NEB Curriculum Objectives Checklist (Unit 8)
Formation of Stationary Waves in Pipes
When a sound wave (longitudinal wave) travels through an air column enclosed inside a pipe, it undergoes reflection at the boundary end. The incident sound wave and the reflected sound wave—having the same amplitude, frequency, and speed, traveling in opposite directions along the same line—superimpose according to the Principle of Superposition. This superposition creates a longitudinal stationary (standing) wave inside the pipe.
1. Reflection at a Closed Boundary
At a rigid or closed end, air molecules are physically restricted from moving back and forth. Therefore, a Displacement Node (zero particle displacement) and a Pressure Antinode (maximum pressure variation) are formed at the closed boundary. Phase changes by $\pi$ radians ($180^\circ$).
2. Reflection at an Open Boundary
At an open end, air molecules are free to vibrate with maximum amplitude into the atmosphere. Therefore, a Displacement Antinode (maximum particle displacement) and a Pressure Node (zero pressure variation) are formed near the open boundary.
Harmonics and Overtones
Fundamental Frequency / Fundamental Tone
The lowest possible frequency produced by a vibrating body or air column is called the fundamental frequency ($f_1$) or fundamental tone.
Overtones
All frequencies higher than the fundamental frequency actually emitted by the vibrating system are called overtones ($f_2, f_3, f_4, \dots$).
Harmonics Definition
Harmonics are integral multiples of the fundamental frequency ($1f_1, 2f_1, 3f_1, 4f_1, \dots$).
• $1f_1$ is the 1st harmonic (Fundamental frequency).
• $2f_1$ is the 2nd harmonic.
• $3f_1$ is the 3rd harmonic, and so on.
Distinction Matrix: Harmonics vs Overtones
| Parameter / Aspect | Harmonics | Overtones |
|---|---|---|
| Definition | Integral multiples of fundamental frequency ($n f_1$). | Higher frequencies produced above fundamental tone. |
| Inclusion of Fundamental | Includes the fundamental frequency ($1st \text{ Harmonic} = f_1$). | Excludes fundamental frequency ($1st \text{ Overtone} > f_1$). |
| Presence in Instruments | May or may not be physically present in a system (e.g., closed pipe lacks even harmonics). | Always represents actual physically present higher frequencies. |
| Numbering Sequence | 1st, 2nd, 3rd, 4th Harmonic... ($f_1, 2f_1, 3f_1, 4f_1 \dots$) | 1st, 2nd, 3rd Overtone... (1st present, 2nd present, etc.) |
Harmonics and Overtones in Closed and Open Organ Pipes
A. Closed Organ Pipe
A pipe open at one end and closed at the other end of length $L$.
Vibration Modes in Closed Organ Pipe
Mathematical Derivations for Closed Pipe:
1. Mode 1: Fundamental Mode (1st Harmonic)
In the simplest mode, one Node (N) forms at the closed end and one Antinode (A) forms at the open mouth.
Length of pipe, \( L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L \)
Fundamental Frequency, \( f_1 = \frac{v}{\lambda_1} = \frac{v}{4L} \) \quad ---(Equation 1)
2. Mode 2: First Overtone (3rd Harmonic)
In this mode, two Nodes and two Antinodes are formed inside the pipe.
Length of pipe, \( L = \frac{3\lambda_2}{4} \implies \lambda_2 = \frac{4L}{3} \)
Frequency, \( f_2 = \frac{v}{\lambda_2} = \frac{3v}{4L} = 3 f_1 \) \quad ---(Equation 2)
Thus, the 1st overtone is equal to the 3rd harmonic.
3. Mode 3: Second Overtone (5th Harmonic)
In this mode, three Nodes and three Antinodes are formed.
Length of pipe, \( L = \frac{5\lambda_3}{4} \implies \lambda_3 = \frac{4L}{5} \)
Frequency, \( f_3 = \frac{v}{\lambda_3} = \frac{5v}{4L} = 5 f_1 \) \quad ---(Equation 3)
Conclusion for Closed Organ Pipe:
Frequency ratio: \( f_1 : f_2 : f_3 : \dots = 1 : 3 : 5 : \dots \)
Key Result: Only ODD harmonics are present in a closed organ pipe. Even harmonics ($2f_1, 4f_1, \dots$) are absent.
B. Open Organ Pipe
A pipe open at both ends of length $L$.
Vibration Modes in Open Organ Pipe
Mathematical Derivations for Open Pipe:
1. Mode 1: Fundamental Mode (1st Harmonic)
Antinodes form at both open ends with one Node at the center.
Length of pipe, \( L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L \)
Fundamental Frequency, \( f_1' = \frac{v}{\lambda_1} = \frac{v}{2L} \) \quad ---(Equation 4)
2. Mode 2: First Overtone (2nd Harmonic)
Three antinodes and two nodes are formed.
Length of pipe, \( L = \lambda_2 \implies \lambda_2 = L \)
Frequency, \( f_2' = \frac{v}{\lambda_2} = \frac{v}{L} = 2 \left(\frac{v}{2L}\right) = 2 f_1' \) \quad ---(Equation 5)
3. Mode 3: Second Overtone (3rd Harmonic)
Four antinodes and three nodes are formed.
