11.3.3.1: Illustrations

Illustration 1: Representations of Two-Dimensional Waves

When we have an oscillating source on the surface of a body of water, a wave is generated that travels out in circular wave fronts in two dimensions. The amplitude of the wave (the actual direction of the waving) is in a direction that is perpendicular to the surface of the water. So how do we represent such a wave?

One way to represent such a wave is in two dimensions where the amplitude of the wave is represented by grayscale. When the wave has a positive amplitude, the color is white, when the amplitude is zero the color is light gray, and when the amplitude is negative the color is black. This is shown in the animation (position is given in centimeters and time is given in seconds). Restart.

Another way to represent a traveling wave in two dimensions is in three dimensions. After all, there are three dimensions to consider: the propagation (which accounts for two dimensions) and the direction of waving. Click the "check to see three-d mode" check box, then click the "set values and play" to see the three-dimensional representation of the wave.

Which representation do you like? In which one is it easier to "see" the wave's motion? While the three-dimensional representation is the more realistic representation, the pure two-dimensional representation that uses grayscale is certainly easier to view and determine the properties of the wave phenomena.

Illustration 2: Molecular View of a Sound Wave

A sound wave is a longitudinal wave. Restart. In a longitudinal wave, the waving of the medium (here the air molecules) is in the direction of the propagation of the wave. In show wave/hide molecules mode, we see a speaker, which is the source of the sound wave, and wave fronts propagating toward the detector (the man's ear). This is the way we usually think of sound waves: originating from a source and propagating toward a detector. But what is really going on in the medium as the sound wave goes by?

In show wave/show molecules mode, we see the individual air molecules, which are the medium in which the wave travels, and the waving of the medium. Consider the motion of the red air molecule. The molecule oscillates back and forth about its equilibrium position. If we were describing the sound wave in terms of the individual molecules, we would call the wave a displacement wave. It turns out that the amplitude of the displacement wave is only on the order of \(10^{-6}\text{ m}\)! The other way to describe the sound wave is in terms of the pressure wave that travels to the right. The pressure wave fluctuates ever so slightly about atmospheric pressure.

Illustration authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.
Script authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.

Illustration 3: Interference in Time and Beats

This animation depicts the superposition of two sound waves. A \(\color{red}{\text{red wave}}\) with \(\lambda = 3.43\text{ m}\) and a frequency of \(f = 100\text{ Hz}\) is added to a \(\color{green}{\text{green wave}}\), and the resulting wave is shown in \(\color{blue}{\text{blue}}\) (position is given in meters and time is given in seconds). Restart.

Now change the frequency of the green wave to \(120\text{ Hz}\). What is the new wavelength of the green wave? You should right-click on the graph to clone it and resize the graph to make it easier to make measurements. The result is \(2.86\text{ m}\). Did you really need to make a measurement to get the wavelength? Since the speed of sound is a constant and \(v =\lambda f\), we also know that \(\lambda = 343/f\). Notice the animation always does the right thing: As you change the frequency of the green wave, the wavelength also changes to maintain the wave speed of \(343\text{ m/s}\).

Now consider what happens when the two waves add together to create the blue wave. Enter several frequencies for the \(\color{green}{\text{green wave}}\) and see the resulting \(\color{blue}{\text{blue waves}}\). What do you see? When the two frequencies are the same, the resulting blue wave looks like the original waves but with twice the amplitude. But the resulting wave is more interesting when the frequencies (and therefore the wavelengths) do not quite match. Consider the resulting wave when the green wave's frequency is \(120\text{ Hz}\). If you were at \(x =20\text{ m}\), you would hear the sound wave getting louder and softer, louder and softer with time. When you hear this pattern, you are hearing beats. The time in between the loud sounds (or conversely the soft sounds) can be measured and is \(0.05\) seconds. This corresponds to a beat frequency of \(20\text{ Hz}\). This is precisely the difference in the frequencies! What happens when the green wave's frequency is now \(80\text{ Hz}\)? We get the same period and therefore the same beat frequency of \(20\text{ Hz}\). Therefore we find that the beat frequency is \(f_{\text{beat}} = | f_{1} - f_{2} |\).

So what is going on? Look at the underlying waves. They go in and out of phase with each other as a function of time. At one instant they exactly add together (constructively interfere) at another time they exactly cancel (destructively interfere). We therefore can say that the phenomenon of beats is due to an interference in time.

Try this out for yourself. You may vary the green wave's frequency between \(50\) and \(150\text{ Hz}\).

Illustration 4: Doppler Effect

In this Illustration we consider what happens when the source of sound is moving either toward or away from a detector at rest (position is given in meters and time is given in milliseconds). At the same time we can consider what happens when the detector is moving toward a sound source at rest. Restart.

What we notice from everyday experience is that if the source of the sound is moving toward us, the frequency we hear increases. If it is moving away, the frequency we hear drops. If we are moving toward a source, the frequency we hear increases and if we are moving away the frequency we hear drops. The reason for the difference between when the observer is moving and the source is moving is due to how the detected sound waves change in each case.

