A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respecti

Q: Why is the first overtone of a closed pipe the 3rd harmonic instead of the 2nd?

A closed pipe has a node at one end and an antinode at the other. This boundary condition only allows odd harmonics ($1, 3, 5, \dots$) to exist. Therefore, the first overtone (the first allowed frequency above the fundamental) is the 3rd harmonic.

Q: What does equal compressibility tell us about the gases?

Compressibility is the inverse of Bulk Modulus ($K = 1/B$). Equal compressibility means both gases have the exact same Bulk Modulus ($B$), which allows us to cancel $B$ out when taking the ratio of the speeds of sound.

Q: Why does the speed of sound depend on density?

Sound is a mechanical wave that propagates through the vibration of particles. A denser gas has more mass per unit volume (higher inertia), making it harder for the particles to accelerate and pass on the wave, which slows down the speed of sound.

Q: Can we assume the ratio of specific heats (gamma) is the same for both gases?

Yes, because the Bulk Modulus of a gas under acoustic wave propagation is the adiabatic Bulk Modulus, $B = \gamma P$. Since the compressibilities (and thus Bulk Moduli) are equal at the same pressure, the term $B$ fully accounts for any differences in $\gamma$.

Q: What would happen to the length of the open pipe if the density of the gas inside it increased?

If $\rho_2$ increases, the speed of sound $v_o$ in the open pipe decreases. To maintain the same frequency with a slower wave, the open pipe would need to be shorter. This matches our formula where $L_o \propto 1/\sqrt{\rho_2}$.

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 organ pipe is

Options:

(A) $\frac{L}{3}$

(B) $\frac{4L}{3}$

(C) $\frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

(D) $\frac{4L}{3} \sqrt{\frac{\rho_2}{\rho_1}}$

Sigma Percentile · Accepted Answer

### The Symphony of Standing Waves

Imagine two musical pipes standing side by side in a physics laboratory. One is closed at one end, holding its breath, while the other is open at both ends, breathing freely.

They are filled with different gases, yet when excited, they sing in perfect unison, vibrating in their **first overtone** with the exact same frequency.

How does the geometry of these pipes adapt to the invisible, microscopic differences in the gases they contain?

This problem is a beautiful dance between **wave mechanics**, **thermodynamics**, and **fluid elasticity**. Let's break down the physics step-by-step to find the elegant relationship between their lengths.

### Unmasking the Elasticity: Compressibility and Bulk Modulus

Before we look at the waves, we must understand the medium. The problem states that the **compressibility** of the gases in both pipes is equal.

What is compressibility? In physics, compressibility ($K$) is a measure of how much a substance's volume decreases under pressure. It is mathematically defined as the reciprocal of the **Bulk Modulus** ($B$):

$$K = \frac{1}{B}$$

Since the compressibilities are equal, their Bulk Moduli must also be identical:

$$B_1 = B_2 = B$$

This is our first major bridge. It tells us that both gases offer the exact same elastic resistance to compression, even though their densities are different.

### The Speed of Sound: A Density Duel

The speed of sound in any fluid depends on its elastic properties and its inertia. This is beautifully captured by the Newton-Laplace formula:

$$v = \sqrt{\frac{B}{\rho}}$$

Since the Bulk Modulus $B$ is constant for both gases, the speed of sound is purely governed by the density of the gas.

For the closed pipe containing gas of density $\rho_1$:

$$v_c = \sqrt{\frac{B}{\rho_1}}$$

For the open pipe containing gas of density $\rho_2$:

$$v_o = \sqrt{\frac{B}{\rho_2}}$$

Notice the profound physical insight here: **the speed of sound is inversely proportional to the square root of the gas density**. A denser gas has more inertia, slowing down the propagation of the sound wave.

### Tuning the Pipes: Harmonics and Overtones

Now, let's look at the geometry of the standing waves inside the pipes.

#### 1. The Closed Organ Pipe

A closed organ pipe of length $L$ has a boundary condition of a **node** at the closed end (where air molecules cannot move) and an **antinode** at the open end (where air molecules vibrate freely).

- The fundamental mode (first harmonic) corresponds to a quarter-wavelength: $L = \frac{\lambda}{4}$, giving $f_1 = \frac{v_c}{4L}$.
- The **first overtone** is the next possible standing wave pattern, which is the **third harmonic** ($3f_1$):

$$f_{3,c} = \frac{3v_c}{4L}$$

#### 2. The Open Organ Pipe

An open organ pipe of length $L_o$ must have **antinodes** at both open ends.

- The fundamental mode (first harmonic) corresponds to a half-wavelength: $L_o = \frac{\lambda}{2}$, giving $f_1 = \frac{v_o}{2L_o}$.
- The **first overtone** is the next possible pattern, which is the **second harmonic** ($2f_1$):

$$f_{2,o} = 2 \left(\frac{v_o}{2L_o}\right) = \frac{v_o}{L_o}$$

### The Master Equation and Final Triumph

We are given that both pipes vibrate in their first overtone with the **same frequency**:

$$f_{3,c} = f_{2,o}$$

Substituting our expressions for the frequencies:

$$\frac{3v_c}{4L} = \frac{v_o}{L_o}$$

Now, let's isolate the length of the open pipe, $L_o$:

$$L_o = \frac{4L}{3} \left(\frac{v_o}{v_c}\right)$$

To find the ratio of the velocities $\frac{v_o}{v_c}$, we substitute our speed of sound formulas:

$$\frac{v_o}{v_c} = \frac{\sqrt{B/\rho_2}}{\sqrt{B/\rho_1}} = \sqrt{\frac{\rho_1}{\rho_2}}$$

Substituting this back into our expression for $L_o$ yields the final, beautiful result:

$$L_o = \frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$$

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

### Deep Physical Intuition

Why does the density ratio appear as $\rho_1/\rho_2$ under the square root?

If the gas in the closed pipe is denser (higher $\rho_1$), the speed of sound $v_c$ inside it is slower. To match its frequency with the open pipe, the open pipe (which has a faster wave speed $v_o$) must be physically longer to compensate for the faster wave propagation.

Thus, the length of the open pipe $L_o$ must scale directly with the square root of the density of the closed pipe's gas. Physics is truly a beautifully self-consistent symphony!

Select Answer:

Visualized Solution

Visualizing the Setup

Connecting Compressibility to Bulk Modulus

The Speed of Sound Formula

First Overtone of a Closed Pipe

First Overtone of an Open Pipe

Setting Up the Master Equation

Expressing $L_{o}$ in terms of $L$

Substituting the Velocity Ratio

The Final Expression for $L_{o}$

Deep Physical Insights

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Standing Waves

Unmasking the Elasticity

The Speed of Sound

Tuning the Pipes

# 1

# 2

The Master Equation and Final Triumph

Deep Physical Intuition

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