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2.3: Interference Effects

  • Page ID
    18451
  • Standing Waves

    Sound can create a variety of interference effects, like any other wave. Among those interference effects are standing waves. These are formed in precisely the way described in Section 1.5, with two traveling waves reflecting back-and-forth between two endpoints. In the case of sound in a gas, it isn't immediately clear what constitutes "fixed" and "free" endpoints, since sound waves in gas do not involve particles of media that can ever be held in place. We have to broaden our notion of what it means to have a fixed boundary to mean simply that whatever physical quantity plays the role of "displacement" for a wave is unchanging. In the case of sound, this would mean pressure or density. We will see in a moment how this can be.

    We are used to talking about one-dimensional standing waves, so it is fair to ask how this can be set up for sound in a gas. If a region of the gas has a pressure higher than the ambient pressure, the gas will naturally expand into the lower-pressure region, so to create a 1-dimensional standing sound wave, we must set up circumstances so that the compressed regions doesn't expand into all three dimensions. We can do this by producing the sound within a hollow pipe. The fixed or free boundary conditions at the ends of the pipe then depend upon whether the end of the pipe is open or closed, but which case is fixed, and which is free?

    Let's look first at a closed end. When a compression propagates its way toward a closed end, the means for the compression to restore itself to equilibrium is restricted – it can't continue forward. The compression grows even greater than the amplitude of the wave, just as the displacement of a string at the point of reflection from a free end grows above the amplitude (see Figure 1.5.5). So in fact a closed end of a hollow pipe represents a "free" end for a standing sound wave.

    The open end is quite different. In fact, it does not behave like an end at all, but rather like a transition point. We know this is true if we follow the energy – a sound wave headed for an open end will transmit energy out of that open end, as the compressions and rarefactions are transmitted to the region of the gas outside the pipe. So why would there be any reflection back into the pipe at all? As in the case of going from slow-to-fast or fast-to-slow medium, the mathematics of the boundary interaction is beyond this course, but the result is the same. The obvious objection here is that the speed of the wave within the air in the pipe is no different from the speed of sound outside the pipe. This is true, and in fact it requires a bit of a revision to that observation. Perhaps a better way of describing the reflection/transmission phenomenon is to say that there is at least partial reflection whenever the wave equation governing the wave changes. One way for the wave equation to change is for the wave velocity to change. Another is for the wave to go from being confined to one dimension to three dimensions. This is exactly what happens when the sound in a pipe emerges from an open end.

    We know that waves in three-dimensions spread the energy very fast, causing the amplitude to drop-off in proportion to the distance. It is therefore not a bad approximation to claim that at the open end (or slightly beyond it) is at a fixed pressure/density (the ambient pressure/density). This means that if we are forced to choose, the open end of the pipe to a good approximation behaves like a fixed end, and the sound wave that reflects back into the pipe is phase-shifted.

    From these considerations, we now know that standing sound waves in a pipe create pressure antinodes at close ends, and pressure nodes at open ends. Armed with this information, we can use all the same machinery regarding standing waves that we learned in Section 1.5. What is interesting about the case of sound is how this can be used to create tones in organ pipes and wind instruments. These devices make use of two special aspects of standing waves in pipes. The first we have already mentioned – an open end allows for some transmission of sound waves, which means that we can hear the tone produced without having to crawl inside the pipe. The second has to do do with the idea of resonance.

    Digression: Resonance

    Resonance is a very important concept in many fields of physics, and we unfortunately don't have time to cover it in great detail here, though it will come up again in both Physics 9C and 9D. The main idea is that if vibrating systems interact with each other, the amount of energy that can be transmitted from one to the other is largely dependent upon how closely the "natural frequencies" (think of a masses on a springs with frequencies that look like \(\omega=\sqrt{\frac{k}{m}}\)) of those systems match. If the frequencies are a good match, then the superposition of the displacements of the two systems leads to constructive interference. If they don't match, then the displacements don't synchronize, and there is very little overall constructive interference. A good analogy is pushing a child on a swing. If you give them a push forward every time they reach the peak of their backswing, then the frequency of your pushes matches the natural frequency of the swing, and energy is transmitted to the swing. If, however, you were to push with a different frequency, then some of the pushes would add energy to the swing, but many others would push the swing forward as it is coming backward, taking energy away from the swing.

