Welcome to Lecture 124 (Master Index Entry: L118) of our comprehensive JEE Physics Master Series! Expanding our rigorous exploration of acoustic wave mechanics in Unit 10: Oscillations and Waves, we present the quintessential acoustic resonance masterclass: Standing Waves in Organ Pipes — Boundary Phase Mechanics, Open vs. Closed Pipes, Harmonics & Overtones, Rayleigh’s End Correction, and Resonance Tube Dynamics. When longitudinal sound waves propagate inside a hollow air column of finite length, reflections at the boundaries create counter-propagating waves. Their superposition establishes acoustic standing waves characterized by alternating displacement and pressure nodes and antinodes. In this lecture, we rigorously differentiate between displacement waves and pressure waves, derive the normal mode frequencies for open and closed columns, evaluate the acoustic boundary inertia via end correction, analyze the classical resonance tube experiment, and master high-frequency JEE Main and Advanced problem archetypes.
Detailed Concept Card: Acoustic Standing Waves in Air Columns
1. Boundary Mechanics: Displacement vs. Pressure Waves
Sound in gases propagates as a longitudinal wave. An acoustic disturbance can be described either by the longitudinal displacement of gas molecules $s(x, t)$ or by the gauge pressure fluctuation $\Delta P(x, t)$ relative to atmospheric equilibrium pressure $P_0$. These two fields are fundamentally related via the bulk modulus $B = \rho v^2$ of the medium:
$$\Delta P(x, t) = -B \frac{\partial s}{\partial x}$$
Because pressure fluctuation depends on the negative spatial derivative (divergence) of displacement, a spatial phase difference of $\frac{\pi}{2}\text{ rad}$ ($90^\circ$ or $\lambda/4$) exists between the displacement wave and the pressure wave. This distinction dictates acoustic boundary conditions:
| Pipe Boundary Type | Physical Boundary Condition | Displacement State ($s$) | Pressure State ($\Delta P$) | Reflection Phase Shift ($\Delta\phi$) |
|---|---|---|---|---|
| Closed End (Rigid Wall) | Air molecules are mechanically constrained and cannot move axially across the solid barrier. | Displacement Node ($s = 0$) | Pressure Antinode ($|\Delta P| = \Delta P_{\max}$) | $\Delta\phi = \pi$ for displacement; $\Delta\phi = 0$ for pressure. |
| Open End (Atmosphere) | Air column connects to external atmosphere where pressure remains clamped at equilibrium $P_0$. Air molecules can surge freely. | Displacement Antinode ($|s| = s_{\max}$) | Pressure Node ($\Delta P = 0$) | $\Delta\phi = 0$ for displacement; $\Delta\phi = \pi$ for pressure. |
Golden JEE Axiom: A Displacement Node is always a Pressure Antinode, and a Displacement Antinode is always a Pressure Node. In standing waves, the distance between any consecutive node and antinode (of the same field) is $\mathbf{\frac{\lambda}{4}}$, while the distance between two consecutive nodes (or two consecutive antinodes) is $\mathbf{\frac{\lambda}{2}}$.
2. Open Organ Pipe (Both Ends Open)
In an open organ pipe of physical length $L$, both ends are open to the atmosphere. Consequently, displacement antinodes must form at both ends $x = 0$ and $x = L$ (or pressure nodes at both boundaries).
The length $L$ of the pipe must accommodate an integer number of half-wavelength loops:
$$L = n \left(\frac{\lambda_n}{2}\right) \implies \lambda_n = \frac{2L}{n}, \quad \text{where } n = 1, 2, 3, 4, \dots$$
The permitted resonant frequencies (normal modes) are therefore given by:
$$f_n = \frac{v}{\lambda_n} = n \left(\frac{v}{2L}\right) = n f_1$$
Where $v$ is the speed of sound in the gas column. The harmonic structure of an open organ pipe features:
- Fundamental Mode ($n = 1$, 1st Harmonic): $\lambda_1 = 2L \implies f_1 = \frac{v}{2L}$. The wave pattern contains 2 displacement antinodes (at the ends) and 1 displacement node (at the center).
- First Overtone ($n = 2$, 2nd Harmonic): $\lambda_2 = L \implies f_2 = 2 \left(\frac{v}{2L}\right) = 2f_1$. Contains 3 displacement antinodes and 2 displacement nodes.
