A closed organ pipe of length L and an open organ pipe contain gases of densities rho_1 and rho_2 re

Q: Why is the first overtone of a closed pipe the 3rd harmonic and not the 2nd?

A closed pipe has a node at one end and an antinode at the other. This boundary condition only allows odd multiples of the fundamental frequency ($f, 3f, 5f...$). Therefore, the first overtone (the first resonance above the fundamental) is the 3rd harmonic ($3f$).

Q: What exactly does "equal compressibility" mean in this context?

Compressibility is the measure of how much a volume of gas decreases under pressure. Mathematically, it is the exact reciprocal of the Bulk Modulus ($B$). If two gases have equal compressibility, they have the exact same Bulk Modulus.

Q: Why do we use the formula $v = \sqrt{B/\rho}$ instead of $v = \sqrt{\gamma RT/M}$?

Both formulas are correct, but $v = \sqrt{B/\rho}$ is the fundamental mechanical definition of wave speed in a fluid. Since the problem explicitly gives us information about compressibility (which relates to $B$) and density ($\rho$), this form is the most direct and efficient path to the solution.

Q: Does the radius of the organ pipe affect the frequency?

In ideal theoretical problems, we ignore the radius. However, in reality, a wider pipe requires an "end correction" because the antinode forms slightly outside the physical end of the pipe, effectively increasing its acoustic length.

Q: What is the difference between an overtone and a harmonic?

A harmonic is any integer multiple of the fundamental frequency (e.g., 1st, 2nd, 3rd). An overtone is any resonant frequency strictly *above* the fundamental. The 1st overtone is the first allowed resonance after the fundamental, which could be the 2nd harmonic (in open pipes) or the 3rd harmonic (in closed pipes).

Question

A closed organ pipe of length $L$ and an open organ pipe contain gases of densities $\rho_1$ and $\rho_2$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is $\frac{x}{3}L\sqrt{\frac{\rho_1}{\rho_2}}$ where $x$ is ...... .
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The Symphony of Organ Pipes

Imagine you are standing in a grand cathedral, surrounded by the majestic sound of a massive pipe organ. The physics behind this awe-inspiring music is rooted in the elegant mathematics of standing waves.

In this problem, we are analyzing two such pipes side by side. One pipe is closed at one end, while the other is open at both ends. They are filled with different gases, but they are singing the exact same note!

Let's dive into the mechanics of how their lengths, gas densities, and overtones perfectly align to create this harmony.

### Decoding the Overtones

When air is blown into an organ pipe, it creates standing sound waves. The boundary conditions dictate the allowed frequencies.

For a **closed organ pipe**, the closed end must be a displacement node (air cannot move), and the open end must be an antinode. Because of this asymmetry, a closed pipe only supports odd harmonics. 
The fundamental frequency is $f_1 = \frac{v}{4L}$. 
The **first overtone** is the next possible standing wave, which is the third harmonic. Therefore, its frequency is:
$$f_c = \frac{3v_1}{4L}$$

On the other hand, an **open organ pipe** has antinodes at both ends. This symmetry allows it to support all integer harmonics. 
Its fundamental frequency is $f_1 = \frac{v}{2L}$. 
The **first overtone** is simply the second harmonic. Its frequency is:
$$f_o = \frac{2v_2}{2L_2} = \frac{v_2}{L_2}$$

### The Speed of Sound Connection

The problem states a fascinating condition: both pipes are vibrating with the same frequency in their first overtone. 
Equating our two expressions, we get our master equation:
$$\frac{3v_1}{4L} = \frac{v_2}{L_2}$$

But what are $v_1$ and $v_2$? The speed of sound in a gas is governed by the Newton-Laplace formula:
$$v = \sqrt{\frac{B}{\rho}}$$
where $B$ is the bulk modulus and $\rho$ is the density of the gas.

The problem mentions that the **compressibility** of the gases is equal. Compressibility is simply the reciprocal of the bulk modulus ($K = \frac{1}{B}$). 
Therefore, equal compressibility means both gases have the exact same bulk modulus $B$.

### The Final Calculation

Let's substitute the wave speeds into our master equation:
$$\frac{3}{4L} \sqrt{\frac{B}{\rho_1}} = \frac{1}{L_2} \sqrt{\frac{B}{\rho_2}}$$

Notice how the bulk modulus $B$ beautifully cancels out from both sides. This leaves us with a pure geometric and density relationship:
$$\frac{3}{4L} \frac{1}{\sqrt{\rho_1}} = \frac{1}{L_2} \frac{1}{\sqrt{\rho_2}}$$

Now, we simply rearrange the terms to isolate the length of the open pipe, $L_2$:
$$L_2 = \frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$$

The problem defines the length of the open pipe as $\frac{x}{3}L \sqrt{\frac{\rho_1}{\rho_2}}$. 
By comparing our derived expression with the given format, the answer reveals itself instantly:
$$x = 4$$

This elegant result shows how the physical properties of the medium and the geometric boundaries of the instrument perfectly balance to produce the same musical pitch.

Enter Numerical Value:

Visualized Solution

$System Setup$

$First Overtone of Closed Pipe$

$First Overtone of Open Pipe$

$Equating Frequencies$

$Speed of Sound in Gas$

$Substituting Wave Speeds$

$Solving for L_{2}$

$Comparing with Given Expression$

$The Way Forward$

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Organ Pipes

Decoding the Overtones

The Speed of Sound Connection

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Overtone of Closed Pipe

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Overtone of Open Pipe

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Frequencies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Speed of Sound in Gas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Wave Speeds

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing with Given Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward