An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed an

Q: Why do we only consider odd harmonics for the closed pipe?

In a pipe closed at one end, the closed end must be a displacement node (zero movement) and the open end must be an antinode (maximum movement). This asymmetric boundary condition only allows an odd number of quarter-wavelengths to fit inside the pipe ($L = (2m-1)\frac{\lambda}{4}$), which restricts the resonant frequencies to odd harmonics only.

Q: Does the length of the pipe change when we close one end?

No, the physical length of the pipe remains exactly $L$. Closing one end simply changes the boundary condition at that end from an open boundary (antinode) to a closed boundary (node).

Q: Why must $f_2$ be strictly greater than $f_1$?

The problem states that the frequency is 'increased to $f_2$'. This phrasing explicitly tells us that the new resonant frequency $f_2$ is higher than the initial resonant frequency $f_1$.

Q: What is the difference between displacement waves and pressure waves in this context?

Displacement waves represent the actual physical movement of air molecules, while pressure waves represent the variation in pressure. A displacement node is always a pressure antinode (where pressure varies the most), and a displacement antinode is a pressure node (where pressure remains atmospheric). Both representations yield the same resonant frequencies.

Q: How would end correction affect the answer in a real experiment?

In a real pipe, the displacement antinode occurs slightly outside the open end by a distance of about $0.6r$ (where $r$ is the radius). This effectively increases the acoustic length of the pipe. For an open pipe, the effective length is $L + 1.2r$, and for a closed pipe, it is $L + 0.6r$. This would slightly shift the exact frequency values, but the fundamental physics and the harmonic relationship ($n=5$) would remain conceptually similar.

Question

An open pipe is in resonance in $2^{	ext{nd}}$ harmonic with frequency $f_1$. Now one end of the tube is closed and frequency is increased to $f_2$ such that the resonance again occurs in $n^{	ext{th}}$ harmonic. Choose the correct option.

Options:

(A) $n = 3, f_2 = \frac{3}{4} f_1$

(B) $n = 3, f_2 = \frac{5}{4} f_1$

(C) $n = 5, f_2 = \frac{5}{4} f_1$

(D) $n = 5, f_2 = \frac{3}{4} f_1$

Sigma Percentile · Accepted Answer

### Introduction to Standing Waves in Air Columns Standing waves in organ pipes are one of the most beautiful demonstrations of wave mechanics in physics. When sound waves travel through an air column, they reflect off the boundaries of the pipe. The superposition of the incident and reflected waves creates a standing wave pattern characterized by nodes (points of zero displacement) and antinodes (points of maximum displacement). In this problem, we explore how a simple change in boundary conditions—closing one end of an open pipe—completely reshapes the allowed resonant frequencies of the system. --- ### Analyzing the Open Pipe Initially, we have a cylindrical tube open at both ends. Because both ends are open to the atmosphere, the air molecules at the boundaries are free to move with maximum amplitude. This means that **both ends of an open pipe must be displacement antinodes**. For a pipe of length $L$, the fundamental frequency (first harmonic) corresponds to a standing wave with an antinode at each end and a single node in the middle. This pattern represents half a wavelength fitting inside the pipe: $$L = \frac{\lambda_0}{2} \implies \lambda_0 = 2L$$ The fundamental frequency is therefore: $$f_0 = \frac{v}{\lambda_0} = \frac{v}{2L}$$ We are given that the open pipe is in resonance in its **second harmonic** with frequency $f_1$. The second harmonic frequency is simply twice the fundamental frequency: $$f_1 = 2 f_0 = 2 \left(\frac{v}{2L} ight) = \frac{v}{L}$$ This is our starting frequency. Now, let's see what happens when we modify the pipe. --- ### The Transformation: Closing One End When we close one end of the pipe, we fundamentally alter the boundary conditions. At the closed end, the rigid wall prevents the air molecules from moving, forcing a **displacement node** at that boundary. The other end remains open, which means it must still be a **displacement antinode**. This asymmetric boundary condition (node at one end, antinode at the other) means that only an odd number of quarter-wavelengths can fit inside the pipe: $$L = n \frac{\lambda_n}{4} \quad ext{where } n = 1, 3, 5, \dots$$ Thus, the allowed resonant frequencies for a closed pipe are given by: $$f_n = n \left(\frac{v}{4L} ight) \quad ext{where } n = 1, 3, 5, \dots$$ Notice that a closed pipe only supports **odd harmonics**. --- ### Solving the Frequency Constraint We are told that the frequency is increased to a new value, $f_2$, such that resonance occurs again in the $n^{ ext{th}}$ harmonic of the closed pipe. This gives us a strict mathematical constraint: $$f_2 > f_1$$ We need to find the smallest odd integer $n$ that satisfies this inequality. Let's test the possible odd harmonics systematically: 1. **First Harmonic ($n = 1$):** $$f_1' = 1 \cdot \left(\frac{v}{4L} ight) = \frac{v}{4L} = \frac{1}{4} f_1$$ Since $\frac{1}{4} f_1 < f_1$, this represents a decrease in frequency, which violates our constraint. 2. **Third Harmonic ($n = 3$):** $$f_3' = 3 \cdot \left(\frac{v}{4L} ight) = \frac{3v}{4L} = \frac{3}{4} f_1$$ Since $\frac{3}{4} f_1 < f_1$, this also represents a decrease in frequency and is not the correct state. 3. **Fifth Harmonic ($n = 5$):** $$f_5' = 5 \cdot \left(\frac{v}{4L} ight) = \frac{5v}{4L} = \frac{5}{4} f_1$$ Since $\frac{5}{4} f_1 > f_1$, this is the first resonant frequency of the closed pipe that is strictly greater than our initial frequency $f_1$! --- ### Conclusion Therefore, the resonance occurs in the **fifth harmonic** ($n = 5$), and the new frequency is: $$f_2 = \frac{5}{4} f_1$$ This perfectly matches **Option (c)**.

Animated Solution for Physics - Waves: An open pipe is in resonance in $2^{nd}$ harmonic with frequency $f_{1}$ . Now one end of the tube is closed and frequency is increased to $f_{2}$ such that the resonance again occurs in $n^{th}$ harmonic. Choose the correct option.

Select Answer:

Visualized Solution

Visualizing the Open Pipe Resonance

Transitioning to a Closed Pipe

Analyzing the Frequency Constraint

Testing the First Harmonic ( $n = 1$ )

Testing the Third Harmonic ( $n = 3$ )

Testing the Fifth Harmonic ( $n = 5$ )

Determining the Correct Option

Deepening Your Understanding

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Standing Waves in Air Columns

Analyzing the Open Pipe

The Transformation

Solving the Frequency Constraint

Conclusion

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