Take a rope or slinky spring. Have two people hold the rope or spring stretched out horizontally. Flick the one end of the rope up and down continuously to create a train of pulses.

Figure 1

Waves have moving peaks (or crests) and troughs. A peak is the highest point the medium rises to and a trough is the lowest point the medium sinks to.

Peaks and troughs on a transverse wave are shown in Figure 5.

Figure 5: Peaks and troughs in a transverse wave.

Definition 3: Peaks and troughs

A peak is a point on the wave where the displacement of the medium is at a maximum. A point on the wave is a trough if the displacement of the medium at that point is at a minimum.

There are a few properties that we saw with pulses that also apply to waves. These are amplitude and wavelength (we called this pulse length).

Definition 4: Amplitude

The amplitude is the maximum displacement of a particle from its equilibrium position.

Investigation : Amplitude

Figure 6

Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the wave above. Use your ruler to measure the distances.

Table 1

Peak/Trough	Measurement (cm)
a
b
c
d
e
f

1. What can you say about your results?
2. Are the distances between the equilibrium position and each peak equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium and trough?

As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position is equal to the distance between the trough and the equilibrium position. This distance is known as the amplitude of the wave, and is the characteristic height of wave, above or below the equilibrium position. Normally the symbol AA is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).

Figure 7

Investigation : Wavelength

Figure 8

Fill in the table below by measuring the distance between peaks and troughs in the wave above.

Table 2

	Distance(cm)
a
b
c
d

1. What can you say about your results?
2. Are the distances between peaks equal?
3. Are the distances between troughs equal?
4. Is the distance between peaks equal to the distance between troughs?

As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no matter which two adjacent peaks you choose. There is a fixed distance between the peaks. Similarly, we have seen that there is a fixed distance between the troughs, no matter which two troughs you look at. More importantly, the distance between two adjacent peaks is the same as the distance between two adjacent troughs. This distance is called the wavelength of the wave.

The symbol for the wavelength is λλ (the Greek letter lambda) and wavelength is measured in metres (mm).

Figure 9

Investigation : Points in Phase

Fill in the table by measuring the distance between the indicated points.

Figure 11

Table 3

Points	Distance (cm)
A to F
B to G
C to H
D to I
E to J

What do you find?

In the activity the distance between the indicated points was the same. These points are then said to be in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They do not have to be peaks or troughs, but they must be separated by a complete number of wavelengths.

We then have an alternate definition of the wavelength as the distance between any two adjacent points which are in phase.

Definition 5: Wavelength of wave

The wavelength of a wave is the distance between any two adjacent points that are in phase.

Figure 12

Points that are not in phase, those that are not separated by a complete number of wavelengths, are called out of phase. Examples of points like these would be AA and CC, or DD and EE, or BB and HH in the Activity.

Imagine you are sitting next to a pond and you watch the waves going past you. First one peak arrives, then a trough, and then another peak. Suppose you measure the time taken between one peak arriving and then the next. This time will be the same for any two successive peaks passing you. We call this time the period, and it is a characteristic of the wave.

The symbol TT is used to represent the period. The period is measured in seconds (ss).

Definition 6: Period (TT)

The period (TT) is the time taken for two successive peaks (or troughs) to pass a fixed point.

Imagine the pond again. Just as a peak passes you, you start your stopwatch and count each peak going past. After 1 second you stop the clock and stop counting. The number of peaks that you have counted in the 1 second is the frequency of the wave.

Definition 7: Frequency

The frequency is the number of successive peaks (or troughs) passing a given point in 1 second.

The frequency and the period are related to each other. As the period is the time taken for 1 peak to pass, then the number of peaks passing the point in 1 second is 1T1T. But this is the frequency. So

or alternatively,

For example, if the time between two consecutive peaks passing a fixed point is 12s12s, then the period of the wave is 12s12s. Therefore, the frequency of the wave is:

The unit of frequency is the Hertz (HzHz) or s-1s-1.

In Motion in One Dimension, we saw that speed was defined as

The distance between two successive peaks is 1 wavelength, λλ. Thus in a time of 1 period, the wave will travel 1 wavelength in distance. Thus the speed of the wave, vv, is:

However, f=1Tf=1T. Therefore, we can also write:

We call this equation the wave equation. To summarise, we have that v=λ·fv=λ·f where

The following video provides a summary of the concepts covered so far.