    If we get air moving near the end of a pipe, like when we blow into a flute, the result is turbulent flow. This adds energy to the system, but the waves created come in a wide variety of frequencies. All of these sound waves travel the length of the pipe, partially reflecting and transmitting when they get there. But as we have seen in our study of standing wave, only those waves that have one of the harmonic wavelengths will exhibit the constructive interference required for a standing wave. Those waves that do have the right frequency exhibit resonance, building energy for the standing wave at that frequency. The result is that of the many sound waves that escape the pipe, those at a resonant frequency of the pipe (determined by its length) have by far the most energy, and are the only sounds heard. It turns out that by far most of the energy goes to the fundamental harmonic, though overtones can also often be heard. To modulate the frequency of the sound that escapes therefore becomes a matter of changing the length of the pipe. An organ designates a pipe for every key on the keyboard; a slide trombone allows the player to physically expand the length of the pipe; valves and holes in other wind instruments also serve this same purpose. Note also that all of these instruments rely upon the turbulent flow to provide the spectrum of sound waves, whether it is through a vibrating reed, vibrating lips, or air forced across an open end. (Note: Air forced into an open end does not induce much turbulence compared to air forced across the opening.)

    Lastly, it should be noted that to produce a sustained tone, one must continually add energy. This is because energy is always exiting the pipe via the transmitted wave. The rate at which energy is added equals the rate of energy escaping via sound waves, while the energy in the standing wave within the pipe remains constant.

    Example \(\PageIndex{1}\)

    A string is plucked above a pipe that is open at one end, and the fundamental harmonic tone is heard coming from the pipe. If the closed end of the pipe is now opened, how must the tension of the string be change in order to excite the new fundamental harmonic?

    Solution

    Open ends of pipes act as fixed points (nodes) for standing sound waves. The fundamental harmonic of the pipe with one end closed fits one quarter of a wave between the ends of the pipe (node to first antinode), while the first harmonic with both ends open fits a half-wavelength (node to node). Therefore opening a closed end shrinks the wavelength of the fundamental harmonic by a factor of 2. The speed of the sound wave is unchanged, so its fundamental harmonic frequency rises a factor of 2. The standing sound wave is being driven by (i.e. is getting energy from) the standing wave in the string, so their frequencies must match for this resonance to occur, which means that when the end of the pipe is opened, the string's standing wave frequency must also rise by a factor of 2. The string has not changed length, so the only other way to change its frequency is to change the speed of the traveling wave on the string. To double the speed of the string wave (and therefore double the frequency), the tension must be quadrupled, since the wave speed on the string is proportional to the square root of the tension, and the linear density of the string cannot be changed.

    Constructive and Destructive Interference

    When we discuss interference of waves like sound and light, the most striking results are the extremes – places and times where the interference is completely constructive (doubling the amplitude and quadrupling the intensity), and especially where the interference is completely destructive. It is quite striking for a tone to come from two different sources, and the result is that nothing is heard! We will return now to our discussion from Section 1.4 in the context of sound.

    We start with the expression we derived for the superposed wave intensity, Equation 1.4.10. It's clear that this value will be zero (i.e. total destructive interference will result) whenever the total phase difference is some odd multiple of \(\pi\):

    \[\text{destructive interference:}\;\;\;\;\;\Delta \Phi = \frac{2\pi}{\lambda}\Delta x\pm \frac{2\pi}{T}\Delta t + \Delta\phi = n\pi\;,\;\;\;\;\;n=\pm 1,\;\pm 3,\;\pm 5 \dots\]

    It's easy enough to see this mathematically, but we need to understand physically what this means. We will do this with a few simplifications. First, it should be clear that two waves out of phase by \(3\pi\) looks exactly like waves that are out of phase by \(\pi\), so for now we will not worry about the infinitude of possibilities beyond \(\pm \pi\). Second, destructive interference is easier to see graphically if representations of the waves are juxtaposed for easy comparison. But the reader is exhorted to keep in mind that the waves can be traveling in any direction – they interfere destructively when a rarefaction of one wave superposes with a compression of another, and density differences in gas are not directional. Certainly the overall interference pattern differs based on the wave directions, but when talking about total destructive interference, we are looking at a single point in space.

