitude and phase:

•Response function: (26) )()(

)()(

22220

2

.)(

)(.cos

2

2200

m

FAAdis•Dispersive amplitude:

(31) .1

.sin 0

m

FAAabs•Absorptive amplitude:

(33) 2

re whe)(. 2

000

m

FPPP•Power absorbed:

SF2003 31

Curve shape of response function and power absorption near – the Lorentzian

2)(but important, is )( term then the~ if ,For 0000

)(2))(( Thus 00022

0

220

2

2

22220

2

)()(4

)(

)()(

)()( So

(34) )()2/()(

)2/()(

220

2

L

L( ) is a Lorentzian: -symmetric about = 0

Figure 15. Lorentzian curve

-maximum at = 0 with P = P 0

-at P = P0/2, = 0 ( /2)

(35) 21

This is known as the full width at half maximum

height, FWHM, and is widely used in spectroscopy to

quantify peak shapes.

SF2003 33

Example: IR absorption spectrum of methane: molecular vibrations.

IR beammethane

IR detector

(wavelength =1/wavenumber: 1/ 3000 cm-1 = 3.33x10-4 cm = 3.33 m = wavelength)

SF2003 34

Example: Electron paramagnetic resonance (EPR) absorption

SF2003 36

Driven oscillators: unity in physics

Mechanical Electrical LCR

m L

b R

k 1/C

F0 V0

mb / LR /

mk /20 LC/12

0

/0Q C

L

RQ

1

20

0max

1

m

QFx 02

0

0max

1QVV

L

QVq C

0

0max

1

m

QFx 0

0

0

0max

1VV

R

V

L

QVq R

m

QFx 0

max 00

max QVVL

QVq L

SF2003 37

Non-sinusoidal driving forces

•In general, driving forces may be non-sinusoidal and the system response can then be

determined using Fourier analysis

•Example: saw tooth waveform

f(t)

t 0 -

)0( 2/)(

)0( 2/)(

tttf

tttf

Even function Fourier series:

1

20 cos)(

nTn

n taatf

0

22 cos)( and dtttfaTn

n

0

10 )(with dttfa

(no offset)00a For , integrate by parts to obtain na . . . 5 3, ,1 /4 2 nnan

SF2003 38

.} . . . 5cos3cos{cos)(251

914 ttttfFourier expansion of saw tooth:

(22) becomes:

. . . 5 3, ,1 cos)/4( 220 nntnxxx

. . . 531 xxxx

ntBntAx nnn sincos and

-giving a set of equations with the principle of superposition giving the total

displacement

Substituting , and equating coefficients for the sine and cosine terms gives us:nx

])()[(

)(4 and

)( 22220

220

2220 nn

n

nA

n

nAB n

nn

Letting , we finally obtain:2222

0 )()( nnD

4

and )(4

2

220

nDB

Dn

nA nn

SF2003 39

•for this lightly damped system ( ), the harmonic dominates250Q 5n

-a saw tooth driving force of frequency produces a dominant (lightly-damped) system

response at (the system selects the Fourier component nearest to its natural

frequency)0

•if we increase the damping, the motion becomes much more complicated.

•we take (Kreyszig, 4th edition, p.495) and use an Excel

spreadsheet to find the Fourier coefficients and the driven waveform

02.0 and 50

•new physical insight:

-a sinusoidal driving force of frequency forces a system response at

SF2003 40

Summary (3)

•Lorentzian lineshape near resonance: (34) )2/()(

)2/()(

220

2

L

(35) 21•FWHM spectroscopic width:

•Lifetime: (36) 1

21

free •Larger Q means sharper resonance: (37)

21

00Q

•Q is also a measure of the size of the resonance: (38) 0

max

A

AQ

•New physical insight on system response to different types of driving force:

-a sinusoidal driving force of frequency forces a system response at

-a saw tooth driving force of frequency produces a dominant (lightly-damped) system

response at (the Fourier component nearest to the natural frequency is favoured

in a lightly-damped system)0

SF2003 41

Appendix A: Mathematics

Useful basic relations

de Moivre's theorem: sincos iei

BABABA sinsincoscos)cos(

BABABA sincoscossin)sin(

xpptCpdt

xd

pxptpCdt

dx

ptCx

22

2

2

)exp(

)exp(

)exp(

SF2003 42

Alternative mathematics for the SHO

(4) )sin( and )cos(

where)sin()cos(

)sin()sin()cos()cos(

(3) )cos(

00

00

0

ABAB

tBtB

tAtA

tAx

qp

qp

We can also start from the trial solution of the differential equation, but this has 4 arbitrary

constants. We use the fact that x is real to reduce the number to 2.

