Problem 2828 - JEE Physics

Question

An open pipe is in resonance in \(2^{\text {nd }}\) harmonic with frequency \(\mathrm{f}_{1}\). Now one end of the tube is closed and frequency is increased to \(\mathrm{f}_{2}\) such that the resonance again occurs in nth harmonic. Choose the correct option. (A) \(\mathrm{n}=5, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\) (B) \(\mathrm{n}=3, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (C) \(\mathrm{n}=5, \mathrm{f}_{2}=(3 / 4) \mathrm{f}_{1}\) (D) \(\mathrm{n}=3, \mathrm{f}_{2}=(5 / 4) \mathrm{f}_{1}\)

Step-by-Step Solution

Answer

The correct option is (B) with n = 3 and the relationship between f1 and f2 as \(f_2 = \frac{3}{4}f_1\).

Step 1: Resonance formula for an open pipe

For an open pipe, the resonance formula says that when a pipe is in resonance, the frequency is given by \[f = \frac{2n}{L} \times v\] where n is the harmonic number, L is the length of the pipe, and v is the speed of sound.

Step 2: Given conditions for the open pipe

We are given that the open pipe is in resonance in the 2nd harmonic with frequency f1. Using the formula from Step 1, we can write the frequency equation for the open pipe: \[f_1 = \frac{4}{L} \times v\]

Step 3: Resonance formula for a closed pipe

For a closed pipe, the resonance formula says that when a closed pipe is in resonance, the frequency is given by \[f = \frac{2n - 1}{4L} \times v\] where n is the harmonic number, L is the length of the pipe, and v is the speed of sound.

Step 4: Given conditions for the closed pipe

We are given that the closed pipe is in resonance in the nth harmonic with frequency f2. Using the formula from Step 3, we can write the frequency equation for the closed pipe: \[f_2 = \frac{2n - 1}{4L} \times v\]

Step 5: Relate f1 and f2

Now we need to find a relationship between f1 and f2 using the equations from Steps 2 and 4. We can do this by dividing the equation of f2 by the equation of f1: \[\frac{f_2}{f_1} = \frac{(2n - 1)/4L}{4/L}\]

Step 6: Simplify the relationship

After simplifying the relationship from Step 5, we have: \[\frac{f_2}{f_1} = \frac{2n -1}{16}\]

Step 7: Check for the correct answer

Now that we have a relationship between f1 and f2, we must check the provided options to find the correct one. For this, we need to plug in the given values of n and f2/f1 into the relationship from Step 6. Let's first plug the values from Option (A): \(n = 5, f_2 = (5/4)f_1\) \[\frac{5}{4} = \frac{2(5) - 1}{16} \implies \frac{5}{4} = \frac{9}{16}\] This is not correct. Now plug the values from Option (B): \(n = 3, f_2 = (3/4)f_1\) \[\frac{3}{4} = \frac{2(3) - 1}{16} \implies \frac{3}{4} = \frac{5}{16}\] This is correct. There is no need to check more options as we have already found the correct answer. The correct option is (B). So, n = 3, and the relationship between f1 and f2 is \[f_2 = \frac{3}{4}f_1\]

Key Concepts

Open Pipe Resonance

Understanding open pipe resonance is crucial in understanding harmonics in sound waves. An open pipe refers to a tube that is open at both ends. When sound waves travel through such a pipe, they create what we call resonant waves. These waves resonate at particular frequencies, known as harmonics.

In an open pipe, both ends of the pipe are antinodes, where the maximum displacement of air particles occurs. The length of the pipe helps in determining the harmonic frequencies. The relationship is given by the formula:

• For open pipes: \( f = \frac{2n}{L} \times v \)

where:

• \( f \) is the frequency.
• \( n \) is the harmonic number (1st, 2nd, etc.).
• \( L \) is the length of the pipe.
• \( v \) is the speed of sound in air.

This equation shows that as the harmonic number increases, so does the frequency. For the 2nd harmonic, for example, the frequency would be double that of the fundamental frequency.

Closed Pipe Resonance

Closed pipe resonance occurs when one end of a pipe is closed, altering the pattern of resonance compared to open pipes. Inside a closed pipe, the closed end becomes a node, where there is no displacement of air molecules, while the open end remains an antinode. This shift affects the harmonics achievable within such a system.

The resonance frequency for a closed pipe follows this formula:

• For closed pipes: \( f = \frac{2n - 1}{4L} \times v \)

This equation is different from the open pipe as it implies that only odd-numbered harmonics can be sustained.

• \( L \) is the pipe length.
• \( v \) is the speed of sound.

This restriction results in a more complex frequency pattern. An important takeaway is that closed pipes exhibit a different set of resonant frequencies, emphasizing the influence of structural differences on resonant behaviors and sound waves.

Frequency Relationships

The relationship between frequencies in different harmonic setups is based on the structure of the pipe and the formula used for calculating harmonics. When transitioning from an open pipe to a closed pipe, the frequencies of resonance differ due to the change in structure and the corresponding harmonic formulas applied.

By comparing the open pipe with a frequency \( f_1 \) in the 2nd harmonic, and the closed pipe’s nth harmonic frequency \( f_2 \), the relationship can be expressed mathematically:

• \( \frac{f_2}{f_1} = \frac{(2n - 1)}{16} \)

For correct resonance, only specific integer values of \( n \) result in matching the formulas with the observed frequencies, as shown in the solution (\( n = 3\), \( \frac{f_2}{f_1} = \frac{3}{4} \)).

Such relationships highlight how frequency patterns shift with physical changes in the system, providing a real-life application of wave physics and demonstrating the dynamic nature of harmonic sounds.