1/ Nodes and anti-nodes A node is a point on a wave where no displacement takes place. In a standing wave, a node is a place where the two waves cancel out completely as 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. A free end of a rope is an anti-node. Definition: Node A node is a point on a wave where no displacement takes place. In a standing wave, a node is a place where the two waves cancel out completely as two waves destructively interfere in the same place. A fixed end of a rope is a node Definition: Anti-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. A free end of a rope is an anti-node. Important: The distance between two anti-nodes is only 1/2λ 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, 1/4 λ. 2/ 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. 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 n to number the anti-nodes. If all of the tubes have a length L and we know the end constraints we can find the wavelength, λ, 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 = 1. Case 1: In the first tube, both ends must be nodes, so we can place one anti-node in the middle of the tube. We know the distance from one node to another is 1/2 λ and we also know this distance is L. So we can equate the two and solve for the wavelength: 1/2 λ = L = > λ = 2L Case 2: In the second tube, one end must be a node and the other must be an anti-node. We are looking at the case with one anti-node we are forced to have it at the end. We know the distance from one node to another is 1/2 λ but we only have half this distance contained in the tube. So : 1/2 (1/2 λ) = L = > λ = 4L 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. 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 λ. 2(1/2 λ) = L => λ = L 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 => λ = 4L/3 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 = 2L Important: 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 = 3. 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 λ. 3 (1/2 λ) = L => λ = 2L/3 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 = 4L/5 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 3/ 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. 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. Exercise: Superposition and Interference E - 1. For each labelled point, indicate whether constructive or destructive interference takes place at that point. E - 2. A ride at the local amusement park is called ”Standing on Waves”. Which position (a node or an antinode) on the ride would give the greatest thrill? E - 3. How many nodes and how many anti-nodes appear in the standing wave below? E - 4. For a standing wave on a string, you are given three statements: A you can have any λ and any f as long as the relationship, v = λ · f is satisfied. B only certain wavelengths and frequencies are allowed C the wave velocity is only dependent on the medium Which of the statements are true: (a) A and C only (b) B and C only (c) A, B, and C (d) none of the above E - 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. What is the wavelength? Xem thêm: Physics 10.V Transverse Wave High School Students Studying the Sciences Physics