Waves & Oscillations

Transcription

1 Physics Waves & Oscillations Lecture 22 Review Spring 2013 Semester Matthew Jones

2 Midterm Exam: Date: Wednesday, March 6 th Time: 8:00 10:00 pm Room: PHYS 203 Material: French, chapters 1-8

3 Review 1. Simple harmonic motion (one degree of freedom) mass/spring, pendulum, water in pipes, RLC circuits damped harmonic motion 2. Forced harmonic oscillators amplitude/phase of steady state oscillations transient phenomena 3. Coupled harmonic oscillators masses/springs, coupled pendula, RLC circuits forced oscillations 4. Uniformly distributed discrete systems masses on string fixed at both ends lots of masses/springs

4 Review 5. Continuously distributed systems (standing waves) string fixed at both ends sound waves in pipes (open end/closed end) transmission lines Fourier analysis 6. Progressive waves in continuous systems dispersion, phase velocity/group velocity reflection/transmission coefficients 7. Waves in two and three dimensions Laplacian operator Rotationally symmetric solutions in 2d and 3d

5 Coupled Discrete Systems The general method of calculating eigenvalues will always work, but for simple systems you should be able to decouple the equations by a change of variables. l l l + = 0 + l = = = 0 = /l, = / 3 4 = + = + = 0 + = 0

6 Forced Oscillations We mainly considered the qualitative aspects We did not analyze the behavior when damping forces were significant Main features: Resonance occurs at each normal mode frequency Phase difference is = 2at resonance Example: driven by the force = Calculate force term applied to normal coordinates = = cos Reduced to two one-dimensional forced oscillators: + = /cos + = /cos

7 Uniformly Distributed Discrete Systems Equations of motion for masses in the middle:! + 2!!" +!# = 0 = l $% + 2 $ % $ %# + $ %" = 0 = & l

8 Uniformly Distributed Discrete Masses Proposed solution: % = ' % cos ' %" + ' %# ' % = + 2 We solved this to determine ' % and ( : - ' %,( = *sin. + 1 ( = 2 sin General solution: 2 % = 01 ( sin ( cos ( (

9 Vibrations of Continuous Systems Amplitude of mass - for normal mode : - ' %,( = *sin. + 1 Frequency of normal mode : ( = 2 sin Solution for normal modes: % = ' %,( cos ( General solution: 2 % = 01 ( sin ( cos ( (

10 Masses on a String First normal mode Second normal mode

11 Lumped LC Circuit < < < ; ; ; ; 6 %" () 6 % () 6 %# () 4 56 % 5 1 * 7 6 % 6 %# 5 1 * 7 6 % 6 %" 5 = % % (6 %" + 6 %# ) = 0 This is the exact same problem as the previous two examples.

12 Forced Coupled Oscillators Qualitative features are the same: Motion can be decoupled into a set of. independent oscillator equations (normal modes) Amplitude of normal mode oscillations are large when driven with the frequency of the normal mode Phase difference approaches /2 at resonance You should be able to anticipate the qualitative behavior when coupled oscillators are driven by a periodic force.

13 Continuous Distributions Limit as. and = $ Boundary conditions specified at = 0 and = 4: Fixed ends: $ 0 = $ 4 = 0 Maximal motion at ends: $B 0 = $B 4 = 0 Mixed boundary conditions Normal modes will be of the form $ %, = ' % sin( % )cos( % % ) or $ %, = ' % cos( % )cos( % % )

14 Properties of the Solutions $ 4, ~sin % 4 = 0 % 4 = - E % = 24 - % = -A 4 F % = -A 24

15 Examples: Boundary Conditions String fixed at both ends: $ 0 = $ 4 = 0 Organ pipe open at one end: $B 0 = $B 4 = 0 Driving end has maximal pressure amplitude Organ pipe closed at one end: $B 0 = 0, $ 4 = 0 Transmission line open at one end: 6 4 = 0 Transmission line shorted at one end: A 4 H! I = 0 HJ

