The air column in a pipe closed at one end is made to vibrate in its second overtone by tuning fork

Q: Why does a closed organ pipe only have odd harmonics?

A closed organ pipe has asymmetric boundary conditions: a node at the closed end and an antinode at the open end (for displacement). This geometry only allows an odd number of quarter-wavelengths to fit inside the pipe, resulting in only odd harmonics ($1^{\text{st}}, 3^{\text{rd}}, 5^{\text{th}}, \dots$).

Q: Why is the open end of a pipe a pressure node?

The open end of the pipe is in direct contact with the atmosphere, which is a massive reservoir of air at constant pressure $p_0$. Any local pressure change is instantly equalized, keeping the pressure variation $\Delta p$ at zero.

Q: How do we know the second overtone is the 5th harmonic?

Overtones are the resonant frequencies above the fundamental frequency. For a closed pipe, the fundamental is the $1^{\text{st}}$ harmonic. The first overtone is the next possible frequency, which is the $3^{\text{rd}}$ harmonic. The second overtone is the next, which is the $5^{\text{th}}$ harmonic.

Q: Why is the pressure variation at the midpoint not zero?

The midpoint of the pipe ($x = L/2$) does not coincide with any of the pressure nodes. For the $5^{\text{th}}$ harmonic, the nodes are at $x = 0, 0.4L, \text{ and } 0.8L$. Since $0.5L$ is not a node, there is a non-zero pressure variation of $\pm \Delta p_0 / \sqrt{2}$.

Q: What is the difference between displacement nodes and pressure nodes?

They are $90^\circ$ out of phase. A displacement node (where air molecules cannot move, such as at a closed wall) corresponds to a pressure antinode (where molecules pile up, causing maximum pressure variation). A displacement antinode corresponds to a pressure node.

Question

The air column in a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency $440	ext{ Hz}$. The speed of sound in air is $330	ext{ m/s}$. End corrections may be neglected. Let $p_0$ denote the mean pressure at any point in the pipe and $\Delta p_0$ the maximum amplitude of pressure variation.
(a) Find the length $L$ of the air column.
(b) What is the amplitude of pressure variation at the middle of the column?
(c) What are the maximum and minimum pressures at the open end of the pipe?
(d) What are the maximum and minimum pressures at the closed end of the pipe?

Sigma Percentile · Accepted Answer

### The Magic of Standing Waves in Pipes

Imagine standing inside a grand cathedral, listening to the deep, resonant notes of a pipe organ.

Have you ever wondered how these magnificent instruments produce such pure, powerful sounds?

The answer lies in the physics of **standing waves** and **acoustic resonance**.

When a sound wave travels down a pipe, it reflects off the ends, creating waves that travel in opposite directions.

Under the right conditions, these waves interfere constructively, forming a stable pattern of nodes and antinodes known as a standing wave.

In this problem, we will explore a classic JEE Advanced question from 1998 that dives deep into the pressure variations inside a closed organ pipe.

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### Boundary Conditions: The Rules of the Game

Before we jump into the mathematics, let us establish the physical constraints of our system.

We are dealing with a cylindrical pipe of length $L$ that is closed at one end and open at the other.

At the **open end**, the air is in direct contact with the outside atmosphere.

Because the atmosphere is vast, any local pressure variation is immediately equalized.

Therefore, the open end must always be a **pressure node**, where the pressure variation is strictly zero:

$$\Delta p = 0$$

Conversely, at the **closed end**, the rigid wall completely blocks the motion of air molecules.

As the molecules rush toward the wall, they pile up, creating maximum compression, and as they pull away, they create maximum rarefaction.

This makes the closed end a **pressure antinode**, where the pressure variation reaches its maximum amplitude:

$$\Delta p = \pm \Delta p_0$$

These boundary conditions are the foundation of our entire analysis.

---

### Deciphering the Harmonic: The Second Overtone

For a pipe closed at one end, the asymmetric boundary conditions mean that only **odd harmonics** can exist.

