# Mathematical Relationship to find number of antinodes/nodes in a standing wave

So I have been learning about closed pipes (nodes at both ends), open pipes (antinodes at both ends) and open/closed pipes (node at one end and antinode on the other).

I have also learnt that for a closed pipe and an open pipe the length of the pipe is equivalent to half a wavelength ($L = \frac{\lambda}{2})$ and for a open/closed pipe the length is equivalent to a quarter wavelength ($L = \frac{\lambda}{4})$.

I was wondering if there is a mathematical relationship to calculate how many nodes and antinodes there are in each one of these 3 pipes (is there a similar mathematical formula for all 3 or is it the same), I tried doing some research and could not find any formula to calculate the number of nodes and antinodes in these 3 types of pipes (standing waves).

There is (in theory) no limit to the number of nodes $N$ or anti-nodes $A$, except that these numbers cannot differ by more than 1. Also, standing waves in the pipe of length $L$ can have other wavelengths $\lambda$ besides those you have stated :

• for a pipe closed at both ends $N=A+1$ and $\lambda=\frac{2L}{A}$

• for a pipe closed at one end $N=A$ and $\lambda=\frac{4L}{N}=\frac{4L}{A}$

• for a pipe open at both ends $A=N+1$ and $\lambda=\frac{2L}{N}$

• So there is no relationship that exists between (for example) the wavelength $\lambda = \frac{4L}{2n}$ and $N=A+1$ for a pipe closed at both ends that we can relate to find the number of nodes/antinodes. For example if L=1.00 and n=1 so the resulting wavelength is $2.00m$ , we can't use this value to find the number of nodes or antinodes? Commented Jul 19, 2016 at 20:52
• Yes, sorry, I will fix that. Commented Jul 19, 2016 at 20:52

The crucial first step, even before writing down any formula, is to visualize the standing waves. Doing this we will see, there is not only one wavelength possible, but many different.

#### Pipe with both ends closed

The first few standing waves (i.e. those with the longest wavelengths $$\lambda$$) in a pipe of length $$L$$ look like this:

So we have $$L = \frac 12\lambda,\ \frac 22\lambda,\ \frac 32\lambda,\ ...$$ which we can write as $$L = \frac{n}{2}\lambda \quad \text{with }n=1,2,3,...,\infty$$ The number $$n$$ can be arbitrarily large. Then we get more and more nodes and anti-nodes between the ends, and hence standing waves with shorter and shorter wavelengths.

### Pipe with both ends open

Now the first few standing waves look like this:

So we have $$L = \frac 12\lambda,\frac 22\lambda,\frac 32\lambda,...$$ or $$L = \frac{n}{2}\lambda \quad \text{with }n=1,2,3,...,\infty$$ which is the same formula as for the pipe with both ends closed.

#### Pipe with the left end closed and the right end open

Then the first few standing waves look like this:

So we have $$L = \frac 14\lambda,\frac 34\lambda,\frac 54\lambda,...$$ which we can write as $$L=\frac{2n-1}{4}\lambda \quad \text{with }n=1,2,3,...,\infty$$ Obviously this formula is different from the previous two cases.