Wave Phenomena Physics 15c. Lecture 12 Multi-Dimensional Waves

Transcription

1 Wve Phenomen Physics 15c Lecture 12 Multi-Dimensionl Wves

2 Gols For Tody Wves in 2- nd 3-dimensions Extend 1-D wve eqution into 2- nd 3-D Norml mode solutions come esily Plne wves Boundry conditions Rectngulr drum Chldni pltes Sound in rectngulr room

3 1-Dimensionl Wves We ve studied 1-D wves extensively Non-dispersive wve eqution is Norml modes re Dispersion reltion is t 2 ξ(x,t) = Ae i(kx±ωt ) ω = c w k 2 ξ(x,t) 2 2 ξ(x,t) = c w x 2 True even for dispersive wves Non-dispersive cse Try to extend nturlly to 2-D or 3-D It s esy

4 1-D to 2-D Let s extend from 1-D to 2-D Simplest wy is to ignore y We just declre tht ξ (x, y, t) is constnt in y Eqution nd solution remins the sme s 1-D 2 ξ(x,y,t) t 2 = c w 2 2 ξ(x,y,t) x 2 ξ(x,t) ξ(x,y,t) e.g. ξ(x,y,t) = ξ 0 e i(kx ωt ) This isn t the whole story We should be ble to send wves in y-direction too Move this wy

5 Isotropy We my define x-y coordintes s we find convenient Physics should not depend on prticulr direction Tht is, unless the medium hs preferred direction Non-directionl medi re clled isotropic Extend the 1-D wve eqution to mke it isotropic 2 ξ(x,y,t) t ξ(x,y,t) = c w x 2 Extend the solution s well + 2 ξ(x,y,t) y 2 Substitute into the eqution nd we find the dispersion reltion: ξ(x,y,t) = ξ 0 e i(k x x+k y y ωt ) ω 2 = c w 2 (k x 2 + k y 2 ) Miniml extension

6 Wvenumber Vector We cn consider (k x, k y ) s 2-D vector Dispersion reltion determines the length of k k = (k x,k y ) Solution cn be written s ω 2 = c w 2 (k x 2 + k y 2 ) = c w 2 (k k) = c w 2 k 2 ξ(x,t) = ξ 0 e i(k x x+k y y ωt ) = ξ 0 e i(k x ωt ) Depends on the dot-product Points on line perpendiculr to k hve sme vlue of k x As t increses, this line moves k points the direction of wve propgtion k x y k x x

7 Norml Modes We now hve plne-wve solutions trveling in ll directions ξ k (x,t) = e i(k x ωt ) ω = ω(k) = c w k The direction is given by the wvenumber vector k Dispersion reltion determines the length of k For ech ω, there re infinite number of norml modes It s convenient to use k to specify norml mode There is one norml mode for every 2-D vector k Any rbitrry wves cn be expressed by liner combintion of these norml modes To show this, we need to expnd Fourier trnsform

8 Fourier Trnsform Any function f(x,y) cn be expressed s f (x,y) = Suppose t t = 0, the wve hd form f(x, y) Fourier integrl breks it into They trvel s + + So we know the complete solution F(k x,k y )e i(k x x+k y y ) dk x dk y where the Fourier trnsform F(k x,k y ) is given by F(k x,k y ) = f (x,y)e i(kx x+ky y ) dx dy (2π) 2 ξ(x,y,t) = e i(k x x+k y y ωt ) + + e i(k x x+k y y ) F(k x,k y )e i(k x x+k y y ωt ) dk x dk y NB: ω depends on k

9 Rectngulr Membrne Imgine rectngulr drum Elstic film is stretched over rigid frme Consider smll piece Δx Δy Mss is m = ρδxδy It s pulled from 4 edges by tension L y T Δy T Δx T Δx T Δy Proportionl to the length of the edge L x Forces re blnced in the x-y plne Let s mke the film vibrte in z

10 Rectngulr Membrne Vibrtion mkes Viewed from the bottom edge, there is force in the z direction F z = T Δy z(x,y,t) +T Δy x T 2 z(x,y,t) ΔxΔy x 2 Sme with the other edges Totl force on this little piece is F z = T 2 z(x,y,t) x 2 z = z(x,y,t) z(x + Δx,y,t) x ΔxΔy +T 2 z(x,y,t) y 2 ΔxΔy T Δy x x + Δx x

11 Wve Eqution The eqution of motion is ρδxδy 2 z(x,y,t) t 2 = T 2 z(x,y,t) x 2 ΔxΔy +T 2 z(x,y,t) y 2 ΔxΔy 2 z(x,y,t) = T 2 z(x,y,t) t 2 ρ x 2 We lredy know the norml modes + 2 z(x,y,t) y 2 2-D wve eqution! z(x,y,t) = e i(k x ωt ) ω = c w k = T ρ k Remining problem: wht hppens t the edges? The film cn t move there z = 0

12 Stnding Wves Edge of the film is fixed z(0,y,t) = z(l x,y,t) = z(x,0,t) = z(x,l y,t) = 0 Similr to string with fixed ends ξ(0,t) = ξ(l,t) = 0 Solution ws stnding wves ξ(x,t) = sin nπ L x sin nπc w L Guess: 2-D stnding wves t z(x,y,t) = sin(k x x)sin(k y y)e iωt Let s see where this brings us

13 Stnding Wves To stisfy the boundry conditions To stisfy the dispersion reltion Frequencies don t come in nice integer rtios Drums don t hve cler pitch Bottom line: z(x,y,t) = sin(k x x)sin(k y y)e iωt sink x L x = 0 sink y L y = 0 k x L x = nπ k y L y = mπ ω = c w π n2 L x 2 + m2 L y 2 ω = c w k 1-D string hd L x ω = c w nπ L z(x,y,t) = sin nπ mπ x sin y e iωt L y

