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Wave Superposition Principle Physics 306: Waves Lecture 5 2/7/2008 Wave Superposition Principle It is quite a common situation for two or more waves to arrive at the same point in space or to exist together along the same direction. We will consider today several important cases of the combined effects of two or more waves. Let us consider where two waves with displacements given by Ψ1 and Ψ2 are present. What is the net displacement? Superposition Principle: Ψ = Ψ1 + Ψ2 The principle is valid for linear PDE’s. The wave equation is: ∂ 2Ψ 1 ∂ 2Ψ = ∂x 2 v 2 ∂t 2 If Ψ1 and Ψ2 are solutions of the above PDE, then Ψ = aΨ1 + bΨ2, (where a and b are constants) is also a solution. This makes “interference” possible Page 1 of 15 Physics 306: Waves Lecture 5 2/7/2008 Constructive (waves combine “in step”) = Destructive (waves combine “out of step”) = Ψ1 and Ψ2 have the same frequency Something in between Get some Ψ with the same frequency = Let us develop the mathematical formalism for combining two waves: Beginning with the wave equation ∂ 2Ψ 1 ∂ 2Ψ = ∂x 2 v 2 ∂t 2 Let Ψ1 and Ψ2 be solutions. Then try the sum of solutions as a possible solution: Ψ = Ψ1 + Ψ2. We get ∂ 2Ψ ∂ 2Ψ 1 ∂ 2Ψ 1 ∂ 2Ψ 1 + 2 = 1 + 2 ∂x 2 ∂x 2 v 2 ∂t 2 v 2 ∂t 2 Page 2 of 15 Physics 306: Waves Lecture 5 2/7/2008 Therefore, Ψ is a solution. In general, we may form a solution as a linear combination of solutions: N Ψ()x,t = ∑Ci Ψi ()x,t i=1 But suppose ∂ 2Ψ 1 ∂ 2Ψ + cΨ 2 = ∂x 2 v 2 ∂t 2 2 2 2 For solutions Ψ1, Ψ2, and Ψ = Ψ1 + Ψ2, the cΨ is c(Ψ1 + Ψ2 + 2Ψ1Ψ2 ). So, LHS ≠ RHS So the principle of superposition is only valid for linear PDEs (where the function and its derivatives appear only in first order.) The Addition of Waves of the Same Frequency As a starting point, let us say that we have two 1-D waves that are harmonic. We can describe these waves as: Ψ(x,t) = Ψ0 sin(ωt − (kx + ε )) If α = kx + ε, then we can write Ψ1 (x,t) = Ψ01 sin(ωt −α1 ) Note, we are and assuming they have the same frequency Ψ2 (x,t) = Ψ02 sin(ωt −α 2 ) Let us suppose these two waves coexist in space. The resultant disturbance is the linear superposition of these waves Ψ = Ψ1 + Ψ2 = Ψ01 sin()ωt −α1 + Ψ02 sin ()ωt −α 2 = Ψ01 (sin()ωt cos ()α1 + cos ()ωt sin ()α1 )+ Ψ02 ()sin ()ωt cos ()α 2 + cos ()ωt sin ()α 2 Therefore, Page 3 of 15 Physics 306: Waves Lecture 5 2/7/2008 Ψ = ()Ψ01 cos()α1 + Ψ02 cos(α 2 ) sin(ωt)+ (Ψ01 sin(α1 )+ Ψ02 sin(α 2 ))()cos ωt independent of time This can be simplified if we define Ψ0 cos(α ) = (Ψ01 cos(α1 )+ Ψ02 cos(α 2 )) (1) and Ψ0 sin(α ) = (Ψ01 sin(α1 )+ Ψ02 sin(α 2 )) (2) If we now square the equations above and add 1 2 2 2 2 Ψ0 = ()cos α + sin α Ψ0 2 2 2 2 = Ψ01 cos α1 + Ψ02 cos α 2 + Ψ01Ψ02 cosα1 cosα 2 2 2 2 2 + Ψ01 sin α1 + Ψ02 sin α 2 + Ψ01Ψ02 sinα1 sinα 2 Ψ 2 = Ψ 2 + Ψ 2 + 2Ψ Ψ cos(α −α ) ⇒ 0 01 02 01 02 2 1 (3) If we divide equation (2) by (1) we get Ψ sinα + Ψ sinα tan()α = 01 1 02 2 Ψ01 cosα1 + Ψ02 cosα 2 Mathematical Aside: It is often convenient to make use of the complex representation of waves when dealing with the superposition of waves. Recall that Ψ1 = Ψ01 sin(ωt − (kx + ε1 )) = Ψ01 sin()ωt +α1 can be written as: i(ωt+α1 ) Ψ1 = Ψ01e where the wave is the imaginary part of this equation. Page 4 of 15 Physics 306: Waves Lecture 5 2/7/2008 Similarly, we can write: i(ωt+α2 ) Ψ2 = Ψ02e Let i(ωt+α ) Ψ = Ψ0e = Ψ1 + Ψ2 iωt What are Ψ0 and α? Factor out e , which cancels. Re(Ψ) = Re(Ψ1 )+ Re(Ψ2 ) Im(Ψ) = Im(Ψ1 )+ Im(Ψ2 ) Im(Ψ) tanφ = Re()Ψ 2 * * * Ψ0 = Ψ0 Ψ0 = (Ψ1 + Ψ2 )(Ψ1 + Ψ2 ) * * * * = Ψ1 Ψ1 + Ψ2 Ψ2 + Ψ2 Ψ1 + Ψ1 Ψ2 2 2 * * = Ψ1 + Ψ2 + Ψ2 Ψ1 + Ψ1 Ψ2 2 2 −iα2 iα1 −iα1 iα2 = Ψ01 + Ψ02 + Ψ02e Ψ01e + Ψ01e Ψ02e 2 2 i()α1 −α2 −i ()α1−α2 = Ψ01 + Ψ02 + Ψ01Ψ02 ()e + e 2 2 2 Ψ0 = Ψ01 + Ψ02 + 2Ψ01Ψ02 cos(α1 −α 2 ) which is identical to Eq. (3), and Ψ sinα + Ψ sinα tan()α = 01 1 02 2 Ψ01 cosα1 + Ψ02 cosα 2 We can generalize this to N waves: Page 5 of 15 Physics 306: Waves