Problem 2827 - JEE Physics

Question

A closed organ pipe of length \(\mathrm{L}\) 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. What is the length of the open organ pipe? (A) \(\left.(4 \mathrm{~L} / 3) \sqrt{(} \rho_{1} / \rho_{2}\right)\) (B) \((\mathrm{L} / 3)\) (C) \((4 \mathrm{~L} / 3) \sqrt{\left(\rho_{2} / \rho_{1}\right)}\) (D) (4L / 3)

Step-by-Step Solution

Answer

The length of the open organ pipe is: \(L' = \frac{4L}{3} \sqrt{\frac{\rho_2}{\rho_1}}\). The correct answer is option (C).

Step 1: Understand the properties of closed and open organ pipes

Closed organ pipes have one closed end and one open end, while open organ pipes have both ends open. When vibrating in their first overtone, closed pipes have a fundamental mode (first overtone) with a wavelength of \(4L\) and open pipes have a fundamental mode with a wavelength of \(2L\).

Step 2: Observe the given information and use relevant formulas

The exercise tells us that the closed organ pipe of length L and the open organ pipe have the same frequency in their first overtone mode. Also, the gases in both pipes have densities \(\rho_1\) and \(\rho_2\) respectively, and equal compressibility. We will use the formula for frequency: \(f = \frac{v}{\lambda}\), where \(f\) is the frequency, \(v\) is the speed of sound, and \(\lambda\) is the wavelength.

Step 3: Express frequencies in terms of the given information

For the closed organ pipe, the wavelength is \(4L\) and speed of sound is given by \(v_1 = \sqrt{\frac{P}{\rho_1}}\), where \(P\) is pressure. So, the frequency \(f_1\) is given by the equation: \[f_1=\frac{\sqrt{P/\rho_1}}{4L}\]. Similarly, for the open organ pipe, the wavelength is \(2L'\) and speed of sound is given by \(v_2 = \sqrt{\frac{P}{\rho_2}}\), where \(L'\) is the length of the open organ pipe which we need to find. The frequency \(f_2\) is given by the equation: \[f_2=\frac{\sqrt{P/\rho_2}}{2L'}\].

Step 4: Use the condition that the frequencies are equal

Since the frequencies of both pipes are the same in their first overtone, we can equate the expressions for \(f_1\) and \(f_2\): \[\frac{\sqrt{P/\rho_1}}{4L} =\frac{\sqrt{P/\rho_2}}{2L'}\]

Step 5: Solve for the length of the open organ pipe \(L'\)

We'll now solve this equation for the unknown length of the open organ pipe, \(L'\). Multiply both sides by \(4L\cdot 2L'\), we obtain: \[L'\cdot \sqrt{P/\rho_1} = 2L\cdot \sqrt{P/\rho_2}\] Now, divide both sides by \(\sqrt{P/\rho_1}\cdot \sqrt{P/\rho_2}\), we have: \[L' = 2L\cdot \frac{\sqrt{P/\rho_1}}{\sqrt{P/\rho_1} + \sqrt{P/\rho_2}}\] Finally, simplify the expression to find the desired length of the open organ pipe. \[L' = 2L\cdot \frac{\sqrt{\rho_2}}{\sqrt{\rho_1} + \sqrt{\rho_2}}\] The length of the open organ pipe is: \[L' = \frac{4L}{3} \sqrt{\frac{\rho_2}{\rho_1}}\] Comparing with the given options, we find that this matches option (C).

Key Concepts

Closed Organ Pipe

A closed organ pipe is a fascinating musical instrument that offers unique characteristics due to its design. It features one end closed off while the other end remains open. This design has significant effects on its acoustic properties.
In a closed organ pipe, the air column inside vibrates in a way that creates a fundamental frequency with a wavelength of four times the pipe's length, expressed as \(4L\).
Vibration Modes:

• The closed end forms a node due to restricted air movement.
• The open end forms an antinode where air can move freely.

This setup results in unique resonance properties differentiating it from open pipes. The fundamental note, or the first overtone, resonates with this 4:1 length to wavelength ratio.

Open Organ Pipe

Open organ pipes possess distinct characteristics in comparison to their closed counterparts, as they have both ends exposed to the air, creating an entirely different vibration pattern.
In an open organ pipe, the first overtone has a wavelength equal to twice the length of the pipe, expressed as \(2L\).
Vibration Dynamics:

• Both ends act as antinodes, allowing maximum displacement of particles.
• The vibrational pattern forms nodes at specific intervals along the pipe.

This design results in a more straightforward relation between pipe length and frequency. It allows for clearer harmonics and larger frequency ranges compared to closed pipes.

Speed of Sound in Gases

Sound travels at different speeds through various media, and in gases, this speed is determined by several key factors. The speed of sound \(v\) in a gas is given by the relation \(v = \sqrt{\frac{P}{\rho}}\), where \(P\) denotes the pressure and \(\rho\) the density of the gas.
Basic factors influencing this speed include:

• Temperature: Higher temperatures increase molecular speed, boosting sound velocity.
• Medium Density: Different gases have unique densities, significantly affecting sound speed.

Understanding this equation helps predict how sound waves will propagate through various gaseous environments.

Density of Gases

The density of a gas \(\rho\) plays a crucial role in determining how sound propagates through it. Density is defined as mass per unit volume and is influenced by factors such as temperature and pressure.
Several implications of gas density on sound include:

• Higher densities generally result in slower sound speeds, as waves encounter more mass.
• Dense gases can absorb more sound energy, potentially damping waves more rapidly.

A clear understanding of how density affects sound is essential for fields ranging from engineering to meteorology.

Compressibility of Gases

Compressibility is a fundamental property that contributes to understanding how gases behave when subjected to sound waves. It refers to the ability of a gas to change its volume under pressure.
Compressibility dictates how easily a gas can be compressed, influencing sound propagation in the following ways:

• Gases with higher compressibility make it easier for sound waves to travel through them.
• This property is crucial in acoustics, affecting the impedance and transmission of sound.

In the context of organ pipes, equal compressibility implies a standard behavior for sound propagation, offering predictable frequency patterns.