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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