University of Cape Town Department of Physics
PHY2014F Vibrations and Waves Part 2 Coupled oscillators Normal modes of continuous systems The wave equation Fourier analysis … covering (more or less) French Chapters 2, 5 & 6
Andy Buffler Department of Physics University of Cape Town
[email protected]
1
Problem-solving and homework Each week you will be given a take-home problem set to complete and hand in for marks ... In addition to this, you need to work through the following problems in French, in you own time, at home. You will not be asked to hand these in for marks. Get help from you friends, the course tutor, lecturer, ... Do not take shortcuts. Mastering these problems is a fundamental aspect of this course. The problems associated with Part 2 are: 2-2, 2-3, 2-4, 2-5, 2-6, 5-2, 5-8, 5-9, 6-1, 6-2, 6-6, 6-7, 6-10, 6-11, 6-14 You might find these tougher: 5-4, 5-5, 5-6, 5-7 2
French page 20
The superposition of periodic motions Two superimposed vibrations of equal frequency
x1 A1 cos(ω0t + φ1 ) =
combination can be written as
x2 A2 cos(ω0t + φ2 ) =
x A cos(ω0t + φ ) =
z1 = A1e j (ω0t +φ1 ) Using complex numbers: z2 = A2 e j (ω0t +φ2 )
{
z= z1 + z2
= ∴ z e j (ω0t +φ1 ) A1 + A2 e j (φ2 −φ1 ) A
β A1
ω0t + φ1
A2
}
Phase difference φ= φ2 − φ1 Then
φ2 − φ1
A2 = A12 + A22 + 2 A1 A2 cos(φ2 − φ1 ) A sin β A2 sin (φ2 − φ1 ) and=
3
Superposed vibrations of slightly different frequency: Beats If we add two sinusoids of slightly different frequency ω1 and ω2 … we observe “beats”… ω1 − ω2 ω1 + ω2 cos ω1t + cos ω2t = 2cos t cos t French 2 2 page 22 cos ω1t cos ω t 2 x1 x2 t x1+x2 t cos ω1t + cos ω2t
Tbeat
2π = ω1 − ω2
ω − ω2 cos 1 t 2 4
Combination of two vibrations at right angles x A1 cos(ω1t + φ1 ) = y A2 cos(ω2t + φ2 ) =
French page 29
???
Consider case where frequencies are equal and let initial phase difference be φ Write
x = A1 cos(ω0t )
y A2 cos(ω0t + φ ) = and
x = A1 cos(ω0t ) y = A2 cos(ω0t )
Case 1 : φ = 0
Case 2 : φ = π 2
A2 y= x A1
Rectilinear motion
x = A1 cos(ω0t ) y= A2 cos(ω0t + π 2) = − A2 sin(ω0t )
x2 y 2 ∴ 2+ 2 = 1 A1 A2
Elliptical path in clockwise direction
5
Combination of two vibrations at right angles …2 Case 3 : φ = π
x = A1 cos(ω0t ) y= A2 cos(ω0t + π ) = − A2 cos(ω0t )
A2 y= − x A1
Case 4 : φ = 3π 2 x = A1 cos(ω0t ) y= A2 cos(ω0t + 3π 2) = + A2 sin(ω0t ) x2 y 2 Elliptical path in ∴ 2+ 2 = −1 A1 A2 anticlockwise direction Case 5 : φ = π 4 x = A1 cos(ω0t ) = y A2 cos(ω0t + π 4) Harder to see … use a graphical approach … 6
Superposition of simple harmonic vibrations at right angles with an initial phase difference of π 4
7
Superposition of two perpendicular simple harmonic motions of the same frequency for various initial phase differences.
8
Abbreviated construction for the superposition of vibrations at right angles … see French page 34. 9
Perpendicular motions with different frequencies: Lissajous figures
See French page 35.
Lissajous figures for ω2 = 2ω1 with various initial phase differences.
φ=
0
π 4
π 2
3π 4
π
10
ω2 : ω1 1:1 Lissajous figures 1:2 1:3 2:3 3:4 3:5 4:5 5:6
φ= 0
π 4 π 2 3π 4
π
11
French page 121
Coupled oscillators
When we observe two weakly coupled identical oscillators A and B, we see: xA t xB t … these functions arise mathematically from the addition of two SHMs of similar frequencies … so what are these two SHMs? These two modes are known as normal modes … which are states of the system in which all parts of the system oscillate with SHM 12 either in phase or in antiphase.
