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S3/ExPhyIII/VL/ExIIIVL2.typ
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235
S3/ExPhyIII/VL/ExIIIVL2.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 2,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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Wenn man N Oszillatoren koppelt, erhlaet man ein System mit wieviel Eigenfrequenzen?
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gekoppelte Oszillatoren und Eigenmoden
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$
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psi_(n) (t) = x_(n) exp(i omega t) \
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==> m/K (dif psi_(n) ) / (dif t) = psi_(n + 1) - 2 psi_n + psi_(n - 1) \
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==> - omega^2 m/k x_n = x_(n + 1) - 2 x_n + x_(n - 1).
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$
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Dabei wird unter festen und periodischen Raendern unterschieden.
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- Randbedinungen selektonieren die Loesungen $omega''$ als Eigenwertproblem
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- Matrizen symmetrisch $==>$ $N$ Eigenwerte
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$
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partial _(t) ^2 overline(psi) = underline(M) overline(psi) \
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vec(psi_1 (t), psi_2 (t), ..., psi_(n) (t)) = overline(psi) (t) = sum _(j = 1) ^(N) underbrace(A_(j) e^(i omega_(j) t), Q_(j) \ "Normalmoden") arrow(e)_(j) + B_(j) e^(- i omega_(j) t) arrow(e)_(j) \
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arrow(psi) (t) = sum _(i = 1) ^(N) Q_(j) (t) arrow(e)_(j) \
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arrow(e)_(j) = e ^(i k a n) hat(e)_(z) .. k = (2 pi)/ lambda.
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$
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Die Dispersionsrelation ist gegeben durch
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$
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omega (k) = sqrt((2 k)/m) (1 - cos (k n))^(1/2).
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$
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Und in linearer Form als
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$
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omega = c k.
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$
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Die erste Brillouin-Zone gibt an, welche Wellen sich ausbreiten koennen. Wie kann man dort sehen, dass dies diskret ist.
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Wir sagen
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$
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psi_1 = psi_(n + 1) \
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e ^( i k n) = e ^( i k (n + 1)a) \
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e ^(i k n a) = 1 \
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==> 2 n a = 2 pi i \
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k = (2 pi)/a i/n
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$
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Gegenueberstellung von Welle und Schwingung
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Notationen
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$
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psi (t) = psi_0 sin (omega t + phi) \
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psi (x, t) = psi_0 sin (k x - omega t + phi) \
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psi (r, t) = psi_0 sin (k * r - omega t + phi).
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$
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Relationen
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$
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v = phi / (2 pi) \
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T = 1/nu \
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lambda = (2 pi)/k \
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I prop abs(psi (x, t))^2.
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$
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= Die Wellengleichung
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$
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(diff^2 psi) / (diff t^2 ) = c^2 (diff^2 psi) / (diff x^2 ) \
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dot.double(psi) = c ^2 psi''
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$
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$
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psi (x, t) = psi_0 sin (k x - omega t) \
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partial _(t) ^2 psi = omega^2 psi_0 sin (k x - omega t) \
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omega/k = c = lambda nu \
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partial _(x) ^2 psi = k^2 psi_0 sin (k x - omega t)
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$
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Als homogene DGL
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$
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dot.double(psi) - c^2 psi'' = underbrace(0, "Quelle").
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$
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Die allgemeine Loesung ist
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$
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psi (x, t) = f_(1) (x + c t) + f_2 (x - c t) , space f_1, f_2 in C^2 (RR).
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$
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In mehereren Dimenisonen
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$
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dot.double(psi) - c^2 arrow(nabla) ^2 psi = 0 , space psi = psi (arrow(r), t).
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$
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Kann ein Sperarationsansatz auch Loesungen liefern?
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$
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psi (arrow(r), t) = u (arrow(r)) e ^(i omega t) \
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u (arrow(r)) ( - omega ^2 ) e ^( i omega t) = c ^2 e ^(i omega t) arrow(nabla) ^2 u (arrow(r)) \
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==> arrow(nabla) ^2 u + k^2 u = 0 , space "stationaere Wellengleichung (Helmholtzgleichung)".
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$
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Die ruecklaufende Welle wird benoetigt um alle Randbedinungen abzudecken.
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Wellen breiten sich mit einer linearen Superposition aus.
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$
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xi (x, t) = sin (x +- v t) = sin (k x - omega t)
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$
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= Skalare Wellen und Vektorwellen
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$
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psi (arrow(r), t) = arrow(A) e ^(i (arrow(k) arrow(r) - omega t))
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$
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Abhaengig von $arrow(A)$ ist die Welle entweder eine Skalarewelle oder eine Vektorwelle.
