Column I shows four systems, each of the same length L, for producing standing waves. The lowest pos

Q: Why are sound waves in pipes always longitudinal?

Sound waves in air columns are longitudinal because air is a fluid (gas) and lacks shear elasticity. Fluids cannot support transverse shear stresses, so waves can only propagate through compressions and rarefactions parallel to the direction of wave travel.

Q: Why is an open end of a pipe considered a displacement antinode?

At the open end of a pipe, the air is in direct contact with the atmosphere, allowing air molecules to move in and out with maximum freedom. This maximum displacement amplitude forms a displacement antinode.

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

They are completely out of phase (by $90^\circ$ or $\pi/2$ radians). A displacement node (where air molecules cannot move, e.g., at a closed end) is a pressure antinode (where pressure variations are maximum). Conversely, a displacement antinode (e.g., at an open end) is a pressure node (where pressure remains constant at atmospheric pressure).

Q: How does clamping the midpoint of a wire change its fundamental frequency?

Clamping the midpoint forces a node to form at $x = L/2$. This divides the wire into two segments of length $L/2$. Since the effective length of the vibrating segment is halved, the fundamental wavelength is halved (from $2L$ to $L$), which doubles the fundamental frequency.

Q: Why does a pipe open at both ends have the same fundamental wavelength formula as a string clamped at both ends?

Both systems have symmetric boundary conditions at their ends. The open pipe has antinodes at both ends, and the clamped string has nodes at both ends. In both cases, the distance between these identical boundary states in the fundamental mode is exactly half a wavelength ($\lambda_f/2 = L$).

Question

Column I shows four systems, each of the same length $L$, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda_f$. Match each system with statements given in Column II describing the nature and wavelength of the standing waves.

List-I:

P. Pipe closed at one end

Q. Pipe open at both ends

R. Stretched wire clamped at both ends

S. Stretched wire clamped at both ends and at mid-point

List-II:

1. Longitudinal waves

2. Transverse waves

3. $\lambda_f = L$

4. $\lambda_f = 2L$

5. $\lambda_f = 4L$

Sigma Percentile · Accepted Answer

### Introduction to Standing Waves

Standing waves, or stationary waves, represent one of the most visually and mathematically elegant phenomena in classical wave mechanics.

Unlike progressive waves that transport energy across space, standing waves store energy within localized regions.

They are formed by the superposition of two identical waves traveling in opposite directions, typically due to reflections at boundaries.

In this problem, we explore four classic systems of length $L$ that support standing waves: two acoustic air columns (organ pipes) and two mechanical strings (stretched wires).

Our objective is to classify the nature of these waves (longitudinal vs. transverse) and determine their fundamental wavelengths $\lambda_f$.

---

### Analyzing the Nature of Waves

Before diving into the mathematics of boundary conditions, let us establish the physical nature of the waves in each medium.

#### Acoustic Air Columns (Pipes)

In organ pipes, the medium of propagation is air.

Sound waves in gases are strictly **longitudinal waves**.

This is because gases lack shear strength and cannot support transverse shear stresses.

As sound travels through the pipe, air molecules oscillate back and forth parallel to the length of the tube.

This parallel oscillation creates alternating regions of high pressure (compressions) and low pressure (rarefactions).

Thus, both the closed pipe (A) and the open pipe (B) support longitudinal waves.

This immediately maps **A** and **B** to **(p)** in Column II.

#### Mechanical Strings (Wires)

In stretched wires, the wave is initiated by plucking or vibrating the string.

The particles of the wire oscillate up and down (or side to side) perpendicular to the direction of wave propagation along the length of the wire.

This perpendicular motion defines a **transverse wave**.

Thus, both the clamped wire (C) and the mid-clamped wire (D) support transverse waves.

This maps **C** and **D** to **(q)** in Column II.

---

### Determining Fundamental Wavelengths

To find the fundamental wavelength $\lambda_f$ for each system, we must apply the appropriate boundary conditions at the ends of the medium of length $L$.