Let's list them out to be absolutely clear:

1. The **fundamental mode** (1st harmonic) has a frequency of $f_1 = \frac{v}{4L}$.
2. The **first overtone** (3rd harmonic) has a frequency of $f_3 = \frac{3v}{4L}$.
3. The **second overtone** (5th harmonic) has a frequency of $f_5 = \frac{5v}{4L}$.

Since our pipe is vibrating in its second overtone, it must be vibrating in its **fifth harmonic**.

This means the length of the pipe $L$ must accommodate exactly five quarter-wavelengths:

$$L = \frac{5\lambda}{4}$$

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### Part (a): Finding the Length of the Air Column

We are given the frequency of the tuning fork $f = 440	ext{ Hz}$ and the speed of sound in air $v = 330	ext{ m/s}$.

First, let us calculate the wavelength $\lambda$ of the sound wave using the fundamental wave relation:

$$\lambda = \frac{v}{f} = \frac{330}{440} = \frac{3}{4}	ext{ m} = 0.75	ext{ m}$$

Now, substituting this wavelength into our harmonic relation, we find the length $L$ of the air column:

$$L = \frac{5}{4} 	imes \frac{3}{4} = \frac{15}{16}	ext{ m} = 0.9375	ext{ m}$$

This is a beautifully clean result, showing that our pipe is just under one meter long!

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### Part (b): The Midpoint Pressure Variation

Now, let us find the amplitude of pressure variation at the exact middle of the column, $x = \frac{L}{2}$.

Using a coordinate system with $x = 0$ at the open end, the pressure variation at any point $x$ is given by:

$$\Delta p(x) = \pm \Delta p_0 \sin(kx)$$

Where the wave number $k$ is:

$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{3/4} = \frac{8\pi}{3}	ext{ rad/m}$$

Substituting $x = \frac{L}{2} = \frac{15}{32}	ext{ m}$ into our sine term, we get:

$$	heta = kx = \left(\frac{8\pi}{3}ight) \left(\frac{15}{32}ight) = \frac{5\pi}{4}	ext{ rad}$$

Now, we evaluate the sine of this angle:

$$\sin\left(\frac{5\pi}{4}ight) = -\frac{1}{\sqrt{2}}$$

Since amplitude represents the maximum magnitude of variation, we write the pressure variation amplitude at the midpoint as:

$$\Delta p_{	ext{mid}} = \pm \frac{\Delta p_0}{\sqrt{2}}$$

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### Parts (c) & (d): Analyzing the Extremes

Finally, let us look at the absolute pressures at the two ends of the pipe.

At the **open end** ($x = 0$), there is no pressure variation.

Therefore, the pressure remains constant at the mean atmospheric pressure $p_0$ at all times.

$$	ext{Maximum Pressure} = 	ext{Minimum Pressure} = p_0$$

At the **closed end** ($x = L$), the pressure variation is at its maximum amplitude, $\pm \Delta p_0$.

This means the absolute pressure oscillates between a maximum during compression and a minimum during rarefaction:

$$	ext{Maximum Pressure} = p_0 + \Delta p_0$$

$$	ext{Minimum Pressure} = p_0 - \Delta p_0$$

This elegant analysis completes our journey through the physics of resonant air columns!

Visualized Solution

Visualizing the Closed Organ Pipe & Boundary Conditions

Identifying the Vibration Mode (Second Overtone)

Calculating the Wavelength of the Sound Wave

Determining the Length of the Air Column (Part a)

Formulating the Pressure Variation Equation

Setting up the Pressure Variation at the Midpoint (Part b)

Calculating the Midpoint Pressure Amplitude

Analyzing Pressures at the Open End (Part c)

Analyzing Pressures at the Closed End (Part d)

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Magic of Standing Waves in Pipes

Boundary Conditions

Deciphering the Harmonic

Part (a)

Part (b)

Parts (c) & (d)

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