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$
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arrow(k) * arrow(r) = phi , space "Normalendarstellung einer Ebene" \
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psi (arrow(r), t) = A e ^(i (arrow(k) arrow(r) - omega t)) , space "ebene Welle" \
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psi (arrow(r), t) = A e ^(i (k r - omega t)) .
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$
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== Kugelwellen
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Wir betrachten Kugelkoordinaten.
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Betrachte eine Loesung
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$
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psi (r, theta, phi, t) = psi (r, t) \
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dot.double(psi) = c^2 arrow(nabla) ^2 psi \
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arrow(nabla) ^2 psi , space "fuer kugelsymmetrisen Fall" \
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arrow(nabla) ^2 psi = 1/r diff / (diff r^2 ) (r psi) \
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==> dot.double(psi)= c^2 1/r diff / (diff r^2 ) (r psi) \
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diff / (diff t^2 ) (r psi) = c^2 diff / (diff r^2 ) (r psi) ==> psi (r, t) = cases(
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1/r f (r - c t) \, space "auslaufen" , 1/r f(r + c t) \, space ""
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)\
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$
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== Zylinderwellen
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$
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psi (rho, t) = A 1/sqrt(rho) e ^(i (k s +- c t)) \
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dot.double(psi ) = e ^2 r 1/rho diff / (diff rho) (rho (diff psi) / (diff rho) ) \, space "loesung durch Besselfunktion"
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$
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Energiedichte
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$
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I prop abs(psi)^2 \
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abs(psi)^2 prop 1/r^2 \, space "Energiesatz" \
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I prop E^2
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$
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= Stehende Welle
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Nur bestimmte Frequenzen erlauben fuer eine stehende Welle.
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#figure(
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image("typst-assets/drawing-2025-11-05-11-18-39.rnote.svg"),
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)
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$
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psi (x, t) = A_0 sin ( - k x - omega t) + A_(r) sin (k x - omega t)
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$
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Was ist die Randbedinungen bei $x = 0$?
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$
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psi (x = 0, t) = - A_0 sin (omega t) - A_(r) sin (omega t) = ^(!) 0 \
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==> A_(0) = - A_(r) = A \, space "Vorzeichenwechsel der Amplitude oder einen Phasensprung" \
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psi (x, t) = A [sin ( - k x - omega t ) + sin ( - k x + omega t)] \
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sin alpha + sin beta = 2 sin ((alpha + beta)/2) cos ((alpha - beta)/2) \
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==> psi (x, t) = - 2 A sin (k x) cos (omega t) ==> "zeitliche und raeumliche Abhaengigkeit werden separiert".
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$
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= Resonator
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Ein Resonator wird erzeugt, wenn zwei Waende gegenuebergestellt werden. Das heisst es gibt Randbedinungen an beiden Enden.
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Eigenwertproblem
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$
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L = n lambda/2 \, space n in NN \
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lambda_(n) = (2 L ) / ( n) \, space k_(n) L = n pi \, space c = lambda nu = nu (2 L) / (n) \
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==> Delta y = (c) / (2 L).
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$
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Falls ein Ende offen ist, dann
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$
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L = (2 n + 1) lambda/4.
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$
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$
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psi (x, t) = u (x) f (t) \
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u'' + k^2 u = 0.
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$
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= Schwingung einer rechteckigen Membran
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Die Kantenlaengen sind $a, b$. Es ergibt sich
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$
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arrow(nabla) ^2 u = - k ^2 u \
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u_(n, m) = u_0 sin ((n pi x) / (a)) sin ((m pi y)/b)) \, space n, m in NN \
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==> k ^2 _(n, m) = pi^2 (n^2 /a^2 + m^2 /b^2 ) \
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omega ^2 _(n, m) = c^2 pi^2 (n^2 /a^2 + m^2 /b^2 ) ==> "Raumfrequenzen"
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$
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= Schallwellen
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Starten mit der idealen Gasgleichung. Es schwingt der Druck
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$
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p (x, t) = p + tilde(p) (x, t) \
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rho (x, t) = rho_0 + tilde(rho) (x, t)
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$
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und die Auslenkung mit der Geschwindigkeit
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$
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xi (x, t) \
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v (x, t) = dot(xi).
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$
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$
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A Delta x rho (diff v) / (diff t) = ^("Taylor") - A (diff p) / (diff x) Delta x \
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(diff v) / (diff t) = - 1/rho (diff p) / (diff x)
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$
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