Animated Solution for Physics - Waves: Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes $A$ and $B$ is

When you blow air across the top of a bottle or play a flute, you are creating music through the physics of standing waves. Organ pipes are classic examples of this phenomenon. The pitch or frequency of the sound produced depends entirely on the length of the pipe and whether its ends are open or closed.

In this problem, we are comparing two identical pipes, each of length $l$. Tube A is open at both ends, while Tube B is closed at one end. Our goal is to find the ratio of their fundamental frequencies. The fundamental frequency is the lowest possible frequency of a standing wave that can form inside the pipe.

The Open-Open Pipe

Let's start with Tube A. Because both ends are open to the atmosphere, the air molecules are free to vibrate maximally at these boundaries. In the language of waves, an open end always forms a displacement antinode.

For the fundamental mode (the simplest wave pattern), there must be exactly one node (a point of zero displacement) right in the middle of the pipe. The distance between two consecutive antinodes is exactly half a wavelength. Therefore, the length of the pipe $l$ accommodates half of the wavelength ${\lambda }_A$:

This means the wavelength is ${\lambda }_A=2l$. Using the universal wave equation $v=n\lambda$, where $v$ is the speed of sound and $n_A$ is the frequency, we can express the fundamental frequency of Tube A as:

The Closed-Open Pipe

Now, let's examine Tube B. One end is closed, acting like a rigid wall. The air molecules cannot move here, so the closed end strictly forms a displacement node. The other end is open, forming an antinode.

The shortest distance between a node and an antinode is one-quarter of a wavelength. Thus, for the fundamental mode of this closed-open pipe, the length $l$ fits exactly a quarter of the wavelength ${\lambda }_B$:

This gives us a wavelength of ${\lambda }_B=4l$. Notice that this wavelength is twice as long as the one in Tube A! Substituting this into the wave equation, the fundamental frequency for Tube B is:

We now have the fundamental frequencies for both tubes. The question asks for the ratio of the fundamental frequency of Tube A to that of Tube B. Let's set up the ratio:

Since both tubes are identical in length and contain the same air, the speed of sound $v$ and the length $l$ are constant. We can simplify the complex fraction by multiplying by the reciprocal:

The $v$ and $l$ terms cancel out beautifully, leaving us with:

The ratio of their fundamental frequencies is $2:1$. This tells us a fascinating musical fact: an open pipe will always sound exactly one octave higher than a closed pipe of the exact same length!