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1.4: Superposition and Interference

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    Combining Similar Waves

    When two or more waves of the same type in the same medium coexist in the same region of space, they combine to create a new wave. The way they combine is a simple process known as superposition. This consists of simply adding the displacements (or whatever the wave function represents) of the two or more waves at the same place and time. For a 1-dimensional wave, this means:



    It's important to emphasize that two waves can only superpose if they are the same type. Many different kinds of waves can travel through the same medium (light, sound, and displacement waves can all travel through water in a lake, for instance), but these cannot superpose with each other.

    It’s easy to show that if the two individual wave functions satisfy the wave equation, then so does the total wave function. It bears repeating with a diagram that this superposition sum involves adding displacements at the same place and time. So if we took a snapshot of two waves, we would determine the total wave by lining them up:

    Figure 1.4.1 – Superposition


    The composite wave is then the combination of all of the points added thus. Of course, these are traveling waves, so over time the superposition produces a composite wave that can vary with time in interesting ways. Here is a simple example of two pulses "colliding" (the "sum" of the top two waves yields the bottom wave).

    Figure 1.4.2 – "Collision" of Pulses


    Notice that even though the resultant wave looks very different from its "parents," the medium somehow "remembers" the original waves, and when they no longer coincide, they continue along as exactly the waves they were before the superposition. That is, the waves do not affect each other, as particles would if they collided – waves don't bounce off each other, for example. They simply create a new wave while they occupy the same space in the medium, and when their individual motions carry them to different parts of the medium, they return to being the waves they were before.

    Interference – Special Cases

    There are some special cases involving superposition that are particularly interesting to examine, and these involve a phenomenon known as interference. There are many degrees of interference possible, all of which fall between the following two extremes:

    • constructive interference: The waves are perfectly aligned and timed so that their crests and troughs coincide, such that the total wave has the maximum possible amplitude (equal to the sum of the amplitudes of the two constituent waves).
    • destructive interference: The waves are perfectly aligned and timed so that the crests of one wave align with the troughs of the other such that leading to a wave that has the minimum possible amplitude (equal to the difference of the amplitudes of the two constituent waves).

    The phrase total destructive interference refers to the case of destructive interference when the resultant wave has zero amplitude, i.e. the two waves totally cancel each other. In the cases we will discuss, we will only talk about this extreme case of destructive interference, so we will typically leave out the word “total,” even though we are still talking about total cancelation.

    Interference – General Case

    We will examine a great many examples of interference in physical phenomena in the sections to come. We therefore need to take some time to develop the mathematics behind this effect. We will do this within the same framework that we have been using – that of harmonic waves. When we look at the physical attributes of interference, what we will be examining is what happens to the intensity of the combined wave. For example, interference in sound will be exhibited in volume, and in light it will be brightness. Both of these are measures of intensity. We need a reference point for intensity, and the one we will use is that of maximal constructive interference. So what we seek is an equation that relates the intensity of two superposed, out-of-phase, but otherwise identical waves to the intensity we would see if they were in phase. That is, we want something that looks like this:

    \[I\left(\Delta\Phi\right) = I_o g\left(\Delta\Phi\right) \]

    The quantity \(I\) is the intensity of the wave as a function of the phase difference of the two (identical) parent waves. If the two waves happen to be in phase, then the combined wave's intensity is \(I_o\) when the two waves are in phase. Note that this is four times the intensity of each individual wave, since the constructive interference adds the amplitudes (which are equal – the waves are identical) and the intensity is proportional to the square of the amplitude.

    The function needs to have the following properties:

    • It has to always be non-negative, since intensity is never a negative number.
    • It has to vanish when the phase difference equals \(\pi\) (modulo \(2\pi\)), since this means the waves totally destructively interfere.
    • It has to equal 1 when when the phase difference is 0 (modulo \(2\pi\)), since this means the waves constructively interfere.

    To find this function, we start with two wave functions that are identical except for their phases and superpose them:

    \[f_{tot}=f_1+f_2=A\cos\left(\Phi_1\right)+A\cos\left(\Phi_2\right)=A\left[\cos\left(\Phi_1\right)+\cos\left(\Phi_2\right)\right] \]

    We want this function to only depend upon the difference in the two phases, so we will write each total phase in terms of deviation from their average phase (which we will call simply \(\Phi\)), and the difference in phase between the two waves, \(\Delta\Phi\):

    \[\Phi = \dfrac{\Phi_1+\Phi_2}{2}\;\;\;\Rightarrow\;\;\;\Phi_1=\Phi-\dfrac{\Delta\Phi}{2},\;\;\;\Phi_2=\Phi+\dfrac{\Delta\Phi}{2}\]

    Plug these into Equation 1.4.3 gives:

    \[f_{tot} = A\left[\cos\left(\Phi-\frac{\Delta \Phi}{2}\right) + \cos\left(\Phi+\frac{\Delta \Phi}{2}\right)\right]\]

    Now we can apply a trigonometric identity:

    \[\cos\left(A-B\right)+\cos\left(A+B\right)=2\cos A \cos B \;\;\;\Rightarrow\;\;\; f_{tot} = 2A\cos\left(\Phi\right)\cos\left(\dfrac{\Delta\Phi}{2}\right)\]

    The phase difference between the two waves can be written in terms of the difference in position, time, and the phase constant, using Equation 1.2.9:

    \[\Delta\Phi=\frac{2\pi}{\lambda}\left(x_1-x_2\right)\pm \frac{2\pi}{T}\left(t_1-t_2\right) + \phi_1-\phi_2 =\frac{2\pi}{\lambda}\Delta x\pm \frac{2\pi}{T}\Delta t + \Delta\phi \]

