A tube, closed at one end and containing air, produces, when excited, the fundamental note of freque

Q: Why is there a node at the closed end and an antinode at the open end?

At the closed end, the rigid boundary physically prevents air molecules from moving, forcing a displacement node. At the open end, the air is free to move in and out of the atmosphere, creating a displacement antinode.

Q: Does the speed of sound change when the tube is opened?

No, the speed of sound $v$ depends only on the properties of the medium (air temperature, pressure, and composition), which remain constant.

Q: What is the difference between displacement waves and pressure waves in organ pipes?

They are $90^\circ$ out of phase. A displacement node corresponds to a pressure antinode, and a displacement antinode corresponds to a pressure node.

Q: Why does doubling the frequency sound like the same note but higher?

In music, doubling the frequency corresponds to an interval of one octave, which sounds highly harmonious because the peaks of the higher wave align perfectly with every second peak of the lower wave.

Q: Would the end correction affect this result if it were not neglected?

Yes, if end correction $e$ is considered, the effective lengths would be $L + 0.6r$ for the closed pipe and $L + 1.2r$ for the open pipe, which would slightly alter the exact $2:1$ ratio.

Question

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency $512	ext{ Hz}$. If the tube is opened at both ends the fundamental frequency that can be excited is (in Hz)

Options:

(A) $1024$

(B) $512$

(C) $256$

(D) $128$

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Sound and Wind

Have you ever wondered how wind instruments like flutes, panpipes, or massive pipe organs produce such rich, beautiful notes?

At the heart of every wind instrument lies a simple, elegant physical system: **the organ pipe**.

By blowing air across the edge of a tube, we create turbulence that excites the air column inside, setting up standing waves.

In this problem, we explore how a simple change in the boundary conditions of a tube—opening a closed end—dramatically alters the pitch of the sound it produces.

### Analyzing the Setup: Closed vs. Open Pipes

To understand standing waves in tubes, we must first look at the boundaries.

At a **closed end**, the air molecules are physically blocked by a rigid wall.

Because they cannot move, the displacement of the air molecules at this boundary must always be zero, creating a **displacement node** ($N$).

At an **open end**, the air is in direct contact with the atmosphere, allowing the molecules to vibrate with maximum freedom.

This creates a **displacement antinode** ($A$) at the open boundary.

### The Closed Pipe: Our Starting Point

Let's begin with the initial state of our tube: closed at one end and open at the other.

For the fundamental mode (the simplest standing wave pattern), we have a node at the closed end and an antinode at the open end.

This pattern represents exactly one-quarter of a full wave cycle:

$$L = \frac{\lambda_c}{4} \implies \lambda_c = 4L$$

Using the fundamental wave relation $v = f \lambda$, where $v$ is the speed of sound in air, we can write the fundamental frequency of the closed pipe ($f_c$) as:

$$f_c = \frac{v}{4L}$$

We are given that this frequency is $512	ext{ Hz}$:

$$\frac{v}{4L} = 512	ext{ Hz}$$

### Opening the Gates: The Open Pipe

Now, let's open the closed end of the tube, making it open at both ends.

Since both ends are now open to the atmosphere, we must have displacement antinodes at both boundaries.

For the fundamental mode, the simplest wave pattern that fits this condition has an antinode at each end and a single node in the center.

This pattern represents exactly half of a full wave cycle:

$$L = \frac{\lambda_o}{2} \implies \lambda_o = 2L$$

Using the wave relation again, the fundamental frequency of the open pipe ($f_o$) is:

$$f_o = \frac{v}{2L}$$

### The Beautiful Symmetry: Comparing the Two

Let's compare our two frequency equations:

$$f_c = \frac{v}{4L}$$

$$f_o = \frac{v}{2L}$$

Notice the stunning mathematical relationship here:

$$f_o = 2 \left( \frac{v}{4L} ight) = 2 f_c$$

By simply opening the closed end of the tube, we have **doubled** its fundamental frequency!

In musical terms, doubling the frequency corresponds to raising the pitch by exactly **one octave**.

### Final Calculation and Conclusion

Now, we substitute our given value of $f_c = 512	ext{ Hz}$ into our relationship:

$$f_o = 2 	imes 512	ext{ Hz} = 1024	ext{ Hz}$$

Thus, the fundamental frequency that can be excited when the tube is opened at both ends is **$1024	ext{ Hz}$**.

This perfectly matches **Option (a)**.

Animated Solution for Physics - Waves: A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency $512 Hz$ . If the tube is opened at both ends the fundamental frequency that can be excited is (in Hz)

Select Answer:

Visualized Solution

Visualizing the Closed Organ Pipe

Fundamental Frequency of a Closed Pipe

Substituting the Given Value

Visualizing the Open Organ Pipe

Fundamental Frequency of an Open Pipe

Relating Open and Closed Pipe Frequencies

Calculating the Final Frequency

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

Analyzing the Setup

The Closed Pipe

Opening the Gates

The Beautiful Symmetry

Final Calculation and Conclusion

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