Coupled oscillators A
B
xA t xB t
13
The double mass-spring oscillator xA
k
xB
m
k
m
Individually we know that mxA = − kx A
and mxB = − kxB
k For both oscillators: ω0 = m Now add a weak coupling force: xA
k
kc
m For mass A:
xB
k
m
mxA = − kx A + kc ( xB − x A )
x A = −ω x + Ω ( xB − x A ) or 2 0 A
2
k kc 2 ω = where ,Ω = m m 2 0
14
The double mass-spring oscillator …2 For mass A:
x A = −ω02 x A + Ω 2 ( xB − x A )
For mass B:
xB = −ω02 xB − Ω 2 ( xB − x A )
… two coupled differential equations … how to solve ? d2 2 ( x x ) + = − ω Adding them: 0 ( x A + xB ) A B 2 dt d2 2 2 Subtract B from A: ( x x ) ω ( x x ) 2 ( x A − xB ) − = − − − Ω A B 0 A B 2 dt q= x A + xB 1 Define two new variables:
Then
d 2 q1 2 = − ω 0 q1 2 dt
q= x A − xB 2 and
… called “normal coordinates”
d 2 q2 2 2 = − ( ω + 2 Ω )q2 0 2 dt
15
The double mass-spring oscillator …3 The two equations are now decoupled …
Write
d 2 q1 2 ω = − s q1 2 dt d 2 q2 2 = − ω f q2 2 dt
ωs2 = ω02
s = “slow”
ω 2f = ω02 + 2Ω 2
f = “fast”
= q1 C cos(ωs t + φ1 ) … which have the solutions: = q2 D cos(ω f t + φ2 ) x A + xB and q= x A − xB Since q= 1 2 xA We can write =
1 2
( q1 + q2 )
xB and =
1 2
( q1 − q2 ) 16
The double mass-spring oscillator …4 C D cos(ωs t + φ1 ) + cos(ω f t + φ2 ) xA = 2 2 C D Then cos(ωs t + φ1 ) − cos(ω f t + φ2 ) = xB 2 2 So xA and xB have been expressed as the sum and difference of two SHMs as expected from observation. … C, D, φ1 and φ2 may be determined from the initial conditions. … when xA = xB ,then q2 = 0 … there is no contribution from the fast mode and the two masses move in phase … the coupling spring does not change length and has no effect on the motion … ωs = ω0 … when xA = −xB ,then q1 = 0 … there is no contribution from the slow mode … the coupling spring gives an extra force … each mass k + 2 kc 2 experiences a force − ( k + 2kc ) x giving ωf = m = ω02 + 2Ω 2 17
The double mass-spring oscillator …5 symmetric mode antisymmetric mode mixed mode
18
The double mass-spring oscillator …6
Amplitude
We now have a system with two natural frequencies, and experimentally find two resonances.
Frequency
19
French page 127
Pitch and bounce oscillator
Two normal modes (by inspection):
d k
k m
x A = − xB
xB
Restoring force = −2kx 2k d 2x 2 ωbounce = ∴m 2 = −2kx m dt
L Pitching
xA
Bouncing x A = xB
xA xB
Centre of mass stationary τ = − Iθ kd ( 12 θ d ) = − 121 mL2θ
2 6kd → θ = 2 θ mL
θ
2 ωpitch
6k d 2 = m L2
20
N=2 1π ω1 2= ω0 sin ω0 = 2 ( 2 + 1)
2π = ω2 2= ω0 sin 2 ( 2 + 1)
3ω0
21
N=3
N=4
22
French page 136
N-coupled oscillators
Tension T
l
fixed 1
fixed 2
3
p−1
p
p+1
N
Each bead has mass m … consider transverse displacements that are small.
α p−1
y 1
2
3
p−1
αp
p
p+1
N
−T sin α p−1 + T sin α p Transverse force on pth particle: Fp = y p − y p−1 y p+1 − y p = −T +T l l for small α 23
N-coupled oscillators …2 d 2 yp y p − y p−1 y p+1 − y p −T +T Fp = m 2 = dt l l ∴
d 2 yp dt 2
0 + 2ω02 y p − ω02 ( y p+1 − y p−1 ) = T where ω = ml 2 0
,
p = 1, 2 … N
… a set of N coupled differential equations. Normal mode solutions: y p = Ap sin ωt Substitute to obtain N simultaneous equations
( −ω
or
2
+ 2ω02 ) Ap − ω02 ( Ap+1 + Ap−1 ) = 0
Ap+1 + Ap−1 Ap
=
−ω 2 + 2ω02
ω02
24
N-coupled oscillators …3 From observation of physical systems we expect sinusoidal “shape functions” of the form Ap = C sin pθ Substitute into
Ap+1 + Ap−1 Ap
=
−ω 2 + 2ω02
ω02
And apply boundary conditions A0 = 0 and AN +1 = 0 nπ … find that θ = n = 1, 2, 3, … N (modes) N +1 There are N modes: pnπ ω sin sin sin ωnt = y pn A= t C pn n n N +1 nπ ω ω = 2 sin and 0 n 2 ( N + 1)
25
N-coupled oscillators …4
ωn = 2ω0 sin
nπ 2 ( N + 1)
For small N:
ωn 2ω0
0 1 2 3
N+1
n 26
N-coupled oscillators …5 In many systems of interest N is very large… and we are only interested in the lowest frequency modes.
ωn 2ω0
linear region
0 n