- Second Overtone ($n = 3$, 3rd Harmonic): $\lambda_3 = \frac{2L}{3} \implies f_3 = 3 \left(\frac{v}{2L}\right) = 3f_1$. Contains 4 displacement antinodes and 3 displacement nodes.
- General $p$-th Overtone: Corresponds to harmonic number $n = p + 1$, with frequency $f = (p + 1)f_1$.
- Harmonic Ratio: All integer harmonics are present: $$f_1 : f_2 : f_3 : f_4 : \dots = 1 : 2 : 3 : 4 : \dots$$ This rich harmonic content gives open pipes a full, melodious acoustic timbre.
3. Closed Organ Pipe (One End Closed, One End Open)
In a closed organ pipe of physical length $L$, the closed end enforces a displacement node ($s = 0$, pressure antinode), while the open end enforces a displacement antinode ($s = s_{\max}$, pressure node).
The length $L$ must accommodate an odd number of quarter-wavelength sections:
$$L = (2n – 1)\left(\frac{\lambda_n}{4}\right) \implies \lambda_n = \frac{4L}{2n – 1}, \quad \text{where } n = 1, 2, 3, 4, \dots$$
The permitted resonant frequencies are given by:
$$f_n = \frac{v}{\lambda_n} = (2n – 1)\left(\frac{v}{4L}\right) = (2n – 1)f_1$$
- Fundamental Mode ($n = 1$, 1st Harmonic): $\lambda_1 = 4L \implies f_1 = \frac{v}{4L}$. Contains 1 displacement node at the rigid base and 1 displacement antinode at the open lip.
- First Overtone ($n = 2$, 3rd Harmonic): $\lambda_2 = \frac{4L}{3} \implies f_2 = 3 \left(\frac{v}{4L}\right) = 3f_1$. Contains 2 displacement nodes and 2 displacement antinodes.
- Second Overtone ($n = 3$, 5th Harmonic): $\lambda_3 = \frac{4L}{5} \implies f_3 = 5 \left(\frac{v}{4L}\right) = 5f_1$. Contains 3 displacement nodes and 3 displacement antinodes.
- General $p$-th Overtone: Corresponds to harmonic number $N = 2p + 1$, with frequency $f = (2p + 1)f_1$.
- Harmonic Ratio: Only odd harmonics are present; all even harmonics are strictly absent: $$f_1 : f_3 : f_5 : f_7 : \dots = 1 : 3 : 5 : 7 : \dots$$
4. Comparative Matrix: Open vs. Closed Pipes of Same Length $L$
| Property | Open Organ Pipe | Closed Organ Pipe | Direct Ratio / Comparison |
|---|---|---|---|
| Fundamental Wavelength ($\lambda_1$) | $\lambda_1 = 2L$ | $\lambda_1 = 4L$ | $\lambda_{\text{closed}} = 2\lambda_{\text{open}}$ |
| Fundamental Frequency ($f_1$) | $f_1 = \frac{v}{2L}$ | $f_1′ = \frac{v}{4L}$ | $f_{\text{open}} = 2 f_{\text{closed}}$ (Octave higher) |
| Harmonics Generated | All integers: $1f_1, 2f_1, 3f_1, 4f_1 \dots$ | Odd only: $1f_1′, 3f_1′, 5f_1′, 7f_1′ \dots$ | Even harmonics forbidden in closed pipe |
| $p$-th Overtone Frequency | $f = (p + 1)\left(\frac{v}{2L}\right)$ | $f = (2p + 1)\left(\frac{v}{4L}\right)$ | Depends on boundary conditions |
| Center of Pipe State ($x = L/2$) | Displacement Node in Fundamental | Pressure and displacement both non-zero | Fundamental open pipe has central node |
| Acoustic Quality (Timbre) | Rich, warm, harmonious | Nasal, hollow, clarinet-like | Open pipes produce richer overtone series |
5. End Correction (Lord Rayleigh’s Correction)
In physical organ pipes, acoustic reflection does not occur precisely at the geometric open plane of the tube. Because of the acoustic inertia of the vibrating air mass immediately outside the lip, the effective boundary expands slightly into the ambient medium. The displacement antinode is located at an effective distance $e$ outside the physical end of the pipe.
