Animated Solution for Physics - Waves: A cylindrical tube open at both ends, has a fundamental frequency $f$, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

Imagine you are standing in a grand concert hall, listening to the majestic sound of a pipe organ. The physics of organ pipes is essentially the physics of trapped sound—standing waves bouncing back and forth, creating beautiful, resonant frequencies. In this problem, we are going to take a standard open organ pipe and perform a fascinating physical transformation on it by dipping it into water.

Let's embark on this journey to see how the mathematics of waves perfectly mirrors physical reality.

Let's break down the open pipe first. An open organ pipe is exactly what it sounds like: a cylindrical tube open at both ends. Because the ends are open to the atmosphere, the air molecules there are free to oscillate with maximum amplitude. In the language of standing waves, an open end always forms an antinode.

If both ends are antinodes, the simplest wave pattern (the fundamental mode) that can fit inside this pipe must have exactly one node (a point of zero displacement) right in the middle.

Geometrically, the distance between two consecutive antinodes is half a wavelength, $\frac{\lambda }{2}$. Therefore, the length of the pipe $L$ is equal to $\frac{\lambda }{2}$, which means the fundamental wavelength is $\lambda =2L$.

Using the wave equation $v=f\lambda$, where $v$ is the speed of sound in air, we can find the fundamental frequency $f$:

This is our baseline. This is the frequency $f$ given in the question.

A Sudden Transformation

Now, the plot twist. We take this open pipe and dip it vertically into water such that exactly half of its length is submerged.

What does the water do? Water is significantly denser than air. To the sound waves traveling through the air column, hitting the water surface is like hitting a solid brick wall. The air molecules at the water surface cannot oscillate freely; they are physically constrained.

Therefore, the water surface acts as a rigid boundary, forcing a node to form at that exact location. The top end of the pipe remains open, so it remains an antinode.

By dipping the pipe in water, we haven't just shortened it; we have fundamentally changed its acoustic nature. It has transformed from an open organ pipe into a closed organ pipe!

Let's analyze our newly formed closed organ pipe.

First, what is the length of the air column? Since half of the original pipe is underwater, the remaining air column has a new length:

Now, we need the formula for the fundamental frequency of a closed pipe. In a closed pipe, the fundamental mode consists of a node at the closed end and an antinode at the open end. The distance between a node and an adjacent antinode is a quarter of a wavelength, $\frac{{\lambda }^{\prime }}{4}$.

So, $L^{\prime }=\frac{{\lambda }^{\prime }}{4}$, which means the new wavelength is ${\lambda }^{\prime }=4L^{\prime }$.

The new fundamental frequency $f^{\prime }$ is:

We have our formula, and we have our new length. Let's bring them together. We substitute $L^{\prime }=\frac{L}{2}$ into our frequency equation:

Watch what happens to the denominator. The $4$ and the $2$ cancel out beautifully:

Look at that result! Does it look familiar? It is exactly the same expression we derived for the original open pipe.

Since $f=\frac{v}{2L}$, we can definitively conclude:

The fundamental frequency of the air column remains absolutely unchanged. The physical act of halving the length was perfectly counterbalanced by the acoustic shift from an open pipe to a closed pipe. This is the elegance of wave mechanics—where physical constraints and algebraic logic dance in perfect harmony.