    There are three terms in the expression for total phase difference that can affect its value: \(\Delta x\), \(\Delta t\), and \(\Delta \phi\). We'll keep things simple by looking at how each one individually can lead to destructive interference. In every case, we will be looking at two sources of sound with identical frequencies. As they are in the same medium, they have identical speeds and therefore wavelengths as well.

    Figure 2.3.1 – Destructive Interference Due to Travel Distance Only

    different_distances.png

    The diagram above represents what happens when the two sound sources are different distances from the receiver of the sound. In particular, one source is one half wavelength farther away from the listener. This diagram represents the compressions and rarefactions with sine/cosine waves (this not a picture – sound waves are not transverse!), moving along the same axis in the same direction (again, not necessarily what is actually happening). With this simple graphic, we can see that everywhere on the \(x\)-axis (so long as both waves have arrived), the superposition of the two waves always gives zero – the listener hears nothing. This is because everywhere on the \(x\)-axis the travel distances of the two waves are just right to create a phase difference of \(\pi\). It is important to note that the other two contributors to phase difference, \(\Delta t\), and \(\Delta \phi\), are both zero in this case: The waves started out of the speakers at the same time, which we can tell because the snapshot shows 2.25 wavelengths have emerged from both speakers – if one had started emitting the tone first, more wavelengths would have come out. The waves also started out of the speakers with the same phase (the leading edge of the sine wave is at the axis, and is on it way up).

    Figure 2.3.2 – Destructive Interference Due to Starting Time Only

    different_starting_times.png

    In this case, the origins of the waves are the same distance from the listener (as before, pick a spot on the \(x\)-axis for the listener that both waves have had time to reach). The lower speaker started its tone one-half period sooner than the upper speaker, which is why an extra half wavelength has emerged from the lower speaker. As with the previous case, both waves emerge from their speakers with the same phase. Once again, we see that the superposition of the two waves leads to total cancelation of the sound.

    Figure 2.3.3 – Destructive Interference Due Phase Constant Only

    different_phase_constants.png

    Our third case involves sound waves that start at the same position in space (so they are equidistant from the listener), and at the same time (the same number of wavelengths have emerged from the speakers), but they start out of their respective speakers out of phase by \(\pi\) radians, as can be seen in the leading edges.

    As we will see when we work with these situations, it is actually quite rare to take into account the times at which two sound sources are turned-on. If we assume they have been emitting sound in the undetermined past, we can easily redefine any difference in time to a difference in phase. That is, if we define \(t=0\) as any the moment when both speakers are already sending out waves, then by definition \(\Delta t=0\), and the difference in the phases of the waves is due to a difference in phase constant. This becomes apparent when one compares the last two figures, and ignores the red parts of the waves.

    Example \(\PageIndex{2}\)

    Two speakers, both pointing in the \(+x\) direction, are placed on the y-axis, separated from each other by a distance of \(2.00\;m\). They emit the same tone, which has a frequency of \(784\;Hz\), in phase with each other. A microphone is placed directly in front of and very close to one of the speakers, and is gradually moved from along the \(x\)-axis farther and farther from the speaker. Assume that the fact that the microphone is a little farther from one speaker than the other does not result in a noticeable intensity difference between the two speakers, so that the sound waves coming from the speakers have the same amplitude when they reach the microphone.

    two_speakers.png

    1. Find the distances from the closer speaker where the microphone detects no sound.
    2. Find the distances from the closer speaker where the sound gets loudest (i.e. constructive interference).
    3. Suppose the tone coming from the speakers has an adjustable frequency, and that it is gradually lowered. Find the frequency below which the microphone has no position on the \(x\)-axis where it measures total silence.
    Solution

    a. The starting phase and time are the same, so the only source of phase difference comes from the difference in distance traveled. From the \(\Delta x\) contribution to the phase difference that causes destructive interference, we therefore have:

    \[\dfrac{2\pi}{\lambda}\Delta x = n\pi \;\;\;\Rightarrow\;\;\; \Delta x = \frac{1}{2}n\lambda\;, \;\;\;\;\;n=1,\;3,\;5,\;\dots \nonumber\]