(5) )exp(*)exp(

*

real] *[ )exp()]*exp([]Re[

)exp()exp(

)exp(

00

00

00

020

2

tiCtiCx

CC

zztiCtiCx

tiCtiCx

ippptCx

Using de Moivre's theorem :)sincos( iei

2Re[C] = Bp = Acos( ) 2Im[C] = -Bq = Asin( ) (6)

SF2003 43

We can express these relations in another way which is often more convenient:

(8) })]{exp(Re[ }){exp()exp(

)exp()sin()cos(

(7) )sin(]Im[

)cos(]Re[

2 where)]exp(Re[)]exp(Re[2

*]Re[]Re[

but , realfor )]exp(*Re[)]exp(Re[

000

00

00

tiAxtiAtiD

iAiAAD

AD

AD

CDtiDtiCx

zz

xtiCtiCx

This is a rotating vector in the complex (Argand) plane. You will find (8) written with only

implicit understanding that the real part is taken, i.e. with 'Re' missing.

SF2003 44

)2/(tan

sincos

)exp( where0)()0(

(13) from )}exp({)(

(8) from cos)0(

21

21

21

21

1

f

f

f

ff

AA

iADiDx

tiiDx

AAx

Initial conditions: e.g. pull out mass to distance A 1, then release at t = 0:

The initial amplitude is given by:

(15)

10

21

211

)/(

)2/(1tan1sec

AA

AAAA

f

f

SF2003 55

•CO2 animation: http://science.widener.edu/svb/ftir/ir_co2.html

•CO2 IR transmission spectrum

•IR characteristic frequencies of chemical functional groups

SF2003 56

Example: Vibrational modes of carbon monoxide (CO) adsorbed on a ruthenium (Ru) surface.

SF2003 58

•If K is very small, frequency behaviour is like two simple oscillators, but the nature of the

vibrations change dramatically with even a small amount of coupling - spectacular interchange of

energy between the two oscillators occurs.

k 2k K

0

0

1

2

Figure 20.

•the increased separation of mode frequencies with increased coupling strength is typical behaviour

•for this simple system, the lower mode frequency is independent of K.

SF2003 61

2221 amplitude)( pendulum, aFor mE

)(sin2)(

)(cos2)(

mod22

av22

mod2av2

1

mod22

av22

mod2av2

1

tmABmE

tmAAmE

B

A

(55) constant 2 2av

2mAEE BA

•complete energy exchange occurs only if m1 = m2

i.e. oscillators are identical

•(2k/m) term will change where m1 ≠ m2 ≠ m

Amod(t)

-1

0

1

xA

Bmod(t)

-1

0

1

xB

EA(t) EB(t)

0

1E

ner

gy

timeFigure 22.

Energy oscillates back and forth between

pendulums at beat frequency ( 1- 2)

SF2003 65

Example: Some vibrational modes of nickel carbonyl, Ni(CO)4

oxygen wag

[ http://newton.ex.ac.uk/research/semiconductors/theory/jones/projects/oxygen/lvms.html]

Ni-CO stretch

SF2003 66

Summary (4)

•The equations of motion are expressed in normal coordinates, and each normal mode has a distinct

frequency

•Three methods of finding the modes and thus solving the system: physical reasoning,

direct uncoupling of equations, or determinant solution of coupled linear equations:

•With coupled oscillators, the general motion is complicated and is a superposition of SHMs

called normal modes or, simply, modes

•Oscillators are coupled and so exchange energy: modes are independent and do not exchange energy

(48) 1211 BAA xaxax

(49) 2221 BAB xaxax (50) )(

4

)(

2

)( 21122211

2221122112 aaaa

aaaa

•Increased separation of mode frequencies with increased coupling strength is typical behaviour

•Forced coupled oscillators have resonance frequencies at the mode frequencies, and not at the

natural frequency of the uncoupled oscillator

•Normal mode analysis is often the basis of understanding spectroscopic measurements