By Lord Rayleigh’s theoretical and experimental formulation, for a cylindrical tube of internal radius $r$ (or diameter $d = 2r$):
$$\mathbf{e \approx 0.6\, r = 0.3\, d}$$
The effective acoustic lengths ($L_{\text{eff}}$) and modified fundamental frequencies become:
- Open Organ Pipe (Two open ends): $$L_{\text{eff}} = L + 2e = L + 2(0.6 r) = \mathbf{L + 1.2\, r}$$ $$f_1 = \frac{v}{2 L_{\text{eff}}} = \frac{v}{2(L + 1.2\, r)}$$
- Closed Organ Pipe (One open end): $$L_{\text{eff}} = L + e = L + 0.6 r = \mathbf{L + 0.6\, r}$$ $$f_1 = \frac{v}{4 L_{\text{eff}}} = \frac{v}{4(L + 0.6\, r)}$$
Crucial JEE Consequence: Because end correction increases the effective length ($L_{\text{eff}} > L$), the actual resonant frequency is always strictly lower than the idealized geometric frequency. Furthermore, wider pipes (larger radius $r$) experience larger downward frequency shifts.
6. The Resonance Tube Experiment
The resonance tube apparatus consists of a vertical glass cylinder partially filled with water, creating an air column of variable length closed at the bottom by the water surface. A vibrating tuning fork of known frequency $f$ is positioned above the open top.
Resonance occurs when the frequency of the air column matches the fork’s frequency $f$:
- First Resonance (Length $l_1$): $$l_1 + e = \frac{\lambda}{4} \quad \text{— (Equation 1)}$$
- Second Resonance (Length $l_2$): $$l_2 + e = \frac{3\lambda}{4} \quad \text{— (Equation 2)}$$
- Third Resonance (Length $l_3$): $$l_3 + e = \frac{5\lambda}{4} \quad \text{— (Equation 3)}$$
Determination of Sound Velocity (Eliminating $e$): Subtracting Equation 1 from Equation 2 eliminates the unknown end correction $e$ entirely:
$$l_2 – l_1 = \frac{3\lambda}{4} – \frac{\lambda}{4} = \frac{\lambda}{2} \implies \mathbf{\lambda = 2(l_2 – l_1)}$$
Substituting $\lambda$ into the wave equation $v = f\lambda$ yields the standard laboratory formula:
$$\mathbf{v = 2 f (l_2 – l_1)}$$
Determination of End Correction $e$: Multiplying Equation 1 by 3 and subtracting Equation 2:
$$3(l_1 + e) – (l_2 + e) = 3\left(\frac{\lambda}{4}\right) – \frac{3\lambda}{4} = 0$$
$$2e = l_2 – 3l_1 \implies \mathbf{e = \frac{l_2 – 3l_1}{2}}$$
Consistency Condition: In an ideal tube without end correction ($e = 0$), $l_2 = 3l_1$. In real tubes with $e > 0$, notice that $\frac{l_2 + e}{l_1 + e} = 3 \implies l_2 = 3l_1 + 2e$. Thus, in reality, $l_2$ is always slightly greater than $3l_1$.
7. Temperature, Molecular Mass & Gas Composition Shifts
Because sound speed in an ideal gas depends on thermodynamic variables ($v = \sqrt{\frac{\gamma R T}{M}}$), the resonant frequency of an organ pipe scales dynamically:
$$f \propto v \propto \sqrt{\frac{\gamma T}{M}}$$
- Temperature Dependence: For a given gas, $f \propto \sqrt{T}$ (where $T$ is in Kelvin). As ambient temperature rises, sound speed increases, sharpening the pitch of the organ pipe: $$\frac{\Delta f}{f} \approx \frac{1}{2} \frac{\Delta T}{T}$$
- Gas Replacement (Helium / Hydrogen): If air ($M \approx 28.8\text{ g/mol}$, $\gamma = 1.4$) is replaced by Helium ($M = 4.0\text{ g/mol}$, $\gamma = 1.67$), sound speed increases by nearly a factor of 3 ($v_{\text{He}} \approx 1007\text{ m/s}$ vs $v_{\text{air}} \approx 343\text{ m/s}$), causing the pitch to rise dramatically by over an octave.
Four Graded Solved Examples
Example 1: Fundamental & Overtone Matching (Basic to Intermediate)
Problem Statement: An open organ pipe of length $L_o = 60.0\text{ cm}$ resonates in its fundamental mode. A closed organ pipe resonates in its third harmonic (first overtone). If both pipes are sounding in unison at the same frequency in the same room, find the physical length $L_c$ of the closed organ pipe. (Neglect end corrections).
