An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of t

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

A closed organ pipe must have a displacement node at the closed end and an antinode at the open end. This boundary condition can only be satisfied when the length of the pipe is an odd multiple of a quarter-wavelength ($L = (2n-1)\frac{\lambda}{4}$). Consequently, only odd harmonics ($1^{\text{st}}, 3^{\text{th}}, 5^{\text{th}}, \dots$) can exist.

Q: Does the speed of sound change when one end of the pipe is closed?

No, the speed of sound $v$ depends only on the properties of the medium (such as temperature, pressure, and molecular mass of the gas inside the pipe). Closing one end of the pipe does not alter the air inside, so $v$ remains constant.

Q: What is the difference between a node and an antinode in terms of pressure and displacement?

A displacement node is a point where air molecules do not move at all, which corresponds to a pressure antinode (maximum pressure variation). Conversely, a displacement antinode is where air molecules vibrate with maximum amplitude, corresponding to a pressure node (zero pressure variation).

Q: How would the frequencies change if the temperature of the air inside the pipe increased?

The speed of sound in a gas is proportional to the square root of absolute temperature ($v \propto \sqrt{T}$). If the temperature increases, the speed of sound $v$ increases, which in turn increases all the resonant frequencies of the pipe.

Q: Is the 'third harmonic' the same as the 'third overtone' for a closed pipe?

No, they are different! For a closed pipe, the harmonics are odd integers: $1^{\text{st}}$ harmonic (fundamental), $3^{\text{rd}}$ harmonic, $5^{\text{th}}$ harmonic, etc. The $3^{\text{rd}}$ harmonic is the $1^{\text{st}}$ overtone, and the $5^{\text{th}}$ harmonic is the $2^{\text{nd}}$ overtone. Therefore, the third harmonic corresponds to the first overtone.

Question

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by $100	ext{ Hz}$ than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

Options:

(A) $200	ext{ Hz}$

(B) $300	ext{ Hz}$

(C) $240	ext{ Hz}$

(D) $480	ext{ Hz}$

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Standing Waves

Imagine a column of air trapped inside a cylindrical tube. When you blow across the opening, you disturb the air molecules, sending pressure waves traveling down the tube.

These waves reflect off the ends of the tube, interfering with newly generated waves. Under the right conditions, this interference produces **standing waves**—stable patterns of vibration that we hear as musical notes.

In this problem, we explore how a simple physical change—closing one end of an open pipe—dramatically alters its harmonic spectrum and resonant frequencies.

### Analyzing the Open Pipe

Let us start with our organ pipe of length $L$, which is initially open at both ends.

Because both ends are open to the atmosphere, the air molecules at the openings are free to vibrate with maximum amplitude. This means we must have **displacement antinodes** at both ends of the pipe.

For the fundamental mode (the simplest standing wave pattern), there is a single displacement node in the exact middle of the pipe. The distance between two successive antinodes is half a wavelength:

$$L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L$$

Using the fundamental wave relation $v = f \lambda$, where $v$ is the speed of sound in air, we find the fundamental frequency $f_1$ of the open pipe:

$$f_1 = \frac{v}{2L}$$

### Analyzing the Closed Pipe

Now, we suddenly close one end of the pipe. The length remains exactly $L$, but the boundary conditions have changed.

At the closed end, air molecules are restricted by the solid barrier, forcing a **displacement node** at this boundary. The open end, however, remains a **displacement antinode**.

Because of this asymmetry, a closed pipe can only support **odd harmonics**. The resonant wavelengths are given by:

$$L = n \frac{\lambda_n}{4} \implies f_n = n \left(\frac{v}{4L}ight) \quad 	ext{for } n = 1, 3, 5, \dots$$

We are interested in the **third harmonic** ($n = 3$) of this closed pipe. Let us denote its frequency as $f_2$:

$$f_2 = 3 \left(\frac{v}{4L}ight) = \frac{3v}{4L}$$

### Setting Up the Master Equation

The problem states that the frequency of the third harmonic of the closed pipe ($f_2$) is higher than the fundamental frequency of the open pipe ($f_1$) by exactly $100	ext{ Hz}$:

$$f_2 - f_1 = 100$$

Let us substitute our algebraic expressions for $f_1$ and $f_2$ into this relation:

$$\frac{3v}{4L} - \frac{v}{2L} = 100$$

To subtract these fractions, we find a common denominator of $4L$:

$$\frac{3v}{4L} - \frac{2v}{4L} = 100$$

$$\frac{v}{4L} = 100$$

### Finding the Final Answer

We have found that the quantity $\frac{v}{4L}$ is equal to $100	ext{ Hz}$.

Now, let us relate this back to the fundamental frequency of the open pipe, $f_1$, which we want to calculate:

$$f_1 = \frac{v}{2L} = 2 \left(\frac{v}{4L}ight)$$

Substituting our value of $\frac{v}{4L} = 100	ext{ Hz}$:

$$f_1 = 2 	imes 100 = 200	ext{ Hz}$$

Thus, the fundamental frequency of the open pipe is **$200	ext{ Hz}$**, which perfectly matches **Option (a)**.

Animated Solution for Physics - Waves: An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by $100 Hz$ than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

Select Answer:

Visualized Solution

Visualizing the Two Pipe Configurations

Fundamental Frequency of the Open Pipe

Third Harmonic of the Closed Pipe

Setting Up the Given Condition

Substituting the Frequency Expressions

Solving for the Common Factor

Relating Back to the Open Pipe Fundamental

Calculating the Final Frequency

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

Analyzing the Open Pipe

Analyzing the Closed Pipe

Setting Up the Master Equation

Finding the Final Answer

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