SF2003 72

Example: N = 4 and n = 2

•as in a standing wave, we have:

(64) from )1(22

n

aN

n

Ln (74)r wavevecto theis where

2n

nn kk

•number of modes: 2 DOF 2 independent modes; N DOF N independent modes

•beyond n = N, equations do not describe any physically new situations

(72) 5

2sin

1sin 2 pC

N

pnCA npn

z

y

(73) )cos( 2222 tAy pp

(71) from 5

sin2 02

•resonances will occur when the system is driven at the mode frequencies

Figure 27.

SF2003 75

Remembering that the wavevector (or wavenumber for a one-dimensional medium) is given by:

kn

2

n

n

L we have An (z) Cn sin(knz)

The displacement in the nth mode is the equation for standing waves:

(76) )cos()sin(),( nnnnn tzkCtzy

-1

0

1

0

n=3

n=2

n=1

An

(z)

LFigure 28.

SF2003 77

Dispersion in waves

-for simple waves this is often just called the velocity

•the velocity of points of constant phase in a wave is called the phase velocity, v

-if is constant non-dispersive waves)/( k

-if dispersive waves and)()/( kfk )(fv

•we can compare the mode frequency for the dispersive beaded string (77) with that of the

non-dispersive continuous string (78)

)(k

k)( a

vkma

Tkaω 0

)(at 22 00 ak

ma

T

ma

T0

(Solid state JS3007)Figure 29.

SF2003 78

Summary (5)

•Understanding systems of many coupled oscillators allows us to deal with real systems such as

polyatomic molecules and solids (Note: SF Mathematica Laboratory Oscillator experiment)

•For N beads on an elastic string, there are N modes and the nth mode frequency is

(71) e wher}1{2

sin2 000

ma

Tω

N

nn

•Amplitude of pth particle in nth mode is (72) 1

sinN

pnCA npn

(73) )cos( nnpnpn tAy•Displacement of pth particle in nth mode is:

•In the continuous limit, we find that the modes evolve into standing waves of the system:

(76) )cos()sin(),( nnnnn tzkCtzy

•Mode frequencies have an upper limit in the discrete system in contrast to the elastic continuum,

and waves in the discrete system disperse:

(77) )sin(2 max210 akak

ma

Tnn (78) : waveelastic 0 vk

ρ

Tkω

SF2003 79

V. Non-linear response

•So far, we have always taken the restoring force to depend linearly on the displacement, .xxF )(

•This works for many systems where are small. For larger values, significant deviations

from linear behaviour may appear in real systems.

xF ,

•Vibrations of non-linear systems are anharmonic and may be chaotic.

V.1 Asymmetric return force

This is the lowest order departure from linearity:

)()( 2xxkxFk

(79) cos)/()1( 020 tmFxxx

The restoring force is now quadratic in x, and is asymmetric i.e.

)(][])()[()( 22 xFxxkxxkxF kk

SF2003 80

The figure shows the force with negative , where the spring gets stiffer on compression.

We still have oscillations, and thus consider a harmonic series:

n

fn tnAx )cos(

displacement

restoring force

Figure 30.

SF2003 81

It is useful to remember that we are considering small departures from linearity and re-write the

expansion as:

(80) ....)2cos(cos0 ttAAx ff

and remembering that and (and coefficients of higher terms) are small. Substituting this

into (79), and neglecting terms in (and higher terms) we obtain, by

equating coefficients of and :

AA and )( , 22

tfcos tf2cos

(81)

thusand

)21(

0

221

0

61

020

2

f

f

AA

A

A

SF2003 82

Note:

•the behaviour of the system now depends on the amplitude of vibration:

bigger amplitude more non-linearity

•higher harmonics appear in the vibration - anharmonicity

•the first harmonic (or fundamental) term still dominates - the amplitude of the second harmonic

is much smaller, as expected

•the average position of the mass changes: 221

0 AAx

•thermal expansion of solids - atoms vibrate harmonically (experience a parabolic potential)

to a first approximation, but expansion of the solid on heating occurs due to the anharmonic

term where the average position of the atoms changes (thus expansion coefficients are small)