Step-by-Step Solution:
1. Open Pipe Fundamental Frequency: For an open pipe of length $L_o$, the fundamental frequency is:
$$f_o = \frac{v}{2L_o}$$
2. Closed Pipe Third Harmonic Frequency: In a closed pipe, only odd harmonics exist. The fundamental is $f_1 = \frac{v}{4L_c}$, and the first overtone corresponds to the third harmonic ($n = 2$ mode, harmonic number 3):
$$f_c = 3\left(\frac{v}{4L_c}\right) = \frac{3v}{4L_c}$$
3. Equating Frequencies: Since both pipes resonate in unison ($f_o = f_c$):
$$\frac{v}{2L_o} = \frac{3v}{4L_c}$$
Cancel wave speed $v$ from both sides:
$$\frac{1}{2L_o} = \frac{3}{4L_c} \implies 4L_c = 6L_o \implies L_c = \frac{6}{4}L_o = 1.5 L_o$$
4. Numerical Evaluation: With $L_o = 60.0\text{ cm}$:
$$L_c = 1.5 \times 60.0\text{ cm} = \mathbf{90.0\text{ cm}}$$
Example 2: Rigorous End Correction & Percentage Error (Intermediate Application)
Problem Statement: An open organ pipe has an internal radius $r = 2.50\text{ cm}$ and a physical length $L = 40.0\text{ cm}$. Take the speed of sound in air as $v = 344\text{ m/s}$.
(a) Calculate the theoretical fundamental frequency neglecting end correction.
(b) Calculate the true fundamental frequency incorporating Rayleigh’s end correction.
(c) Determine the percentage error incurred if end correction is neglected.
Step-by-Step Solution:
1. Part (a) — Idealized Frequency: Neglecting end correction ($L_{\text{ideal}} = 40.0\text{ cm} = 0.400\text{ m}$):
$$f_{\text{ideal}} = \frac{v}{2L} = \frac{344}{2 \times 0.400} = \frac{344}{0.800} = \mathbf{430.0\text{ Hz}}$$
2. Part (b) — Actual Frequency with End Correction: An open pipe has two open ends, so each end contributes an acoustic extension $e = 0.6 r$:
$$e = 0.6 \times 2.50\text{ cm} = 1.50\text{ cm} = 0.0150\text{ m}$$
The effective acoustic length is:
$$L_{\text{eff}} = L + 2e = 40.0\text{ cm} + 2(1.50\text{ cm}) = 43.0\text{ cm} = 0.430\text{ m}$$
The true fundamental frequency is:
$$f_{\text{actual}} = \frac{v}{2L_{\text{eff}}} = \frac{344}{2 \times 0.430} = \frac{344}{0.860} = \mathbf{400.0\text{ Hz}}$$
3. Part (c) — Percentage Error: The relative discrepancy relative to the true physical frequency is:
$$\text{Percentage Error} = \frac{f_{\text{ideal}} – f_{\text{actual}}}{f_{\text{actual}}} \times 100\% = \frac{430.0 – 400.0}{400.0} \times 100\% = \frac{30.0}{400.0} \times 100\% = \mathbf{7.50\%}$$
Example 3: Resonance Tube Parameter Inversion (Advanced JEE Pattern)
Problem Statement: In a resonance tube experiment conducted with a tuning fork of frequency $f = 500\text{ Hz}$, the first resonance is obtained when the water surface is at a depth $l_1 = 16.0\text{ cm}$ below the open top, and the second resonance occurs at $l_2 = 50.0\text{ cm}$.
(a) Determine the speed of sound $v$ in air.
(b) Calculate the end correction $e$ of the resonance tube.
(c) Find the internal radius $r$ of the tube using Rayleigh’s relation.
(d) Predict the air column length $l_3$ required to observe the third resonance.
Step-by-Step Solution:
1. Part (a) — Speed of Sound: The difference between two successive resonances corresponds precisely to a half wavelength ($\lambda/2$):
$$\frac{\lambda}{2} = l_2 – l_1 = 50.0\text{ cm} – 16.0\text{ cm} = 34.0\text{ cm} = 0.340\text{ m}$$
$$\lambda = 2 \times 0.340\text{ m} = 0.680\text{ m}$$
Using $v = f\lambda$:
$$v = 500\text{ Hz} \times 0.680\text{ m} = \mathbf{340.0\text{ m/s}}$$
2. Part (b) — End Correction $e$: From the first resonance condition:
$$l_1 + e = \frac{\lambda}{4} = \frac{68.0\text{ cm}}{4} = 17.0\text{ cm}$$
$$e = 17.0\text{ cm} – l_1 = 17.0\text{ cm} – 16.0\text{ cm} = \mathbf{1.00\text{ cm}}$$
Cross-verification: $e = \frac{l_2 – 3l_1}{2} = \frac{50.0 – 3(16.0)}{2} = \frac{50.0 – 48.0}{2} = 1.00\text{ cm}$. Perfectly verified.
