# 0.4 Transverse waves  (Page 9/10)

 Page 9 / 10
$\begin{array}{ccc}\hfill \frac{1}{2}\lambda +\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \frac{2}{4}\lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{3}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{3}L\hfill \end{array}$

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

$\begin{array}{ccc}\hfill \frac{1}{2}\lambda & =& L\hfill \\ \hfill \lambda & =& 2L\hfill \end{array}$
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 casewhere $n=3$ .

Case 1 : Both ends are open and so they must be anti-nodes. We can have threenodes 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 thetube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pairof 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 $\lambda$ .

$\begin{array}{ccc}\hfill 3\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& \frac{2}{3}L\hfill \end{array}$

Case 2 : We want to have three nodes inside the tube. The left end must be anode 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 ofdistances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and anadjacent 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 anode to anti-node distance:

$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)+\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{5}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{5}L\hfill \end{array}$

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

$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& L\hfill \end{array}$

## 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 tryto 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 oftheir 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 areboth 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 twowaves. 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. Inthis 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 ofthe 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 thetrough, a smaller peak will appear. And if the trough is deeper then a less deep trough will appear. This is destructive interference .

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