Animation 1 depicts what happens when the source of the sound wave and the detector of the sound wave are both stationary. Notice that the wavelength of the sound wave is \(1.7\text{ m}\) and its period is \(0.5\text{ ms}\), which corresponds to a frequency of \(200\text{ Hz}\).

When the observer is moving, as in Animation 2, the sound waves emitted from the source are undisturbed. The wavelength does not change as observed from the moving observer. He/she just comes across more/less wave fronts per time \(([vt ± v_{D}t]/\lambda t)\) when moving toward/away from the source, and consequently sees a change in frequency.

For the case in which the source is moving, shown in Animation 3, the frequency (time in between wave fronts) and wavelength change. The wave fronts are emitted much closer together/farther apart \((\lambda '=vT -/+ v_{S}T=[v -/+ v_{S}]/f)\) as the source is moving toward/away from us. Animation 4 represents the sound wave of a source moving according to a linear restoring force (simple harmonic motion).

We may write both these cases together, with \(v_{S}\) as the velocity of the source and \(v_{D}\) as the velocity of the observer or detector, as

\[f'=f[v\pm v_{D}]/[v-/+ v_{S}]\nonumber\]

Hence when the source is stationary and the observer/detector is moving \(f' = f [v ± v_{D}]/v\), and when the detector/observer is stationary and the source is moving \(f' = f v/[v -/+ v_{S}]\). Here the upper signs indicate a velocity towards and the lower signs represent a velocity away.

When the source is moving at the speed of sound, the emitted wave travels forward at the same speed as the source. The sound waves build up, causing a sonic boom as shown in Animation 5. For things like supersonic airplanes, we get a double boom-a boom from the front of the plane and a boom from the back. The resulting waves pile up, all on top of each other, and create at first a huge increase in pressure and then a huge decrease in pressure before the return to normal atmospheric pressure.

Illustration 5: The Location of a Supersonic Airplane

If an airplane is flying faster than the speed of sound, it will produce a shock wave called a sonic boom. In the animation, consider an airplane that is flying from point \(A\) to point \(B\). A listener, the ear, is located at point \(C\). We consider how the airplane's speed and the position of the listener affect when the sound from the airplane's engines is heard by the listener. Restart.

In the animation, you can change the airplane's speed, which we call \(v\). The speed of sound is fixed at \(343\text{ m/s}\) and we will represent it as \(v_{S}\) (shown as \(v_s\) in the animation). Press "play" to begin the animation. In addition,

• You can drag the ear across the screen to change its location.
• The program draws sound wave paths to the listener.
• The animation pauses when the sonic boom arrives at the listener; the animation can be resumed by clicking the right mouse button.
• The color of the paths of the sound waves changes to blue when those sound waves reach the listener. The order in which the sound from different paths arrives at the listener is shown as numbers located at the point that the sound was produced.
• Press "reset" for default values.

Figure \(\PageIndex{1}\)

Consider a sound generated by the airplane when it is at some point \(A\) traveling toward some point \(B\) along the straight path \(AB\). The listener hears the sound as the airplane flies toward point \(B\) \((AB > AC)\). \(DC\) is the path of a subsequent sound generated at some point \(D\) and traveling to the listener at point \(C\).

Consider a few time intervals \((\Delta t = |\Delta\mathbf{x}|/v)\). The time it takes sound to travel from \(A\) to \(C\) is \(AC/v_{S}\), the time it takes the plane to move from \(A\) to \(D\) is \(AD/v\), and the time it takes sound to travel from \(D\) to \(C\) is \(DC/v_{S}\).

Now, how does the time interval \(AC/v_{S}\) compare to the time interval \(AD/v + DC/v_{S}\)? In other words, which event happens first: the sound emitted at \(A\) reaches \(C\) or the sound emitted at \(D\) reaches \(C\)?

First consider an airplane traveling at less than or equal to the speed of sound.  \(AD/v + DC/v_{S} > AC/v_{S}\) because the path \(ADC\) is longer than the path \(AC\). The best you can do is when the time interval for \(AD\) is the smallest it can be, which is when \(v = v_{S}\). In this case comparing the two time intervals is equivalent to comparing the two paths. Clearly, \(ADC > AC\). When \(v << v_{S}\), the situation is worse and the time interval for the path \(ADC\) is even longer. Therefore, you would hear the sound from the airplane when it was at \(A\) before you heard it from when it was at point \(D\).

Now consider what you will hear if a supersonic airplane flies over you \((v > v_{S})\). Again, what you hear is dependent on whether \(AD/v + DC/v_{S}\) is greater than, less than, or the same as \(AC/v_{S}\). If \(v\) is large enough, the extra path difference, \(AD\), accounts for a smaller and smaller time interval, and since \(DC < AC\) we may hear the sound emitted at \(D\) before hearing the sound emitted at \(A\). Try it in the applet above. Set \(v\) and move the ear around. Notice when you "hear" the sounds from the airplane by looking at the numbers that show the ordering of the events.

Illustration authored by Fu-Kwun Hwang and Mario Belloni.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.