3. Part (c) — Internal Radius $r$: Using $e = 0.6 r$:
$$r = \frac{e}{0.6} = \frac{1.00\text{ cm}}{0.6} = \frac{10}{6}\text{ cm} \approx \mathbf{1.67\text{ cm}}$$
4. Part (d) — Third Resonance Length $l_3$: The third resonance condition satisfies:
$$l_3 + e = \frac{5\lambda}{4} = 5(17.0\text{ cm}) = 85.0\text{ cm}$$
$$l_3 = 85.0\text{ cm} – e = 85.0\text{ cm} – 1.0\text{ cm} = \mathbf{84.0\text{ cm}}$$
(Notice also that $l_3 = l_2 + \frac{\lambda}{2} = 50.0\text{ cm} + 34.0\text{ cm} = 84.0\text{ cm}$).
Example 4: Gas Inversion, Temperature Scaling & Coupled Modes (JEE Advanced Mastery)
Problem Statement: A cylindrical tube of physical length $L = 85.0\text{ cm}$ closed at one end is filled with air at temperature $T_1 = 300\text{ K}$. It is observed to resonate in its 3rd harmonic (first overtone). The tube is then thoroughly evacuated and refilled with an ideal monatomic gas (Helium, $M_{\text{He}} = 4.00\text{ g/mol}$, $\gamma_{\text{He}} = 5/3$) and heated to temperature $T_2$. In this helium environment, the pipe resonates in its fundamental mode at the exact same frequency as before. For air, take $M_{\text{air}} = 28.8\text{ g/mol}$ and $\gamma_{\text{air}} = 7/5$. Neglect end corrections.
Determine the required absolute temperature $T_2$ of the Helium gas.
Step-by-Step Solution:
1. Initial Frequency in Air: The closed pipe resonates in its 3rd harmonic ($N = 3$, first overtone):
$$f_{\text{initial}} = 3 \left(\frac{v_{\text{air}}}{4L}\right) = \frac{3}{4L} \sqrt{\frac{\gamma_{\text{air}} R T_1}{M_{\text{air}}}}$$
2. Final Frequency in Helium: The closed pipe now resonates in its fundamental mode ($N = 1$):
$$f_{\text{final}} = 1 \left(\frac{v_{\text{He}}}{4L}\right) = \frac{1}{4L} \sqrt{\frac{\gamma_{\text{He}} R T_2}{M_{\text{He}}}}$$
3. Equating the Two Frequencies:
$$\frac{3}{4L} \sqrt{\frac{\gamma_{\text{air}} R T_1}{M_{\text{air}}}} = \frac{1}{4L} \sqrt{\frac{\gamma_{\text{He}} R T_2}{M_{\text{He}}}}$$
Cancel the common factor $\frac{1}{4L}\sqrt{R}$:
$$3 \sqrt{\frac{\gamma_{\text{air}} T_1}{M_{\text{air}}}} = \sqrt{\frac{\gamma_{\text{He}} T_2}{M_{\text{He}}}}$$
4. Squaring Both Sides:
$$9 \left(\frac{\gamma_{\text{air}} T_1}{M_{\text{air}}}\right) = \frac{\gamma_{\text{He}} T_2}{M_{\text{He}}}$$
5. Isolating $T_2$:
$$T_2 = 9 T_1 \times \left(\frac{\gamma_{\text{air}}}{\gamma_{\text{He}}}\right) \times \left(\frac{M_{\text{He}}}{M_{\text{air}}}\right)$$
6. Substituting Given Values:
$$\frac{\gamma_{\text{air}}}{\gamma_{\text{He}}} = \frac{7/5}{5/3} = \frac{7}{5} \times \frac{3}{5} = \frac{21}{25} = 0.84$$
$$\frac{M_{\text{He}}}{M_{\text{air}}} = \frac{4.00}{28.8} = \frac{1}{7.2} = \frac{10}{72} = \frac{5}{36}$$
Substitute into $T_2$ with $T_1 = 300\text{ K}$:
$$T_2 = 9 \times 300 \times \frac{21}{25} \times \frac{5}{36}$$
Simplify factors:
$$T_2 = (9 \times 300) \times \frac{21 \times 5}{25 \times 36} = 2700 \times \frac{105}{900} = 3 \times 105 = \mathbf{315\text{ K}}$$
Physical Insight: Even though Helium has a much smaller molecular mass (which drastically accelerates wave speed by a factor of $\approx 2.6$), shifting from the 3rd harmonic down to the fundamental mode requires an exact 3-fold frequency reduction, balanced precisely at $T_2 = 315\text{ K}$ ($42^\circ\text{C}$).
