Normal Modes, Wave Motion and the Wave Equation Hilary Term 2011 Lecturer: F Hautmann Part A: Normal modes ( 4 lectures) Part B: Waves ( 8 lectures) printed lecture notes slides will be posted on lecture webpage: http:// www-thphys.physics.ox.ac.uk/people/francescohautmann/cp4p/ suggested problem sheets also on webpage
References Textbooks covering aspects of this course include [1] French: Vibrations and Waves, MIT Introductory Physics Series [2] Coulson and Jeffrey: Waves, Longman
A. Normal modes 1 Systems of linear ordinary differential equations 2 Solution by normal coordinates and normal modes 3 Applications to coupled oscillators
B. Waves Partial differential equations (PDEs). The wave equation. Traveling waves. Stationary waves. Dispersion. Phase and group velocities. Reflection and transmission of waves.
Introduction to Normal Modes Consider a physical system with N degrees of freedom whose dynamics is described by a set of coupled linear ODEs. To determine the normal modes of the system means to find a set of N coordinates (normal coordinates) describing the system which evolve independently like N harmonic oscillators. The frequencies of such harmonic motion are the normal frequencies of the system. normal modes describe collective motion of the system general solution expressible as linear superposition of normal modes
SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS more than 1 unknown function: y 1 (x),y 2 (x),...,y n (x) set of ODEs that couple y 1,...,y n physical applications: systems with more than 1 degree of freedom. dynamics couples differential equations for different variables. Example. System of first-order differential equations: y 1 = F 1 (x,y 1,y 2,...,y n ) y 2 = F 2 (x,y 1,y 2,...,y n ) y n = F n (x,y 1,y 2,...,y n )
Systems of linear ODEs with constant coefficients can be solved by a generalization of the method seen for single ODE: General solution = PI + CF Complementary function CF by solving system of auxiliary equations Particular integral PI from a set of trial functions with functional form as the inhomogeneous terms
Example. Solve d 2 x dt 2 + dy +2x = 2sint+3cost+5e t dt dx dt + d2 y y = 3cost 5sint e t dt2 To find CF Set x = Xe αt, y = Ye αt given x(0) = 2; y(0) = 3 ẋ(0) = 0; ẏ(0) = 4 (α2 +2)X + αy αx + (α 2 1)Y = 0 α4 = 2 α 2 = ± 2 α = ±β, ±iβ (β 2 1/4 ) and Y/X = (α 2 +2)/α so the CF is ( ) x y ( ) ( ) β = X a 2+ e βt β +X 2 b 2+ 2 ) +X c ( iβ 2 2 e βt e iβt +X d ( iβ 2 2 ) e iβt
To Find PI Set (x,y) = (X,Y)e t X Y +2X = 5 X +Y Y = 1 X = 1 Y = 2 ( ) ( ) x 1 = e t y 2 Have 2sint+3cost = Re( 13e i(t+φ) ), where cosφ = 3/ 13, sinφ = 2/ 13. Similarly 3cost 5sint = Re( 34e i(t+ψ) ), where cosψ = 3/ 34, sinψ = 5/ 34 Set (x,y) = Re[(X,Y)e it ] and require X +iy +2X = X +iy = 13e iφ ix Y Y = ix 2Y = iy = 13e iφ +i 34e iψ 34e iψ ix = 2i 13e iφ 34e iψ so x = Re(2 13e i(t+φ) +i 34e i(t+ψ) ) = 2 13(cosφcost sinφsint) 34(sinψcost+cosψsint) = 2[3cost+2sint] 5cost 3sint = cost+sint Similarly y = Re( 13ie i(t+φ) 34e i(t+ψ) ) = 13( sinφcost cosφsint) 34(cosψcost sinψsint) = 2cost 3sint 3cost+5sint = cost+2sint.
4 For the initial-value problem ( ) ( ) ( ) 1 1 2 PI(0) = + = 2 1 3 ( ) ( ) ( ) 2 2 0 CF(0) = = 3 3 0 ; PI(0) = ; CF(0) = ( ) ( ) ( ) 1 1 0 + = 2 2 4 ( ) ( ) ( ) 0 0 0 = 4 4 0 Therefore the solution satisfying the initial data is ( ) x y = ( ) ( ) ( ) 1 1 1 cost+ sint+ e t. 1 2 2
Normal Modes Coupled differential equations - e.g. coupled pendula m m mg mx = mg x l +k(yx) my=mg y l k(yx)
x+g x l k m x+ k m y=0 y+g y l k m y+ k m x =0 Coupled first order linear differential equations Solution I - Matrix method : d 2 dt + g 2 l + k m k m k m x d 2 dt + g 2 l + k y = 0 0 m CF : Try x y =Re X Y eit X,Y (complex) constants 2 + g l + k m k m k m X 2 + g l + k Y = 0 0 m
2 + g l + k m k m k m X 2 + g l + k Y = 0 0 m A. =0 Det[A]=0 Eigenvalue equation 2 + g l + k m k m k m 2 + g l + k m =0 2 + g l + k m 2 k m 2 =0 2 + g l + k m = ± k m
2 + g l + k m = ± k m Eigenvalue equation 1 2 = g l or 2 2 = g l +2 k m Eigenvalues x y = 1 : =Re X Y eit + k m k m k m + k m X 1 Y 1 = 0 0 2 + g l + k m k m k m 2 + g l + k m X 1 =Y 1 = A 1 e i 1 X Y Eigenvector equation = 0 0 Eigenvectors x y = 1 1 A 1cos 1 t + 1 ( ) 1st Normal mode
2 + g l + k m = ± k m Eigenvalue equation 1 2 = g l or 2 2 = g l +2 k m Eigenvalues x y = 2 =Re X Y eit k m k m k m k m X 2 Y 2 = 0 0 2 + g l + k m k m k m 2 + g l + k m X 2 = Y 2 = A 2 e i 2 x y = 1 1 A 2cos 2 t + 2 X Y Eigenvector equation = 0 0 Eigenvectors ( ) 2nd Normal mode
mx = mg x l +k(yx) my=mg y l k(yx) General solution given by a superposition of the two independent (normal mode)solutions: x y = 1 1 A cos t + 1 ( 1 1)+ 1 1 A cos t + 2 2 2 ( ) Normal Modes x+y=2a 1 cos( 1 t + 1 ) xy=2a 2 cos( 2 t + 2 ) } N 1D linear differential Equations N normal modes
Summary method of solution for single ODEs extended to systems of coupled differential equations General solution = PI + CF coupled pendula: 2 linear ODEs in x(t),y(t) 2 normal frequencies ω 1, ω 2 at which system can oscillate as a whole. x+y and x y oscillate independently at frequencies ω 1 and ω 2 (normal modes) any generic motion of the system is linear superposition of normal modes : GS = c 1 NM1 + c 2 NM2