[Physics] Why are the closed and open ends of an organ pipe nodes and anti-nodes
acousticsharmonicsresonancewaves
Here is a diagram of a wave in an organ pipe you'll find in most physics books
Waves in air are longitudinal (not traversal), so what do the curves represent?
Why are the open ends always anti nodes and the closed ends always nodes?
Best Answer
There are two ways to describe a sound wave. One is in terms of displacement of the medium and the other is in terms of pressure. This simple diagram shows that tthe two descriptions are $90^\circ$ out of phase with one another.
Note that at a compression $C$ where the pressure is a maximum the displacement of the particle is zero and the same is true at a rarefaction $R$.
Since there are two ways of describing a sound waves then there are two ways of showing a standing wave pattern in a pipe. One in terms of a displacement wave and the other in terms of a pressure wave.
Note theses are graphs and the displacement of the particles is along the direction of wave propagation.
Your graphs are for displacement waves and so at a closed end there is no displacement (a node) and at the open end (or just beyond) there is maximum displacement (an antinode).
The opposite is true of the pressure wave where the open end stays at atmospheric pressure (a node) and the pressure changes most at the closed end (an antinode).
Open organ pipe is the one with two open ends, and instead of the formula you mention you need to use $$L=n\frac{v}{2f_n}$$ where $f_n$ is the frequency of the ${n^{th}}$ mode, and $n=1,2,3,...$ your formula is for a closed organ pipe (with one open and one closed end).
EDIT
Because the number of half-wavelengths ($\lambda /2$) need to be an integral multiple in case of a open pipe. This is because both the ends of an open organ pipe are pressure nodes (or displacement antinodes), and the difference between two successive nodes (or antinodes) is $\frac{\lambda}{2}$. Therefore, to meet the resonance-condition, the number of half wavelengths between the ends need to be an integral value, therefore $L=n\frac{v}{2f_n}$ as $v=f\lambda$.
If you start with the same fundamental frequency for both open and closed, say $f_0$, you'll notice that
for the open pipe, the harmonics are:
$$f_0, 2f_0, 3f_0, ...$$
While for the closed,
$$f_0, 3f_0, 5f_0, ...$$
And you'll notice that the closed pipe has greater spacing of harmonics compared to the open pipe.
That is for a given single note. You have to lower the note of the closed pipe by $\frac{1}{2}f_0$ (1 octave) to produce the same spacing of harmonics as the open pipe.
Best Answer
There are two ways to describe a sound wave. One is in terms of displacement of the medium and the other is in terms of pressure. This simple diagram shows that tthe two descriptions are $90^\circ$ out of phase with one another.
Note that at a compression $C$ where the pressure is a maximum the displacement of the particle is zero and the same is true at a rarefaction $R$.
Since there are two ways of describing a sound waves then there are two ways of showing a standing wave pattern in a pipe. One in terms of a displacement wave and the other in terms of a pressure wave.
Note theses are graphs and the displacement of the particles is along the direction of wave propagation.
Your graphs are for displacement waves and so at a closed end there is no displacement (a node) and at the open end (or just beyond) there is maximum displacement (an antinode).
The opposite is true of the pressure wave where the open end stays at atmospheric pressure (a node) and the pressure changes most at the closed end (an antinode).