Practice Worksheet: 10 NTA-Pattern Questions
Section A: Single Correct Questions (Q1 to Q5)
Q1. An open organ pipe has a fundamental frequency of $300\text{ Hz}$. If one of its ends is suddenly closed tightly with a rigid cap without altering its length, what will be the frequency of its first overtone?
(A) $150\text{ Hz}$
(B) $300\text{ Hz}$
(C) $450\text{ Hz}$
(D) $600\text{ Hz}$
Q2. A tube open at both ends has length $L_1$, while a tube closed at one end has length $L_2$. When both tubes resonate in their respective second overtones, their frequencies are identical. Assuming identical speed of sound and negligible end corrections, the ratio $L_1 / L_2$ is:
(A) $6/5$
(B) $5/6$
(C) $3/5$
(D) $5/3$
Q3. A closed organ pipe of physical length $L$ resonates in its fundamental mode. The displacement amplitude at the open end is $s_0$, and the pressure amplitude at the closed end is $\Delta P_0$. At a distance $x = L/3$ from the closed end inside the pipe, the pressure fluctuation amplitude is:
(A) $\Delta P_0$
(B) $\frac{\sqrt{3}}{2}\Delta P_0$
(C) $\frac{1}{2}\Delta P_0$
(D) Zero
Q4. In a resonance column apparatus, a tuning fork produces resonance at water levels $l_1 = 18.0\text{ cm}$ and $l_2 = 57.0\text{ cm}$. The end correction $e$ of the glass tube is:
(A) $1.0\text{ cm}$
(B) $1.5\text{ cm}$
(C) $2.0\text{ cm}$
(D) $3.0\text{ cm}$
Q5. An organ pipe is tuned to fundamental frequency $f_0$ at temperature $T_1 = 27^\circ\text{C}$ ($300\text{ K}$). If the ambient air temperature increases to $127^\circ\text{C}$ ($400\text{ K}$), neglecting thermal expansion of the pipe material, the new fundamental frequency will be:
(A) $\frac{4}{3} f_0$
(B) $\frac{2}{\sqrt{3}} f_0$
(C) $\frac{\sqrt{3}}{2} f_0$
(D) $\frac{16}{9} f_0$
Section B: Multi-Correct Questions (Q6 to Q8)
Q6. Consider acoustic standing waves established in a closed organ pipe of physical length $L$. Which of the following statements is/are correct?
(A) At the closed boundary, the displacement of air molecules is always zero, while the pressure variation is maximum.
(B) At the open boundary, the gauge pressure fluctuation is zero, while the particle displacement amplitude is maximum.
(C) The distance between any two adjacent displacement nodes in any harmonic is $\lambda/2$.
(D) The frequency ratio of consecutive permitted normal modes is $1 : 2 : 3 : 4$.
Q7. For an open organ pipe of physical length $L$ and internal radius $r$, which of the following statements is/are correct?
(A) Both ends act as displacement antinodes and pressure nodes.
(B) Incorporating Rayleigh’s end correction, its effective acoustic length is $L_{\text{eff}} = L + 1.2\, r$.
(C) The fundamental mode has a displacement node at the exact midpoint of the pipe.
(D) If the pipe is cut into two equal halves, the fundamental frequency of each open half is double the original fundamental frequency.
Q8. In a resonance tube experiment with an air column closed by water, two consecutive resonance lengths are obtained at $l_1$ and $l_2$ ($l_2 > l_1$) using a tuning fork of frequency $f$. Which of the following expressions is/are physically correct?
(A) The speed of sound in air is $v = 2f(l_2 – l_1)$.
(B) The acoustic end correction is $e = \frac{l_2 – 3l_1}{2}$.
(C) The wavelength of the sound wave in the tube is $\lambda = 2(l_2 – l_1)$.
(D) If a third resonance is attained at length $l_3$, then $l_3 = 2l_2 – l_1$.