    The difference in distances traveled by the two waves us found using the Pythagorean theorem, so putting this in above gives:

    \[\Delta x = \sqrt{x^2+\left(2.00\;m\right)^2} - x = \frac{1}{2}n\lambda \]

    Doing the algebra gives:

    \[x_n=\frac{1}{n\lambda}\left(4.00m^2-0.250n^2\lambda^2\right)\nonumber\]

    We are given the frequency of the sound, so we can find its wavelength:

    \[\lambda = \frac{v}{f} = \dfrac{344\frac{m}{s}}{784Hz}=0.439m\nonumber\]

    Notice that although the value of \(n\) is not restricted, when it gets too high, the value of \(x\) will become negative. Plugging in all of the values of \(n\) that give positive values of \(x\) yields five possible values. Calling the \(n^{th}\) value of \(x\) "\(x_n\)" we have:

    \[x_1=9.00m,\;\;\;x_3=2.71m,\;\;\;x_5=1.27m,\;\;\;x_7=0.531m,\;\;\;x_9=0.022m\nonumber\]

    b. Constructive interference occurs when the path difference is an even number of half-wavelengths (i.e. some number of full wavelengths). We can get our answer directly from part (a) simply by taking even values of \(n\) instead of odd values. One again, the number of values of \(n\) is limited by the restriction that sign of \(x\) must be positive.

    \[x_2=4.34m,\;\;\;x_4=1.84m,\;\;\;x_6=0.858m,\;\;\;x_8=0.259m\nonumber\]

    c. The value of \(\Delta x\) clearly gets smaller as \(x\) gets larger (the hypotenuse gets closer and closer to equaling the \(x\) value as \(x\) gets larger), so the largest possible value of \(\Delta x\) is just the separation of the speakers. If the speakers are separated by less than one-half wavelength, then \(\Delta x\) can never get as big as a half wavelength, and no totally destructive interference is possible. These speakers are separated by \(2.00\;m\), so the wavelength of the sound must be shorter than \(4.00m\) for there to be any instance of total destructive interference. This wavelength corresponds to a frequency of:

    \[f = \frac{v}{\lambda} = \dfrac{344\frac{m}{s}}{4.00m} = 86Hz \nonumber\]

    Frequencies lower than this create wavelengths that are so long that \(\Delta x\) is never large enough to cause total destructive interference.

     

    Beats

    Up to now, all of our cases of interference involve waves that have identical frequencies. But a very interesting phenomenon emerges when two sound waves interfere that have slightly different frequencies (actually any two different frequencies will do, but the phenomenon is easier to detect when the frequencies are within 1 or 2 hertz, for reasons we will see). We'll start with another over-simplified diagram like those we have used above. This time, the two waves emerging from the speakers will have different frequencies, and therefore different wavelengths (though for simplicity, we will assume they have the same amplitude).

    Figure 2.3.4 – Superposing Two Sound Waves of Different Frequencies

    speakers_with_different_frequencies.png

    The diagram is of course a snapshot at a moment in time. If, at this moment in time, a listener is positioned at the left vertical line, then the superposition of the two waves results in constructive interference, which means that the sound heard is loud. If, on the other hand, someone is positioned at the right vertical line, then destructive interference results in silence. But now suppose that one remains at the right vertical line for a short time after this snapshot is taken. Both waves are traveling at the same speed (they are both in the same medium), so the rarefactions that coincide at the left vertical line will soon be at the right vertical line. That is, a person listening at the right vertical line will, at one moment, hear silence, and then a short time later, a loud tone. This pattern will in fact repeat itself with regularity, and this pulsing of the sound is referred to as beats.