#### System A: Pipe Closed at One End

For a pipe closed at the left end ($x=0$) and open at the right end ($x=L$):

1. At the closed end, air molecules are physically blocked by the rigid wall, forcing their displacement to be zero. This creates a displacement **node** ($N$).
2. At the open end, air molecules are free to move with maximum freedom, creating a displacement **antinode** ($A$).

The fundamental mode of vibration represents the simplest possible wave pattern that satisfies these boundary conditions: a single node at one end and a single antinode at the other.

The distance between a node and its adjacent antinode is exactly one-quarter of a wavelength:

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

Therefore, System A matches with **(p)** and **(t)**.

$$\mathbf{A \rightarrow p, t}$$

#### System B: Pipe Open at Both Ends

For a pipe open at both ends ($x=0$ and $x=L$):

1. Both open ends must be displacement **antinodes** ($A$) because the air is free to vibrate at these boundaries.
2. To connect two antinodes, there must be at least one displacement **node** ($N$) in the middle ($x=L/2$).

The distance between two adjacent antinodes is exactly half a wavelength:

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

Therefore, System B matches with **(p)** and **(s)**.

$$\mathbf{B \rightarrow p, s}$$

#### System C: Stretched Wire Clamped at Both Ends

For a wire rigidly clamped at both ends ($x=0$ and $x=L$):

1. Both clamped ends are fixed in space, meaning their displacement must be zero. This creates **nodes** ($N$) at both ends.
2. The fundamental mode of vibration consists of a single vibrating loop with a maximum displacement **antinode** ($A$) in the middle ($x=L/2$).

The distance between two adjacent nodes is exactly half a wavelength:

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

Therefore, System C matches with **(q)** and **(s)**.

$$\mathbf{C \rightarrow q, s}$$

#### System D: Stretched Wire Clamped at Both Ends and at Mid-point

For a wire clamped at both ends ($x=0, L$) and also supported/clamped at its midpoint ($x=L/2$):

1. Rigid clamps force **nodes** ($N$) to form at $x=0$, $x=L/2$, and $x=L$.
2. This effectively divides the wire into two independent vibrating segments, each of length $L/2$.
3. Each segment vibrates in its own fundamental mode, with an antinode in the middle of each segment.

For each segment of length $L/2$, the distance between its boundary nodes is half of its fundamental wavelength:

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

Therefore, System D matches with **(q)** and **(r)**.

$$\mathbf{D \rightarrow q, r}$$

---

### Summary of Matches

By systematically applying physical principles and boundary conditions, we have established the complete set of matches:

* **A $\rightarrow$ p, t** (Longitudinal wave, $\lambda_f = 4L$)
* **B $\rightarrow$ p, s** (Longitudinal wave, $\lambda_f = 2L$)
* **C $\rightarrow$ q, s** (Transverse wave, $\lambda_f = 2L$)
* **D $\rightarrow$ q, r** (Transverse wave, $\lambda_f = L$)

List-I

List-II

Select Matching Pairs:

Visualized Solution

Understanding Standing Wave Systems

Nature of Waves in Pipes

Nature of Waves in Stretched Wires

Analyzing System A: Closed Pipe

Calculating $λ_{f}$ for Closed Pipe

Analyzing System B: Open Pipe

Analyzing System C: Clamped Wire

Analyzing System D: Mid-clamped Wire

Final Matrix Match Results

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves

Analyzing the Nature of Waves

# Acoustic Air Columns (Pipes)

# Mechanical Strings (Wires)

Determining Fundamental Wavelengths

# System A

# System B

# System C

# System D

Summary of Matches

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Visualized Solution

Understanding Standing Wave Systems

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Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λf​ for Closed Pipe

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Analyzing System C: Clamped Wire

Analyzing System D: Mid-clamped Wire

Final Matrix Match Results

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves

Analyzing the Nature of Waves

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