Section C: Numerical Value Questions (Q9 to Q10)
Q9. An open organ pipe and a closed organ pipe have the same physical length $L = 80\text{ cm}$. Take the speed of sound in air as $v = 320\text{ m/s}$. Neglecting end corrections, find the difference in their fundamental frequencies (in $\text{Hz}$).
Q10. A cylindrical pipe of internal diameter $d = 6.0\text{ cm}$ ($r = 3.0\text{ cm}$) and physical length $L = 48.2\text{ cm}$ is open at both ends. Sound travels through air inside the tube at speed $v = 340\text{ m/s}$. Taking Rayleigh’s end correction into account, find the fundamental frequency of the pipe (in $\text{Hz}$, rounded to the nearest integer).
Comprehensive Solutions & Answer Key
Answer Key:
Q1: (C) Q2: (A) Q3: (B) Q4: (B) Q5: (B)
Q6: (A, B, C) Q7: (A, B, C, D) Q8: (A, B, C, D)
Q9: 100 Q10: 328
Solution to Q1:
For the open pipe, fundamental frequency:
$$f_o = \frac{v}{2L} = 300\text{ Hz} \implies \frac{v}{L} = 600\text{ Hz}$$
When one end is closed, the pipe becomes a closed organ pipe of the same length $L$.
Its fundamental frequency is $f_{c, 1} = \frac{v}{4L} = \frac{300}{2} = 150\text{ Hz}$.
In a closed pipe, only odd harmonics exist: 1st harmonic ($150\text{ Hz}$), 3rd harmonic ($3 \times 150 = 450\text{ Hz}$), 5th harmonic ($750\text{ Hz}$), etc.
The first overtone corresponds to the 3rd harmonic:
$$f_{\text{first overtone}} = 3 f_{c, 1} = 3 \times 150\text{ Hz} = \mathbf{450\text{ Hz}}$$
Hence, choice (C) is correct.
Solution to Q2:
1. For an open pipe of length $L_1$, the harmonics are $f_n = n \left(\frac{v}{2L_1}\right)$.
The fundamental is $n = 1$, the first overtone is $n = 2$, and the second overtone is $n = 3$ (third harmonic):
$$f_{\text{open}} = 3\left(\frac{v}{2L_1}\right)$$
2. For a closed pipe of length $L_2$, the harmonics are $f_N = N \left(\frac{v}{4L_2}\right)$ where $N$ is odd.
The fundamental is $N = 1$, the first overtone is $N = 3$, and the second overtone is $N = 5$ (fifth harmonic):
$$f_{\text{closed}} = 5\left(\frac{v}{4L_2}\right)$$
3. Equating both frequencies:
$$3\left(\frac{v}{2L_1}\right) = 5\left(\frac{v}{4L_2}\right)$$
$$\frac{3}{2L_1} = \frac{5}{4L_2} \implies \frac{L_1}{L_2} = \frac{3 \times 4}{2 \times 5} = \frac{12}{10} = \mathbf{\frac{6}{5}}$$
Hence, choice (A) is correct.
Solution to Q3:
Let the closed end be at $x = 0$ and the open end at $x = L$.
At $x = 0$, there is a pressure antinode, so the spatial pressure wave distribution is:
$$\Delta P(x) = \Delta P_0 \cos(kx)$$
For the fundamental mode of a closed pipe, the length contains a quarter wavelength:
$$L = \frac{\lambda}{4} \implies k = \frac{2\pi}{\lambda} = \frac{2\pi}{4L} = \frac{\pi}{2L}$$
At a distance $x = L/3$ from the closed end:
$$kx = \left(\frac{\pi}{2L}\right)\left(\frac{L}{3}\right) = \frac{\pi}{6} = 30^\circ$$
The pressure fluctuation amplitude is:
$$\Delta P(L/3) = \Delta P_0 \cos\left(\frac{\pi}{6}\right) = \mathbf{\frac{\sqrt{3}}{2}\Delta P_0}$$
Hence, choice (B) is correct.
Solution to Q4:
For a resonance tube apparatus:
$$e = \frac{l_2 – 3l_1}{2}$$
Given $l_1 = 18.0\text{ cm}$ and $l_2 = 57.0\text{ cm}$:
$$e = \frac{57.0 – 3(18.0)}{2} = \frac{57.0 – 54.0}{2} = \frac{3.0}{2} = \mathbf{1.5\text{ cm}}$$
Hence, choice (B) is correct.
Solution to Q5:
The speed of sound in air is proportional to the square root of absolute temperature: $v \propto \sqrt{T}$.