14 Stnding Wves z(x,y,t) = sin nπ x L x mπ sin y e iωt L y (n,m) = (1,1) Node lines split the film into n m rectngles (n,m) = (3,4) (n,m) = (1,2) (n,m) = (2,2) z = 0 on these lines

15 Stnding Wves vs. Norml Modes Stnding wves nd norml modes don t look relted z(x,y,t) = sin(k x x)sin(k y y)e iωt z(x,y,t) = e i(k x ωt ) They re in fct. Just not esy to see For 1-D wves, 2 norml modes They move in opposite directions hve sme ω Adding them gives stnding wves For 2-D wves, infinite norml modes hve sme ω They move in ll directions in the x-y plne Try dding 4 of them with mixed signs e i(kx±ωt ) e i(k x x+k y y ωt ) e i( k x x+k y y ωt ) e i(k x x k y y ωt ) + e i( k x x k y y ωt ) = 4 sin(k x x)sin(k y y)e iωt

16 Chldni Plte Ernst Chldni ( ) Squre plte is held t the center Vibrtion of the plte = 2-D wves Don t worry bout the wve eqution Norml mode solution is s usul Boundry conditions re No force t the edge 1-D nlog is pipe with both ends open Guess the solution: k x = nπ k y = mπ Tht sounds esy enough z(x,y,t) = cos(k x x)cos(k y y)e iωt z(x,y,t) = cos nπ x mπ y cos e iωt

17 Chldni Plte z(x,y,t) = cos nπ x mπ y cos e iωt Extr condition: fixed in the middle It cn t move, nd cn t hve slope z center = cos nπ 2 z x center = nπ sin nπ 2 cos mπ 2 e iωt = 0 cos mπ 2 e iωt = 0 z = mπ y center cos nπ 2 Obvious solution is cos nπ 2 n nd m must be both odd sin mπ 2 e iωt = 0 = cos mπ 2 = 0

18 Chldni Plte z(x,y,t) = cos nπ x mπ y cos e iωt Think bout ngulr frequency ω Sme for n m If n m we hve two stnding-wve solutions with the sme frequency Liner combintion is lso stnding wves Suppose n nd m re both even Center is not node If we subtrct (m, n) from (n, m), we find cos nπ x mπ y cos mπ x cos This stisfies ll the conditions! nπ y cos ω = c w π Exmple: (2, 4) e iωt n 2 + m 2

19 Chldni Ptterns Both n nd m re even z center = z x center = z y center = cos nπ 2 cos mπ 2 nπ sin nπ 2 mπ cos nπ 2 (2, 4) (4, 2) cos mπ 2 cos nπ 2 e iωt = 0 mπ cos 2 + mπ mπ sin 2 cos nπ 2 sin mπ 2 + nπ cos mπ 2 sin nπ 2 e iωt = 0 e iωt = 0

20 Chldni Ptterns Mke liner combintions from (odd, odd) solutions Both sum nd difference stisfy the constrints (of course) (1, 3) (3, 1) sum 2 solutions for ech (odd, odd) pir 1 solution for ech (even, even) pir All of them hve 4-fold symmetry (= 90 rottion) diff.

21 3-D Plne Wves Esy to extend wve eqution to 3-D 2 ξ t = c 2 2 ξ 2 w x + 2 ξ 2 y + 2 ξ 2 z 2 = c 2 2 ξ = w x, y, z Norml mode is Dispersion reltion This sounds ll fmilir ξ = e i(k x x+k y y +k z z ωt ) = e i(k x ωt ) ω = c w k

22 Isotropy nd Reltivity Wve eqution nd the norml mode solutions for n isotropic medium contin only dot-products of vectors.k.. sclr products 2 ξ t 2 = c 2 ( )ξ w ξ = e i(k x ωt ) They re invrint under rottion of coordinte xes For EM wves in vcuum, is Lorentz sclr k x ωt = k x kct It s invrint under Lorentz trnsformtion Form of equtions re constrined by the symmetry of the physicl system

23 Stnding Wve in Box Imgine rectngulr room with rigid wlls Sound in this room mkes stnding wves ξ(x,y,z,t) = sin(k x x)sin(k y y)sin(k z z)e iωt Boundry conditions k x L x = n x π Dispersion reltion ω = c w π k y L y = n y π n 2 x 2 L x + n 2 y 2 L y + n 2 z 2 L z k z L z = n z π L z L x These frequencies resonte in the room L y Cn we express stnding wves using norml modes? Try it!

24 Concert Hll Acoustics Stnding wves in room is Bd Thing Only prticulr frequencies exist For ech frequency, there re node plnes You cn t her the frequency if you sit on node plne Rel wlls re not completely rigid Stone wlls (e.g. church) come close Sound bsorbers (soft stuff) my be ttched to the wlls Think of them s termintion resistors for sound Complete bsorption mkes the room sound ded You wnt some echo for musicl enjoyment Concert-hll coustics is combintion of rt nd science Look up Wllce Clement Sbine ( ) nd his work on Fogg Lecture Hll

25 Royl Festivl Hll, London

26 1/20 th Acoustic Model

27 Summry Wves in 2- nd 3-dimensions Wve eqution nd norml modes esily extended 2 ξ t = c 2 2 ξ 2 w x + 2 ξ 2 y + 2 ξ 2 z 2 = c 2 2 ξ ξ = e i(k x x+k y y +k z z ωt ) = e i(k x ωt ) w Norml modes re plne wves Isotropy nd (for EM wves) Lorentz invrince Boundry conditions in 2-D nd 3-D Rectngulr drum, Chldni plte, sound in room Nturl extension of the 1-D problems such s string Next: sphericl wves, shock wves