If you look carefully at the pictures of waves you will notice that they look very much like sine or cosine functions. This is correct. Waves can be described by trigonometric functions that are functions of time or of position. Depending on which case we are dealing with the function will be a function of tt or xx. For example, a function of position would be:

where AA is the amplitude, λλ the wavelength and φφ is a phase shift. The phase shift accounts for the fact that the wave at x=0x=0 does not start at the equilibrium position. A function of time would be:

where TT is the period of the wave. Descriptions of the wave incorporate the amplitude, wavelength, frequency or period and a phase shift.

We have seen that when a pulse meets a fixed endpoint, the pulse is reflected, but it is inverted. Since a transverse wave is a series of pulses, a transverse wave meeting a fixed endpoint is also reflected and the reflected wave is inverted. That means that the peaks and troughs are swapped around.

Figure 25: Reflection of a transverse wave from a fixed end.

If transverse waves are reflected from an end, which is free to move, the waves sent down the string are reflected but do not suffer a phase shift as shown in Figure 26.

Figure 26: Reflection of a transverse wave from a free end.

What happens when a reflected transverse wave meets an incident transverse wave? When two waves move in opposite directions, through each other, interference takes place. If the two waves have the same frequency and wavelength then standing waves are generated.

Standing waves are so-called because they appear to be standing still.

We can now look closely how standing waves are formed. Figure 27 shows a reflected wave meeting an incident wave.

Figure 27: A reflected wave (solid line) approaches the incident wave (dashed line).

When they touch, both waves have an amplitude of zero:

Figure 28: A reflected wave (solid line) meets the incident wave (dashed line).

If we wait for a short time the ends of the two waves move past each other and the waves overlap. To find the resultant wave, we add the two together.

Figure 29: A reflected wave (solid line) overlaps slightly with the incident wave (dashed line).

In this picture, we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a solid line:

Figure 30

The important thing to note in this case is that there are some points where the two waves always destructively interfere to zero. If we let the two waves move a little further we get the picture below:

Figure 31

Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like.

Figure 32

In this case the two waves have moved half a cycle past each other but because they are completely out of phase they cancel out completely.

When the waves have moved past each other so that they are overlapping for a large region the situation looks like a wave oscillating in place. The following sequence of diagrams show what the resulting wave will look like. To make it clearer, the arrows at the top of the picture show peaks where maximum positive constructive interference is taking place. The arrows at the bottom of the picture show places where maximum negative interference is taking place.

Figure 33

As time goes by the peaks become smaller and the troughs become shallower but they do not move.

Figure 34

For an instant the entire region will look completely flat.

Figure 35

The various points continue their motion in the same manner.

Figure 36

Eventually the picture looks like the complete reflection through the xx-axis of what we started with:

Figure 37

Then all the points begin to move back. Each point on the line is oscillating up and down with a different amplitude.

Figure 38

If we look at the overall result, we get a standing wave.

Figure 39: A standing wave

If we superimpose the two cases where the peaks were at a maximum and the case where the same waves were at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing wave as it contains all the oscillations of the individual points. To make the concept of the envelope clearer let us draw arrows describing the motion of points along the line.

Figure 40

Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the motion. We cannot draw the up and down arrows for every single point!

A node is a point on a wave where no displacement takes place at any time. In a standing wave, a node is a place where two waves cancel out completely as the two waves destructively interfere in the same place. A fixed end of a rope is a node. An anti-node is a point on a wave where maximum displacement takes place. In a standing wave, an anti-node is a place where the two waves constructively interfere. Anti-nodes occur midway between nodes. A free end of a rope is an anti-node.

Figure 41

Definition 8: Node

A node is a point on a standing wave where no displacement takes place at any time. A fixed end of a rope is a node.

Definition 9: Anti-Node

An anti-node is a point on standing a wave where maximum displacement takes place. A free end of a rope is an anti-node.

There are many applications which make use of the properties of waves and the use of fixed and free ends. Most musical instruments rely on the basic picture that we have presented to create specific sounds, either through standing pressure waves or standing vibratory waves in strings.

The key is to understand that a standing wave must be created in the medium that is oscillating. There are restrictions as to what wavelengths can form standing waves in a medium.

For example, if we consider a rope that can move in a pipe such that it can have

Each of these cases is slightly different because the free or fixed end determines whether a node or anti-node will form when a standing wave is created in the rope. These are the main restrictions when we determine the wavelengths of potential standing waves. These restrictions are known as boundary conditions and must be met.

In the diagram below you can see the three different cases. It is possible to create standing waves with different frequencies and wavelengths as long as the end criteria are met.

Figure 42

The longer the wavelength the less the number of anti-nodes in the standing waves. We cannot have a standing wave with no anti-nodes because then there would be no oscillations. We use nn to number the anti-nodes. If all of the tubes have a length LL and we know the end constraints we can find the wavelength, λλ, for a specific number of anti-nodes.