    As the waves propagate along, the values of \(x\) and \(t\) will change, but as the two waves are identical (traveling in the same direction with the same speed), the differences in \(x\) and \(t\) don’t change for a given phase. Therefore the factor in Equation 1.4.6 that includes the phase difference is a constant. Putting that constant together with the \(2A\) gives us the amplitude of the new conglomerate wave (with the time-varying phase being the average of the phases of the two waves):

    \[f_{tot} = A_{new}\cos\left(\Phi\right),\;\;\;\;\;\;A_{new}\equiv 2A\cos\left(\dfrac{\Delta\Phi}{2}\right)\]

    The intensity is proportional to the square of the amplitude, so the intensity of this combined wave is:

    \[I\propto A_{new}^2 = 4A^2\cos^2\left(\dfrac{\Delta\Phi}{2}\right)\]

    The intensity of each individual wave is proportional to \(A^2\). If the waves were in phase, the total amplitude would double, which means that the total in-phase intensity \(I_o\) is proportional to \(4A^2\). The intensity of the (out of phase) combined wave is therefore:


    Notice that this relationship between total intensity and phase difference exactly matches the three criteria we outlined above.

    Example \(\PageIndex{1}\)

    Two identical harmonically-oscillating devices attached to a taut string are turned on simultaneously and in phase, vibrating at a frequency of \(20Hz\). One source is located at the origin, and the other is positioned at \(x=25cm\). Each source vibrates with a maximum displacement of the string of \(8.0cm\). The string has a density of \(35g\cdot m^{-1}\), and is stretched with a tension of \(5.2N\). The sources are allowed to vibrate long enough for their respective waves to superpose.

    1. Find the amplitude of the combined wave.
    2. Find the power of the combined wave.
    3. Find the distance in the \(+x\) direction that the source not at the origin must be moved for the power output to be maximized, and determine this maximum power output.

    a. The amplitude of the combined wave is given by Equation 1.4.8, which requires that we calculate the phase difference \(\Delta \Phi\):

    \[\Delta\Phi =\frac{2\pi}{\lambda}\Delta x\pm \frac{2\pi}{T}\Delta t + \Delta\phi \nonumber\]

    The sources start at the same moment in time and in phase, so \(\Delta t = 0\) and \(\Delta \phi = 0\). Therefore the only source of phase difference is the separation of the sources, \(\Delta x\). We still need the wavelength, which we can get from the frequency and wave speed. The latter we get from the string density and tension. Putting all this together gives:

    \[\lambda=\dfrac{v}{f},\;\;\; v=\sqrt{\dfrac{F}{\mu}}\;\;\;\Rightarrow\;\;\; \lambda = \dfrac{1}{f}\sqrt{\dfrac{F}{\mu}}= \dfrac{1}{20Hz}\sqrt{\dfrac{5.2N}{0.035kg\cdot m^{-1}}} = 61cm \nonumber\]

    Plugging this in above gives:

    \[\Delta\Phi =\frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{61cm}\left(25cm\right) = 0.82\pi \nonumber\]

    The amplitude of the combined wave is therefore:

    \[A_{combined}=2A\cos\left(\dfrac{\Delta\Phi}{2}\right) = 2\left(8.0cm\right)\cos\left(0.41\pi\right) = 4.46cm\nonumber\]

    Notice that the amplitude of the combined wave is actually smaller than that of each individual wave. This is because the separation is such that there is some partially destructive interference going on.

    b. The power of the combined wave can be found directly from Equation 1.3.9:

    \[P = \frac{1}{2}\sqrt{\mu F} \left(2\pi f\right)^2A^2 = \frac{1}{2}\sqrt{\left(0.035kg\cdot m^{-1}\right)\left(5.2N\right)}\left(2\pi\right)^2\left(20Hz\right)^2\left(0.045m\right)^2 = 6.8W\nonumber\]

    c. The power of a one-dimensional wave is the same as its intensity. From Equation 1.4.10 it is clear that the maximum intensity occurs when the cosine function equals \(\pm1\). This occurs when the argument is some integer multiplied by \(\pi\). The argument in the case above is \(\frac{\Delta \Phi}{2}=\frac{0.82\pi}{2}\), so the source must be shifted enough for the total phase to change by an additional \(1.18\pi\), to give: \(\frac{\Delta \Phi}{2}=\frac{1.18\pi+0.82\pi}{2}=\pi\). This phase difference is related to the change in position through the wavelength:

    \[\Delta \Phi = \frac{2\pi}{\lambda} \Delta x \;\;\;\Rightarrow\;\;\; \Delta x = \Delta \Phi\left(\frac{\lambda}{2\pi}\right) =\left(1.18\pi\right)\frac{61cm}{2\pi}=36cm \nonumber\]

    Notice that the position where the second source creates maximum constructive interference is exactly one wavelength away from the first source: \(25cm+36cm=61cm=\lambda\). This is because with only spatial separation in play, the phase difference changes by \(2\pi\) every time the spatial separation changes by \(\lambda\).

    The maximum possible intensity is \(I_o\), and since we know the intensity of the combined wave and the phase difference, we can compute this maximum from Equation 1.4.10:

    \[I_o = \dfrac{I}{\cos^2\left(\frac{\Delta \Phi}{2}\right)} = \dfrac{6.8W}{\cos^2\left(\frac{0.82\pi}{2}\right)} = 87W \nonumber\]

    A simpler approach would be to note that the intensity is proportional to the square of the amplitude, and since changing the position of the second source increases the amplitude from \(4.46cm\) to \(16cm\) (double the individual wave amplitude), which is an increase of a factor of 3.59, the intensity increases by a factor of \(3.59^2=12.9\), giving the same result (except for some rounding errors along the way).


    This page titled 1.4: Superposition and Interference is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Tom Weideman directly on the LibreTexts platform.

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