16 Fourier Analysis Normal modes satisfying $ 0 = $ 4 = 0: $ %, = ' % sin - cos 4 % % General solution: Initial conditions: L $, = 0 ' % sin - 4 L %3 $,0 = 0 ' % sin - 4 %3 L $B,0 = 0 ' % % sin - 4 %3 cos % sin % cos % % L = 0 M % sin - 4 %3 L = 0 * % sin - 4 %3

17 Fourier Analysis Fourier sine transform: L N = 0 M % sin - 4 %3 I M % = 2-7 N()sin 4 4 Fourier cosine transform: L A = 0 M % cos - 4 M % = 2 4 %3 I - 7 A()cos 4 5 5

18 Fourier Analysis M % = ' % cos % * % = ' % % sin % Solve for amplitudes: Solve for phase: ' % = M % + * % % tan % = * % M % %

19 Fourier Analysis Suggestion: don t simply rely on these formulas use your knowledge of the boundary conditions and initial conditions. Example: If you are given $B,0 = 0 and $ 0 = $ 4 = 0 then you know that solutions are of the form $, = 0' % sin - cos 4 % If you are given $B,0 = 0 and $B 0 = 0,$ 4 = 0 then solutions are of the form $, = 0 ' % cos - 4 QHH % cos %

20 Progressive Waves Far from the boundaries, other descriptions are more transparent: $, = F ± A The Fourier transform gives the frequency components: L ' = 1 7 ()cos 5 2 "L L M = 1 7 ()sin 5 2 L "L L 7 M sin( )5 "L () = 1 7 ' cos( ) "L 2 Narrow pulse in space wide range of frequencies Pulse spread out in space narrow range of frequencies

21 Properties of Progressive Waves Power carried by a wave: String with tension & and mass per unit length? S = 1 2? ' A = 1 2 T ' Impedance of the medium: T =?A = &/A Important properties: Impedance is a property of the medium, not the wave Energy and power are proportional to the square of the amplitude

22 Reflections Wave energy is reflected by discontinuities in the impedance of a system Reflection and transmission coefficients: The wave is incident and reflected in medium 1 The wave is transmitted into medium 2 U = V W V X V W + V X Wave amplitudes: Y = XV W V W + V X ' Z = ['! ' J = \'!

23 Reflected and Transmitted Power Power is proportional to the square of the amplitude. Reflected power: S Z = [ S! Transmitted power: S J = \ S! You should be able to demonstrate that energy is conserved: ie, show that ]^ = ] _ + ]`

24 Dispersion Wave speed is sometimes a function of frequency. Phase velocity: A = EF = a (constant) ( Group velocity: A b = Ha (function of frequency) H( Energy that is carried by a pulse propagates with the group velocity In optics, A = /-( ) and A b = A (evaluated at the average wavenumber of the pulse)

25 Waves in Two and Three Dimensions Wave equation: c d = 1 When the function only depends on the radius, 5e = 0) then this can be written: c d = f@d = Polar coordinates (2d) c d = fd = d Spherical coordinates (3d)

26 Waves in Two Dimensions Wave equation in polar coordinates: c d = f@f f@d = Bessel s d Let g = f where = A d + d(g) = 0 Solutions: h (g)~ i jkl m"i/n m and o (g)~ i lpq m"i/n m

27 Waves in Three Dimensions Wave equation in spherical coordinates: c d = fd = d When rs t = rj s ω d this is ω fd + A d = 0 Solutions are of the form: d f, = ' v!(z f cos

28 Boundary Conditions in Two and Three Dimensions When a boundary condition imposes the restriction that d w, = 0 then the function must have a node at f = w. Analogous to the 1-dimensional case: This imposes the requirement that w is a root of the equation F w = 0 which implies that % = a x y = g %/w where g % are roots of F g = 0.

29 That s all for now Study these topics make sure you understand the examples and assignment questions. Send if you would like specific examples discussed before the exam next Wednesday. Next topics: waves applied to optics.