Waves and Superposition in 1­D

If you drop two pebbles into different parts of a pond at the same time, the ripples or wavefronts fromeach disturbance spread out and pass through each other. At some points and times there are largerripple peaks, at other deeper ripple troughs, and every combination in between. This is calledsuperposition. When two ripples (or two troughs) coincide, we have constructive interference. When aripple and a trough coincide, we have destructive interference. Superposition occurs whenever you havetwo or more sources of waves but the interference effects are most long lasting, and therefore mostnoticeable, when the waves involved have exactly the same peak­to­peak distance or wavelength l.Understanding when constructive or destructive interference occurs is easiest to understand in onedimensional, straight line, cases. Consider the case below in Figure 1 where a person is listening to amonotone sound from two speakers. If the listener is much farther away from the speakers than thespeakers are from each other in the vertical direction, we can treat the problem as one­dimensional.

Typically we graph the displacement amplitude of wave as a function of the distance from each speakeras show in Figure 2. Note that the listener is assumed to be on the right side of the graph. The mainadvantage of the graph is that a glance will tell you if the waves interference constructively ordestructively, i.e. if the peaks in the red wave line up with the peaks or with the troughs in the bluewave.

The difference in the peaks of the top and bottom waves, given in terms of the unit circle, is called thephase difference. The phase difference, and any interference effects, only exists to the right of the bluespeaker because that is the only region where both waves are overlapping. Two factors contribute to thephase difference Df and determine if the listener hears constructive or destructive interference from thespeakers. These are the difference in phase constant f20 – f10 between the waves and the phase

difference caused by distance the red wave has to travel before it overlaps with the blue wave. Themathematical expression that determines the phase difference is given below:

It doesn’t matter which speaker you label as 1 and which as 2 since only differences are important.However the top speaker in a diagram is usually treated as speaker 1. The physical separation Dx = x2 –x1 is easy to read from the graph. For example, Dx = ½l in Figure 2. As we’ve seen, it takes a littlelonger to determine the phase constant f0 from a displacement versus position graph. However, if youhave practiced with the online quizzes, you should be pretty good by now. Recall that you find thephase constant by finding the vector whose y­component matches the initial value of the displacementand whose behaviour as it as rotates counterclockwise matches the subsequent behaviour of the wave.The upper red wave is a maximum on the way to zero. It would be represented by a vector at ½p on areference circle. The lower blue wave is zero but decreasing to a minimum; it would be represented by avector at p on a reference circle. The position of the two vectors is shown in Figure 3. The difference inphase constants is f20 – f10 = p – ½p = ½p. We would say that the lower blue wave is ahead by, orleads by, ½p. One can equally say that the red wave lags, or follows, by ½p. A slightly quicker way tosee this from the graph in Figure 2 is to note that the upper red wave would have to shift ¼l to the leftfor the two speakers to have the same initial phase and that ¼l is equivalent to one quarter of the wayaround the reference circle or ½p.

To construct the reference diagram for the phase difference Df just to the right of the blue bottomspeaker, we need to account for the extra distance the red wave has to travel. The distance is ½l in thepositive direction which is equivalent to one half of the way around the reference circle. So the redvector must be rotated counterclockwise by p as shown in Figure 4. So to the right of both speakers, thephase difference is Df = f2 – f1 = p – 3/2p = –½p. The bottom blue wave, our reference, lags behindthe top red wave by ½p or, if you wish, leads by 3/2p , however you wish to look at it.

You get the same answer if you plug the values into the equation for the phase difference given above,

. Again there is a quicker way to see this from the graph in Figure 2. Pick a peak on the red wave andnote how far the nearest blue peak on the left is away as a fraction of l. In Figure 2, each blue peak is3/4l to the left of each red peak, so this indicates Df = ¾ × 2p = 3/2p. Examining Figure 2 again, one can see that that if the blue speaker is free to move, we could getconstructive interference, peak aligned with peak, if it moved 3/4l to the right or 1/4l to the left. Similarlywe could get destructive interference, peak aligned with trough, if the blue speaker moved 1/4l to theright or 3/4l to the left. In summary you should be able to answer the following questions about any 1D superposition graphsimilar to Figure 2:

What is the phase constant difference f20 – f10 between the two waves?What is the separation of the two speakers as a fraction of a wavelength and as an angle?What is the phase difference Df between the two waves?How far must the bottom speaker be moved to the right (or left) to get constructive interference?How far must the bottom speaker be moved to the right (or left) to get destructive interference?

Practice with the online superposition quizzes.