Consequently, fundamental frequency scales as:
$$f \propto v \propto \sqrt{T}$$
Given $T_1 = 27 + 273 = 300\text{ K}$ and $T_2 = 127 + 273 = 400\text{ K}$:
$$\frac{f_2}{f_0} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \mathbf{\frac{2}{\sqrt{3}}}$$
$$f_2 = \frac{2}{\sqrt{3}} f_0$$
Hence, choice (B) is correct.
Solution to Q6:
Let us evaluate each statement:
(A) At the rigid closed boundary, air molecules cannot displace axially ($s = 0$, node), while the pressure gradient is zero, causing maximum compression/rarefaction ($|\Delta P| = \Delta P_{\max}$, antinode). (A) is correct.
(B) At the open boundary, pressure must equal atmospheric pressure $P_0$ ($\Delta P = 0$, pressure node), so particles can freely oscillate with maximum amplitude ($s = s_{\max}$, displacement antinode). (B) is correct.
(C) In any standing wave, the separation between two successive displacement nodes is always $\lambda/2$. (C) is correct.
(D) In a closed organ pipe, only odd harmonics exist ($1 : 3 : 5 : 7$), NOT $1 : 2 : 3 : 4$. (D) is incorrect.
Hence, choices (A, B, C) are correct.
Solution to Q7:
(A) Both open ends communicate directly with the atmosphere, ensuring displacement antinodes and pressure nodes. (A) is correct.
(B) Each open end introduces an acoustic expansion $e \approx 0.6 r$, so for two open ends, $L_{\text{eff}} = L + 2e = L + 1.2 r$. (B) is correct.
(C) The fundamental mode of an open pipe has displacement antinodes at $x = 0$ and $x = L$. By symmetry, the single displacement node occurs at the exact center $x = L/2$. (C) is correct.
(D) If an open pipe of length $L$ ($f_1 = v/2L$) is cut in half, each new pipe has length $L’ = L/2$. Its new fundamental frequency is $f_1′ = \frac{v}{2(L/2)} = \frac{v}{L} = 2f_1$ (double the original). (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
(A) Successive resonance lengths satisfy $l_2 – l_1 = \lambda/2 \implies \lambda = 2(l_2 – l_1)$. Hence $v = f\lambda = 2f(l_2 – l_1)$. (A) is correct.
(B) Since $l_1 + e = \lambda/4$ and $l_2 + e = 3\lambda/4$, subtracting gives $e = \frac{l_2 – 3l_1}{2}$. (B) is correct.
(C) As derived above, $\lambda = 2(l_2 – l_1)$. (C) is correct.
(D) The third resonance satisfies $l_3 – l_2 = \lambda/2 = l_2 – l_1 \implies l_3 = 2l_2 – l_1$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
Given $L = 80\text{ cm} = 0.80\text{ m}$ and $v = 320\text{ m/s}$:
1. Fundamental frequency of the open pipe:
$$f_{\text{open}} = \frac{v}{2L} = \frac{320}{2 \times 0.80} = \frac{320}{1.60} = 200\text{ Hz}$$
2. Fundamental frequency of the closed pipe:
$$f_{\text{closed}} = \frac{v}{4L} = \frac{320}{4 \times 0.80} = \frac{320}{3.20} = 100\text{ Hz}$$
3. The difference in their fundamental frequencies is:
$$\mathbf{\Delta f = f_{\text{open}} – f_{\text{closed}} = 200\text{ Hz} – 100\text{ Hz} = 100\text{ Hz}}$$
Answer: 100
Solution to Q10:
Given internal diameter $d = 6.0\text{ cm} \implies r = 3.0\text{ cm}$.
Physical length $L = 48.2\text{ cm}$.
For an open organ pipe with two open ends, Rayleigh’s end correction is:
$$2e = 2(0.6 r) = 1.2 r = 1.2 \times 3.0\text{ cm} = 3.6\text{ cm}$$
The effective acoustic length is:
$$L_{\text{eff}} = L + 2e = 48.2\text{ cm} + 3.6\text{ cm} = 51.8\text{ cm} = 0.518\text{ m}$$
Speed of sound $v = 340\text{ m/s}$.
The fundamental frequency is:
$$f = \frac{v}{2 L_{\text{eff}}} = \frac{340}{2 \times 0.518} = \frac{340}{1.036} \approx 328.185\text{ Hz} \approx \mathbf{328\text{ Hz}}$$
Answer: 328