Examples

•optical frequency doubling using lasers - lasers can produce very intense electromagnetic

fields where the induced electric polarisation becomes non-linear:

linear optics (lamp: continuous, broadband) 1 W, 3eV, 2 mm spot 3x105 W/m2

nonlinear optics(10 nJ, 130 fs pulses) 1 W, 3eV, 20 m spot 3x1014 W/m2

SF2003 83

....}coscos{...)2,( 220

)2(0

)1(0 tEtEP

)}2cos1({}cos{)2( 20

)2(

21

022

0)2(

0 tEtEP

Interaction of electromagnetic radiation with matter induces a polarisation:

The polarisation generates a radiated electromagnetic wave of intensity 2~ PI

Linear reflection, transmission and absorption is described by the first term:

lity susceptibilinear theis re whe}cos{)( )1(0

)1(0 tEP

(more common to use the permittivity, or dielectric function, ))1(1

Nonlinear reflection is described by the second term:

-frequency doubling: by using clever physics (phase matching in special crystals),

30% efficiency is easily obtained (i.e. 1/3rd of the laser beam intensity is at 2 )

SF2003 84

V.2 Symmetric return force

•The return force has a cubic dependence on x:

)()( 3xxkxFk

(82) cos)/()1( 022

0 tmFxxx

•The restoring force is now symmetric i.e. )(][])()[()( 33 xFxxkxxkxF kk

The figure shows a soft system, with . 0

displacement

restoring force

Figure 31.

SF2003 85

•We can use the same approach as before:

(83) ....)3cos(cos ttAx ff

where and (and coefficients of higher terms) are small. Note that because we have a

symmetric return force, and even harmonics must be zero. Substituting (83) into (82), and

neglecting higher order terms we find: 0A

)1( 2432

02 Af

(84) 2321 A

Note:

•as before, the behaviour depends on the amplitude, anharmonic vibration occurs, and the

first harmonic dominates

•the average position of the mass does not change

•the frequency of vibration varies with amplitude

SF2003 86

Example: the simple pendulum, where the return force has a dependence and

leading to a soft non-linear response:

xsin

....)!3/(sin 3xxx

x

mg

l

mgsin x

Figure 32.

mglT

x21

21

small angles

angles) (small sin quereturn tor mglxxmglT

2 inertia ofmoment mlI

xl

gxT

dt

xdI sin

2

2

1

Tutorial Q.1

An object, of mass 2 kg, hangs from a spring of negligible mass. The spring extends by 2.5

cm when the object is attached. The top of the spring is then oscillated up and down in

SHM with an amplitude of 1 mm. If Q = 15 for the system, find

(a)

(b) the amplitude of the forced oscillation at .0

0

PY2P10 Oscillations Tutorial (JMcG)

Tutorial Q.2

Show that the FWHM of the resonance power absorption curve defines the phase angle

range .1tan

Tutorial Q.3

Show that the frequencies of the normal modes of oscillation in the

vertical direction in the figure are given by

[Hint: the gravitational force can be ignored because it does not

contribute to the restoring force]

k

k

m

m

mk 2/)53(2 A

B

Ax

Bx

2

Tutorial Q.4

Two identical simple pendulums, each of mass 0.25 kg, are connected by a light spring of

stiffness 1.00 Nm-1. By clamping one pendulum, the period of the other can be measured

and is found to be 2.00 s. Find the periods of the two modes of the system when the clamp

is removed.

Tutorial Q.5

k kk

m M

x y

2221

1211yaxayyaxax

0))(( 12212

222

11 aaaa

The system shown above has general equations of motion of the form

which have solutions satisfying

ija kDetermine values for in terms of the masses, m and M, and the spring constant, .

Hence find expressions for the two mode frequencies. Show that your expression for the mode frequencies

reduces to when , and explain what happens as /3or /2 mkmk Mm M

3

Three beads of mass m are spaced at intervals of length a on an elastic string stretched

between two supports with tension T. The frequencies of the normal modes of vibration of

this system are given by

where n is the normal mode number, N is the number of beads and .

(a) Write down the normal mode frequencies, in rad s-1, if .

(b) Find the wave numbers of the normal modes, , in terms of a.

)1(2sin2 0

N

nn

maT /0

1-0 s rad 12

nk

Tutorial Q.6