One Node

Let's work out the longest wavelength we can have in each tube, i.e. the case for n=1n=1.

Figure 43

Case 1: In the first tube, both ends must be anti-nodes, so we must place one node in the middle of the tube. We know the distance from one anti-node to another is 12λ12λ and we also know this distance is L. So we can equate the two and solve for the wavelength:

1 2 λ = L λ = 2 L 1 2 λ = L λ = 2 L

(19)

Case 2: In the second tube, one end must be a node and the other must be an anti-node. Since we are looking at the case with one node, we are forced to have it at the end. We know the distance from one node to another is 12λ12λ but we only have half this distance contained in the tube. So :

1 2 1 2 λ = L λ = 4 L 1 2 1 2 λ = L λ = 4 L

(20)

Case 3: Here both ends are closed and so we must have two nodes so it is impossible to construct a case with only one node.

Two Nodes

Next we determine which wavelengths could be formed if we had two nodes. Remember that we are dividing the tube up into smaller and smaller segments by having more nodes so we expect the wavelengths to get shorter.

Figure 44

Case 1: Both ends are open and so they must be anti-nodes. We can have two nodes inside the tube only if we have one anti-node contained inside the tube and one on each end. This means we have 3 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between the left side and the middle and another half wavelength between the middle and the right side so there must be one wavelength inside the tube. The safest thing to do is work out how many half wavelengths there are and equate this to the length of the tube L and then solve for λλ.

2 ( 1 2 λ ) = L λ = L 2 ( 1 2 λ ) = L λ = L

(21)

Case 2: We want to have two nodes inside the tube. The left end must be a node and the right end must be an anti-node. We can have one node inside the tube as drawn above. Again we can count the number of distances between adjacent nodes or anti-nodes. If we start from the left end we have one half wavelength between the end and the node inside the tube. The distance from the node inside the tube to the right end which is an anti-node is half of the distance to another node. So it is half of half a wavelength. Together these add up to the length of the tube:

1 2 λ + 1 2 ( 1 2 λ ) = L 2 4 λ + 1 4 λ = L 3 4 λ = L λ = 4 3 L 1 2 λ + 1 2 ( 1 2 λ ) = L 2 4 λ + 1 4 λ = L 3 4 λ = L λ = 4 3 L

(22)

Case 3: In this case both ends have to be nodes. This means that the length of the tube is one half wavelength: So we can equate the two and solve for the wavelength:

1 2 λ = L λ = 2 L 1 2 λ = L λ = 2 L

(23)

Tip:

If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember to check if your answers make sense!

Three Nodes

To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an additional node. Below is the diagram for the case where n=3n=3.

Figure 45

Case 1: Both ends are open and so they must be anti-nodes. We can have three nodes inside the tube only if we have two anti-nodes contained inside the tube and one on each end. This means we have 4 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pair of anti-nodes. We count how many gaps there are between adjacent anti-nodes to determine how many half wavelengths there are and equate this to the length of the tube L and then solve for λλ.

3 ( 1 2 λ ) = L λ = 2 3 L 3 ( 1 2 λ ) = L λ = 2 3 L

(24)

Case 2: We want to have three nodes inside the tube. The left end must be a node and the right end must be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number of distances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and an adjacent anti-node is only half the distance between adjacent nodes. So starting from the left end we count 3 nodes, so 2 half wavelength intervals and then only a node to anti-node distance:

2 ( 1 2 λ ) + 1 2 ( 1 2 λ ) = L λ + 1 4 λ = L 5 4 λ = L λ = 4 5 L 2 ( 1 2 λ ) + 1 2 ( 1 2 λ ) = L λ + 1 4 λ = L 5 4 λ = L λ = 4 5 L

(25)

Case 3: In this case both ends have to be nodes. With one node in between there are two sets of adjacent nodes. This means that the length of the tube consists of two half wavelength sections:

2 ( 1 2 λ ) = L λ = L 2 ( 1 2 λ ) = L λ = L

(26)

If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can have quite different results.

If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum of their efforts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in this case, the two waves have been trying to do the same thing, and so add together constructively. This is called constructive interference.

Figure 46

If one wave is trying to form a peak and the other is trying to form a trough, then they are competing to do different things. In this case, they can cancel out. The amplitude of the resulting wave will depend on the amplitudes of the two waves that are interfering. If the depth of the trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of the trough, a smaller peak will appear. And if the trough is deeper then a less deep trough will appear. This is destructive interference.

Figure 47