    The math that governs this phenomenon results from a straightforward application of trigonometric identities. We start with a wave function for each wave. As we are talking about sound for which the "displacement" is pressure, we will represent the wave function with the variable \(P\left(x,t\right)\). Note we are also keeping things simple by staying in one dimension:

    \[P_1\left(x,t\right) = P_o\cos\left(\frac{2\pi}{\lambda_1}x-\frac{2\pi}{T_1}t+\phi_1\right)\;,\;\;\;\;\;P_2\left(x,t\right) = P_o\cos\left(\frac{2\pi}{\lambda_2}x-\frac{2\pi}{T_2}t+\phi_2\right)\]

    We are interested in what is happening to the sound at a single point in space (i.e. we want to show it gets it gets louder and softer), and any point will do, so for simplicity we'll choose the origin, \(x=0\). This reduces our two functions to:

    \[P_1\left(t\right) = P_o\cos\left(-\frac{2\pi}{T_1}t+\phi_1\right)\;,\;\;\;\;\;P_2\left(t\right) = P_o\cos\left(-\frac{2\pi}{T_2}t+\phi_2\right)\]

    These functions are oscillating with different frequencies, so they are out-of-sync. If we wait long enough, their displacements will – for an instant – match. If we wait still longer, these matching displacements will be at their peaks. We will define the moment that this occurs as "\(t=0\)." This has the effect of simply setting the two phase constants equal to zero. We discussed above how we can redefine the phase constant to get times synchronized (\(\Delta t = 0\)) – this is just doing the same process in reverse. With the time variable defined so that the phase constants are zero, we now have a simple pair of functions to work with (note that we can drop the minus signs on the times, since \(\cos\left(x\right)=\cos\left(-x\right)\)). Superposing these and replacing the periods with frequencies gives:

    \[P_{tot}\left(t\right)=P_1\left(t\right)+P_2\left(t\right)=P_o\cos\left(2\pi f_1 t\right)+P_o\cos\left(2\pi f_2 t\right)\]

    Now apply a trigonometric identity:

    \[\cos X + \cos Y = 2\cos\left(\dfrac{X-Y}{2}\right)\cos\left(\dfrac{X+Y}{2}\right)\;\;\;\Rightarrow\;\;\; P_{tot}\left(t\right)=2P_o\cos\left[2\pi\left(\dfrac{f_1-f_2}{2}\right)t\right]\cos\left[2\pi\left(\dfrac{f_1+f_2}{2}\right)t\right]\]

    This looks like a complicated mess, but there is a reasonable way to interpret it. If we identify the second cosine function as the variation of the pressure that defines the time portion of the sound wave (often referred to as the carrier wave), the tone produced has a frequency that is the average of the two individual frequencies. The first cosine function can then be combined with \(2P_o\), and together they can be treated as a time-dependent amplitude. The amplitude is directly related to the intensity, which is a measure of the loudness of the sound, so a harmonically-oscillating amplitude would manifest itself as regularly-spaced increases and decreases in volume (beats). It's a bit easier to see how this works with a diagram.

    Figure 2.3.5 – Graph of Pressure vs. Time at a Fixed Point Exhibiting Beats

    beats_graph.png

    The blue function shown represents the fluctuations in pressure (at position \(x=0\) as a function of time as the sound wave passes through that point. Notice that the peaks have varying heights, reflecting the time-varying amplitude. The red "envelope" function traces just the amplitude of the wave. The intensity is proportional to the square of the amplitude, so the sound is loud everywhere that a red bump occurs, and silence is heard when the red function crosses the axis.

    Alert

    The volume of the sound doesn't drop to zero every time the blue graph crosses the axis! That is simply when the pressure happens to be passing through the equilibrium point, and there is still energy in a wave when the displacement of the medium passes through the equilibrium point (in the form of kinetic energy of the medium). But when the amplitude of oscillations goes to zero, the energy is zero, and in the case of sound, this means silence.

