An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fo

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

Because one end is closed (node) and the other is open (antinode), the wave pattern inside must always fit an odd number of quarter-wavelengths. Even harmonics would require either nodes at both ends or antinodes at both ends, which violates the physical boundary conditions of a closed pipe.

Q: What is the difference between a displacement wave and a pressure wave in this context?

They are $90^\circ$ out of phase. A displacement node (where air molecules don't move, like at the closed wall) is a pressure antinode (where pressure variations are maximum). Conversely, a displacement antinode (at the open end) is a pressure node (where pressure remains equal to atmospheric pressure).

Q: Does the diameter of the pipe affect the resonant frequency?

Yes, in real life. Due to 'end correction', the antinode at the open end actually forms slightly outside the tube by a distance of approximately $0.6r$, where $r$ is the radius of the pipe. This makes the effective length of the pipe slightly longer than its physical length.

Q: Why is the speed of sound given as 330 m/s? Does it change?

The speed of sound in air depends on temperature ($v \propto \sqrt{T}$). $330\text{ m/s}$ is a standard value near $0^\circ\text{C}$. At room temperature ($20^\circ\text{C}$), it is closer to $343\text{ m/s}$.

Q: Can a tuning fork resonate with a pipe of any length if we change the temperature?

Yes! Since the speed of sound changes with temperature, the wavelength $\lambda = v/f$ will also change. By heating or cooling the air inside the pipe, you can adjust the wavelength to achieve resonance for a fixed physical length.

Question

An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency $264	ext{ Hz}$, if the length of the column in cm is (Speed of sound in air = $330	ext{ m/s}$)

Options:

(A) 31.25

(B) 62.50

(C) 93.75

(D) 125

Sigma Percentile · Accepted Answer

### Introduction to Standing Waves in Air Columns

Imagine blowing across the top of an empty glass bottle. You hear a clear, distinct musical note.

What you are experiencing is the physics of **standing waves** in a closed organ pipe.

When sound waves travel down the bottle, they reflect off the bottom. The incoming and reflected waves interfere with each other.

At most frequencies, this interference is chaotic and quickly dies out. But at specific, special frequencies, the waves reinforce each other, creating a stable, high-amplitude vibration known as **resonance**.

Let's dive into how we can mathematically predict these resonant lengths for a given frequency.

---

### Analyzing the Boundary Conditions

To solve this problem, we must first understand the boundary conditions of a pipe closed at one end.

At the closed end, the air molecules are physically blocked by a solid wall. They cannot move back and forth.

Therefore, the closed end must always be a **displacement node** (where displacement is zero).

At the open end, the air is completely free to move and escape into the atmosphere.

Therefore, the open end must always be a **displacement antinode** (where displacement is at its maximum).

This simple physical constraint dictates the entire mathematical structure of the standing waves inside the pipe.

---

### The Master Equation for Resonance

The simplest wave pattern that can satisfy these boundary conditions is one where we have a node at the closed end and an antinode at the open end, with no other nodes or antinodes in between.

This is called the **fundamental mode** or the **first harmonic**.

For this mode, the length of the pipe $L$ is exactly equal to one-quarter of a wavelength:

$$L_1 = \frac{\lambda}{4}$$

The next possible pattern that satisfies the boundary conditions must have a node at the closed end, an antinode at the open end, and one additional node and antinode in between.

This is the **first overtone** (or **third harmonic**), where the length of the pipe is:

$$L_2 = \frac{3\lambda}{4}$$

Generalizing this, the resonant lengths $L$ for a closed pipe must always be an **odd multiple of a quarter wavelength**:

$$L = (2n-1)\frac{\lambda}{4} \quad 	ext{for } n = 1, 2, 3, \dots$$

Using the fundamental wave relationship $v = f\lambda$, we can express the wavelength as $\lambda = \frac{v}{f}$. Substituting this into our length formula gives:

$$L = (2n-1)\frac{v}{4f}$$

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### Step-by-Step Calculation

Now, let's substitute our given values into this master equation:
* Speed of sound, $v = 330	ext{ m/s}$
* Frequency of the tuning fork, $f = 264	ext{ Hz}$

Let's calculate the fundamental length ($n=1$):

$$L_1 = (2(1)-1)\frac{330}{4 	imes 264} = 1 	imes \frac{330}{1056} = 0.3125	ext{ m}$$

Converting this to centimeters:

$$L_1 = 0.3125 	imes 100 = 31.25	ext{ cm}$$

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

Now, let's find the next resonant length, which is the first overtone ($n=2$):

$$L_2 = (2(2)-1)\frac{v}{4f} = 3 	imes L_1$$

$$L_2 = 3 	imes 31.25	ext{ cm} = 93.75	ext{ cm}$$

This matches **Option (c)** perfectly!

What about the second overtone ($n=3$)?

$$L_3 = 5 	imes L_1 = 5 	imes 31.25	ext{ cm} = 156.25	ext{ cm}$$

Since $156.25	ext{ cm}$ is larger than any of the options provided, we have found all the possible correct choices within the given options.

Therefore, the correct options are **(a)** and **(c)**.

Animated Solution for Physics - Waves: An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency $264 Hz$ , if the length of the column in cm is (Speed of sound in air = $330 m/s$ )

Select Answer:

Visualized Solution

Visualizing the Closed Organ Pipe

The Resonance Condition

Expressing Length in terms of Frequency

Substituting the Given Values

Calculating the Fundamental Mode ( $n = 1$ )

Converting to Centimeters

Calculating the First Overtone ( $n = 2$ )

Checking the Second Overtone ( $n = 3$ )

Selecting the Correct Options

Exploring End Correction

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves in Air Columns

Analyzing the Boundary Conditions

The Master Equation for Resonance

Step-by-Step Calculation

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