Lecture 5 2/7/2008 N Ψ = ∑Ψ j j=1 N i()ωt+α j = ∑Ψ0 j e j=1 ⎡ N ⎤ iα j iωt = ⎢∑Ψ0 j e ⎥e ⎣ j=1 ⎦ iα iωt = Ψ0e ⋅e where N iα iα j Ψ0e = ∑Ψ0 j e j=1 is defined as the source complex amplitude of the superpositional wave. The intensity of the resultant wave is given by: 2 iα iα * Ψ0 = ()Ψ0e (Ψ0e ) N N N 2 = ∑Ψ0i + 2∑∑Ψ0i Ψ0 j cos()α j −α1 i=1 j>=i i 1 and N ∑Ψ0i sinα i i=1 tanα = N ∑Ψ0i cosα i i=1 We can also use equations (1) and (2) to write Ψ as Ψ = Ψ0 cosα sinωt + Ψ0 sinα cosωt = Ψ0 sin()ωt +α What does this last equation say? It shows us that a single wave results from the superposition of the original two. This new wave is harmonic and of the same frequency as the original wave, although its amplitude and phase are different. An important consequence of this is that we can superposition any number of harmonic waves having a given frequency, and get a resultant wave which is harmonic as well. Page 6 of 15 Physics 306: Waves Lecture 5 2/7/2008 Interference The intensity of a wave is proportional to the square of its amplitude. Thus from Eq. (3), we see that the intensity of resulting wave is not just the sum of flux densities of individual original waves, but that there is an additional term: 2Ψ01Ψ02 cos(α 2 −α1 ) This is an interference term, which is a function of the difference in phase between the two original waves. The crucial factor is δ, where δ = α2 – α1 When δ = 0, ± 2π, ± 4π, … → resultant amplitude is max δ = ± π, ± 3π, … → resultant amplitude is min Recall that α = -kx + ε 2π where k = . λ So we can write: 2π δ = α −α = ()()kx + ε − kx + ε = ()()x − x + ε + ε 2 1 1 1 2 2 λ 1 2 1 2 where x1 and x2 are the distances from the sources of the two waves to the point of observation and λ is the wavelength. Let us suppose the two waves are initially in phase: ε1 = ε2, then 2π δ = ()x − x λ 1 2 If we define the index of refraction Page 7 of 15 Physics 306: Waves Lecture 5 2/7/2008 wavelength in vacuum λ n = 0 , λ Then we can write: 2π δ = n()x2 − x1 λ0 this quantity is known as the optical path difference (OPD) ≡ Λ δ = k0Λ Definitions: waves for which ε1 - ε2 = constant are coherent waves. Random and Coherent Sources From a previous lecture, we define the source intensity as ∝ time-average of the amplitude of wave squared: 2 I ∝ Ψ0 For two waves, using Eq. (3) for the resultant wave, we have time-average I = I1 + I 2 + 2 Ψ1Ψ2 cos(α 2 −α1 ) If the square of the waves 1 and 2 have random phase (≡ incoherent), then: cos(α 2 −α1 ) = 0 i.e. the time-average will be zero. Then, I = I1 + I2, Or, more generally: N I = ∑ I i = NI1 i=1 Page 8 of 15 Physics 306: Waves Lecture 5 2/7/2008 if you have N randomly phased sources of equal amplitude and frequency. (resultant intensity arising from N coherent sources is determined by the sum of the individual intensities) [What does this mean? If you have two light bulbs that emit light with random phase, the result will have an intensity equal to the sum of the intensities of each bulb → no interference effect is observed.] If α1 - α2 = constant → coherent waves then cos(α 2 −α1 ) ≠ 0 But varies from 1 and -1, so I varies between I = I1 + I 2 + 2 I1I 2 (cos = 1) and I = I1 + I 2 − 2 I1I 2 (cos = -1) If I1 = I2, (5) then I varies between 4I1 ⇒ constructive interference and 0 ⇒ destructive interference If you have N sources that are in phase with αi = αj and equal amplitude, then 2 iα iα * I = Ψ0 = (Ψ0e )(Ψ0e ) N N N 1 2 = ∑Ψ0i + 2∑∑Ψ0i Ψ0 j cos()α j −α1 i=1 j>=i i 1 N 2 ⎛ ⎞ 2 2 2 I = ⎜∑Ψ0i ⎟ = ()NΨ0i = N Ψ0i ⎝ i=1 ⎠ Page 9 of 15 Physics 306: Waves Lecture 5 2/7/2008 if each amplitude is the same = Ψ01. So here, the amplitudes are added first and then squared to determine the resulting intensity. Standing Waves We already covered this briefly. Here we assume that we have two harmonic waves traveling with some frequency and traveling in opposite directions. Let us assume the two waves have equal amplitude.
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