    We can ask how frequently the beats occur. A beat occurs every time a bump in the red graph occurs. Notice it doesn't matter whether the bump is up or down in the red cosine function, since the intensity is the square of the amplitude. Therefore there are two beats are heard for every full period of the red cosine function; one for the bump at the peak, and one for the bump at the trough. This means that the frequency of beats is twice the frequency of the red cosine function:

    \[f_{beat} = 2\left(\dfrac{f_1-f_2}{2}\right) = f_1-f_2\]

    [Note: We have assumed that \(f_1\) is the greater of the two frequencies, as negative frequencies have no meaning. One can remove this assumption by defining beat frequency in terms of the absolute value.] We can measure the time spans between peaks for the blue function. This is the period of the carrier wave, which we have already stated is the average of the frequencies of the two sound waves:

    \[f_{carrier} = \dfrac{f_1+f_2}{2}\]

    We only really hear the beats clearly when the two individual frequencies are close together – if there is too great of a difference, then the beat frequency is very high, and the beats come too frequently. In this case it’s impossible for the human ear to tell when one beat begins and the other ends. In that case, it sounds to the human ear more like a mixture of two sounds – one at the average frequency and one at the much lower "beat frequency."

    Example \(\PageIndex{3}\)

    A standing wave of sound persists within a hollow pipe that is open at both ends. A bug within the pipe walks along its length at a speed of \(4.0\frac{cm}{s}\). As it does so it passes through nodes and antinodes of the standing wave, hearing the tone get alternately loud and silent, with the time between moments of silence equaling 2.0s. From what we have learned about standing waves, we can compute the frequency of the tone. The distance between the nodes is one-half wavelength of the traveling sound wave, and based on the ant's speed and the time between nodes, we can determine this distance. Then with the wavelength and the speed of sound, we get the frequency:

    \[\text{distance between nodes}=v_{bug}t=\frac{\lambda}{2}\;\;\;\Rightarrow\;\;\;\lambda=2\left(4.0\frac{cm}{s}\right)\left(2.0s\right)=16cm\;\;\;\Rightarrow\;\;\;f=\frac{v}{\lambda}=\dfrac{344\frac{m}{s}}{0.16m}=2150Hz\nonumber\]

    When asked about the experience, the bug claimed it had no idea that it was walking through loud and silent regions. It described the experience as hearing periodic "beats" every 2.0 seconds in the tone. Re-derive the frequency of the sound from the perspective of the bug.

    Solution

    This problem is challenging because it's not immediately clear how beats can occur, when beats require two sound waves of different frequencies. There are two sound waves here – one reflected by each end of the pipe – but they both have the same frequency... or do they? They have the same frequency from our perspective, but not from the bug's! The bug is moving toward one end of the pipe and away from the other, so one of the waves is doppler-shifted to a higher frequency, and the other wave is doppler-shifted to a lower frequency. These waves of different frequencies interfere to provide beats from the ant's perspective. So now that we see how this makes sense, we need to do some math. Both doppler shifts involve a stationary source (the ends of the pipe) and a moving receiver (the bug):

    \[f_1=\left(\dfrac{v+v_{ant}}{v}\right)f\;\;\;\;\;\;f_2=\left(\dfrac{v-v_{ant}}{v}\right)f\nonumber\]

    The frequency \(f_1\) is what is heard by the ant from the wave coming from in front of it, and \(f_2\) is the frequency of the sound coming from behind it, and \(f\) is the frequency of the sound from the perspective of the stationary ends of the pipes (i.e. the frequency of the standing wave). The beat frequency heard by the ant is simply the difference of these two frequencies, so:

    \[f_{beat} = f_1-f_2 = \left(\dfrac{v+v_{ant}}{v}\right)f - \left(\dfrac{v-v_{ant}}{v}\right)f = 2\dfrac{v_{ant}}{v}f\nonumber\]

    We know that the bug hears silence every 2.0 seconds, so the beat frequency is \(0.50Hz\). Plugging this, the speed of the ant, and the speed of sound into this equation gives:

    \[f=\dfrac{v}{2v_{ant}}f_{beat} = \dfrac{344\frac{m}{s}}{2\left(0.040\frac{m}{s}\right)}\left(0.5Hz\right)=2150Hz\nonumber\]

    While we have worked this out for some specific numbers, it is of course true in general. If the bug was moving faster, then the pulses at the antinodes would come more frequently. From the bug's perspective the doppler shifts would greater separate the two individual sound frequencies, leading to a higher difference and faster beat frequency, and this identically equals the rate of passing through antinodes.