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Standing Waves and Boundary Conditions

Reflection of a Transverse Wave from a Fixed End

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 1: Reflection of a transverse wave from a fixed end.
Figure 1 (PG10C5_023.png)

Reflection of a Transverse Wave from a Free 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 2.

Figure 2: Reflection of a transverse wave from a free end.
Figure 2 (PG10C5_024.png)

Standing Waves

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.

Investigation : Creating Standing Waves

Tie a rope to a fixed object such that the tied end does not move. Continuously move the free end up and down to generate firstly transverse waves and later standing waves.

We can now look closely how standing waves are formed. Figure 3 shows a reflected wave meeting an incident wave.

Figure 3: A reflected wave (solid line) approaches the incident wave (dashed line).
Figure 3 (PG10C5_025.png)

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

Figure 4: A reflected wave (solid line) meets the incident wave (dashed line).
Figure 4 (PG10C5_026.png)

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 5: A reflected wave (solid line) overlaps slightly with the incident wave (dashed line).
Figure 5 (PG10C5_027.png)

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 6
Figure 6 (PG10C5_028.png)

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 7
Figure 7 (PG10C5_029.png)

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

Figure 8
Figure 8 (PG10C5_030.png)

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 9
Figure 9 (PG10C5_031.png)

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

Figure 10
Figure 10 (PG10C5_032.png)

For an instant the entire region will look completely flat.

Figure 11
Figure 11 (PG10C5_033.png)

The various points continue their motion in the same manner.

Figure 12
Figure 12 (PG10C5_034.png)

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

Figure 13
Figure 13 (PG10C5_035.png)

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

Figure 14
Figure 14 (PG10C5_036.png)

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

Figure 15: A standing wave
Figure 15 (PG10C5_037.png)

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 16
Figure 16 (PG10C5_038.png)

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!

Note: Interesting Fact :

Standing waves can be a problem in for example indoor concerts where the dimensions of the concert venue coincide with particular wavelengths. Standing waves can appear as `feedback', which would occur if the standing wave was picked up by the microphones on stage and amplified.

Nodes and Anti-nodes

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 17
Figure 17 (PG10C5_039.png)
Definition 1: 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 2: 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.

Tip:

The distance between two anti-nodes is only 12λ12λ because it is the distance from a peak to a trough in one of the waves forming the standing wave. It is the same as the distance between two adjacent nodes. This will be important when we work out the allowed wavelengths in tubes later. We can take this further because half-way between any two anti-nodes is a node. Then the distance from the node to the anti-node is half the distance between two anti-nodes. This is half of half a wavelength which is one quarter of a wavelength, 14λ14λ.

Wavelengths of Standing Waves with Fixed and Free Ends

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

  • both ends free to move (Case 1)
  • one end free and one end fixed (Case 2)
  • both ends fixed (Case 3).

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 18
Figure 18 (PG10C5_040.png)

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 19
Figure 19 (PG10C5_041.png)

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
(1)

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
(2)

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 20
Figure 20 (PG10C5_042.png)

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
(3)

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
(4)

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
(5)
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 21
Figure 21 (PG10C5_043.png)

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
(6)

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
(7)

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
(8)

Superposition and Interference

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 22
Figure 22 (PG10C5_044.png)

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 23
Figure 23 (PG10C5_045.png)

Superposition and Interference

  1. For each labelled point, indicate whether constructive or destructive interference takes place at that point.
    Figure 24
    Figure 24 (g10c14p_waves1.png)
    Figure 25
    Figure 25 (g10c14p_waves2.png)
    Click here for the solution.
  2. A ride at the local amusement park is called "Standing on Standing Waves". Which position (a node or an antinode) on the ride would give the greatest thrill?
    Click here for the solution.
  3. How many nodes and how many anti-nodes appear in the standing wave below?
    Figure 26
    Figure 26 (PG10C5_047.png)
    Click here for the solution.
  4. For a standing wave on a string, you are given three statements:
    1. you can have any λλ and any ff as long as the relationship, v=λ·fv=λ·f is satisfied.
    2. only certain wavelengths and frequencies are allowed
    3. the wave velocity is only dependent on the medium
    Which of the statements are true:
    1. A and C only
    2. B and C only
    3. A, B, and C
    4. none of the above
    Click here for the solution.
  5. Consider the diagram below of a standing wave on a string 9 m long that is tied at both ends. The wave velocity in the string is 16 m··s-1-1. What is the wavelength?
    Figure 27
    Figure 27 (PG10C5_048.png)
    Click here for the solution.

Summary

  1. A wave is formed when a continuous number of pulses are transmitted through a medium.
  2. A peak is the highest point a particle in the medium rises to.
  3. A trough is the lowest point a particle in the medium sinks to.
  4. In a transverse wave, the particles move perpendicular to the motion of the wave.
  5. The amplitude is the maximum distance from equilibrium position to a peak (or trough), or the maximum displacement of a particle in a wave from its position of rest.
  6. The wavelength (λλ) is the distance between any two adjacent points on a wave that are in phase. It is measured in metres.
  7. The period (TT) of a wave is the time it takes a wavelength to pass a fixed point. It is measured in seconds (s).
  8. The frequency (ff) of a wave is how many waves pass a point in a second. It is measured in hertz (Hz) or s-1s-1.
  9. Frequency: f=1Tf=1T
  10. Period: T=1fT=1f
  11. Speed: v=fλv=fλ or v=λTv=λT.
  12. When a wave is reflected from a fixed end, the resulting wave will move back through the medium, but will be inverted. When a wave is reflected from a free end, the waves are reflected, but not inverted.

Exercises

  1. A standing wave is formed when:
    1. a wave refracts due to changes in the properties of the medium
    2. a wave reflects off a canyon wall and is heard shortly after it is formed
    3. a wave refracts and reflects due to changes in the medium
    4. two identical waves moving different directions along the same medium interfere
    Click here for the solution.
  2. How many nodes and anti-nodes are shown in the diagram?
    Figure 28
    Figure 28 (PG10C5_049.png)
    Click here for the solution.
  3. Draw a transverse wave that is reflected from a fixed end.
    Click here for the solution.
  4. Draw a transverse wave that is reflected from a free end.
    Click here for the solution.
  5. A wave travels along a string at a speed of 1,5m·s-11,5m·s-1. If the frequency of the source of the wave is 7,5 Hz, calculate:
    1. the wavelength of the wave
    2. the period of the wave
    Click here for the solution.
  6. Water waves crash against a seawall around the harbour. Eight waves hit the seawall in 5 s. The distance between successive troughs is 9 m. The height of the waveform trough to crest is 1,5 m.
    Figure 29
    Figure 29 (seawall.png)
    1. How many complete waves are indicated in the sketch?
    2. Write down the letters that indicate any TWO points that are:
      1. in phase
      2. out of phase
      3. Represent ONE wavelength.
    3. Calculate the amplitude of the wave.
    4. Show that the period of the wave is 0,67 s.
    5. Calculate the frequency of the waves.
    6. Calculate the velocity of the waves.
    Click here for the solution.

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