How can standing waves be produced in an open organ pipe even though both of its ends are open? Can someone explain with more clarity?
[Physics] How to standing waves be produced in an open organ pipe if the ends are open
acousticsboundary conditionsresonancewaves
Related Solutions
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 $v=f\lambda$.
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).
Best Answer
This is an attempt to explain, in a purely intuitive way why sound waves reflect from the end of an open pipe, and therefore can produce a standing wave.
Consider a pressure wave travelling up the pipe. I've drawn just a single maximum of the pressure wave to keep the diagram uncluttered:
Call the pressure maximum $P_1$ (I haven't marked $P_1$ on the diagram because it doesn't show up against the dots and I can't change the text colour in Google Draw) and the pressure minimum is $P_0$. The pressure wave moves because air flows away from the high pressure $P_1$ and towards the low pressure $P_0$. However because the air is in a tube it can only flow along the tube (green arrows) it can't flow sideways (red arrows) because the tube is in the way. This restriction to the flow will restrict the pressure difference between $P_1$ and $P_0$.
Now consider what happens when the pressure wave reaches the end of the tube:
Now the walls of the tube are no longer restricting the flow, so the air can flow away from the pressure maximum in all directions. The result is that the flow away from the maximum will be greater than it was in the tube and the pressure will fall to a lower value than the minimum pressure in the tube:
But now the air will start flowing back into the low pressure area, and because the pressure there is less than $P_0$ the rebound will produce a pressure greater than $P_1$:
And this is the last step in our argument. The next wave along travelling towards the end of the tube at a pressure $P_1$ meets a region just outside the tube with a higher pressure $P > P_1$, and the result is a reflected pressure wave back down the tube.
And that's how a pressure wave reflects from the end of an open tube.