Length of pipe, \( L = \frac{3\lambda_3}{2} \implies \lambda_3 = \frac{2L}{3} \)
Frequency, \( f_3' = \frac{v}{\lambda_3} = \frac{3v}{2L} = 3 f_1' \) \quad ---(Equation 6)
Conclusion for Open Organ Pipe:
Frequency ratio: \( f_1' : f_2' : f_3' : \dots = 1 : 2 : 3 : 4 : \dots \)
Key Result: BOTH EVEN AND ODD harmonics are present in an open organ pipe. Therefore, sound from an open organ pipe is richer and more musical than a closed organ pipe of the same length.
Note: For equal physical lengths, \( f_{open} = 2 \times f_{closed} \).
End Correction in Pipes and Resonance Tube Experiment
1. Concept and Physical Cause of End Correction
In actual practice, air particles at the open boundary of an organ pipe are not completely confined laterally. Due to inertia, the air particles outside the brim of the pipe also participate in vibration. Consequently, the displacement antinode is not formed exactly at the open end plane, but at a small distance \( e \) outside the boundary brim.
This additional distance \( e \) added to the physical length of the pipe is called End Correction.
Lord Rayleigh's Formula
Lord Rayleigh demonstrated theoretically and experimentally that end correction depends directly on the internal radius \( r \) of the pipe:
- For a Closed Pipe (1 open end): Effective length \( L_{eff} = L + e = L + 0.6r \)
- For an Open Pipe (2 open ends): Effective length \( L_{eff} = L + 2e = L + 1.2r \)
End Correction Diagram
Corrected Frequency Formulas
\( f = \frac{v}{4(L + e)} = \frac{v}{4(L + 0.6r)} \)
\( f = \frac{v}{2(L + 2e)} = \frac{v}{2(L + 1.2r)} \)
2. Determination of Speed of Sound using Resonance Tube Apparatus
The Resonance Tube is an experimental setup consisting of a vertical glass tube open at the top and connected to a water reservoir. By raising or lowering the water reservoir, the length of the air column inside the tube (which acts as a closed organ pipe) is adjusted.
Resonance Tube Setup
Experimental Step-by-Step Derivation:
Let a tuning fork of known frequency \( f \) be vibrated over the mouth of the tube.
\( l_1 + e = \frac{\lambda}{4} \) \quad ---(1)
\( l_2 + e = \frac{3\lambda}{4} \) \quad ---(2)
Subtracting Equation (1) from Equation (2):
Therefore, the velocity of sound in air at room temperature \( t\,^\circ\text{C} \) is:
Determining End Correction from Experiment:
Multiplying equation (1) by 3 and subtracting equation (2):
\( 3(l_1 + e) - (l_2 + e) = \frac{3\lambda}{4} - \frac{3\lambda}{4} = 0 \implies 3l_1 + 3e - l_2 - e = 0 \implies 2e = l_2 - 3l_1 \)
\( \mathbf{e = \frac{l_2 - 3l_1}{2}} \)
Transverse Waves in Stretched Strings & Vibration Laws
1. Velocity of Transverse Waves along a Stretched String
When a stretched string is plucked transversely, transverse stationary waves are set up. The speed \( v \) of the transverse wave depends on the tension \( T \) in the string and its linear mass density \( \mu \) (mass per unit length):
Expressing Linear Mass Density (\( \mu \)) in terms of Radius and Density:
\( \mu = \frac{\text{Mass}}{\text{Length}} = \frac{\text{Volume} \times \text{Density}}{\text{Length}} = \frac{(\pi r^2 L) \times \rho}{L} = \pi r^2 \rho \)
Where \( r = \) radius of wire cross-section, \( \rho = \) density of wire material.
\( \therefore v = \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{r} \sqrt{\frac{T}{\pi \rho}} \)
2. Modes of Vibration and Overtones in Stretched String
Since both ends of a fixed string of length \( L \) are clamped rigid, Nodes (N) are always formed at both boundaries \( x = 0 \) and \( x = L \).
Vibration Modes of Fixed String
Fundamental Mode (n=1)
\( L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L \)
\( f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \)
1st Overtone (n=2)
\( L = \lambda_2 \implies \lambda_2 = L \)
\( f_2 = \frac{2}{2L}\sqrt{\frac{T}{\mu}} = 2f_1 \)
2nd Overtone (n=3)
\( L = \frac{3\lambda_3}{2} \implies \lambda_3 = \frac{2L}{3} \)
\( f_3 = \frac{3}{2L}\sqrt{\frac{T}{\mu}} = 3f_1 \)
Frequency ratio: \( f_1 : f_2 : f_3 : \dots = 1 : 2 : 3 : 4 : \dots \) (All harmonics present)
3. Laws of Transverse Vibration of Fixed String
From the fundamental frequency equation \( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \frac{1}{2L r}\sqrt{\frac{T}{\pi \rho}} \), the following five laws are derived:
Fundamental frequency is inversely proportional to resonating length when \( T \) and \( \mu \) are constant.
\( f \propto \frac{1}{L} \implies f L = \text{constant} \)
Fundamental frequency is directly proportional to the square root of tension when \( L \) and \( \mu \) are constant.
\( f \propto \sqrt{T} \implies \frac{f}{\sqrt{T}} = \text{constant} \)
Fundamental frequency is inversely proportional to square root of linear mass density when \( L \) and \( T \) are constant.
\( f \propto \frac{1}{\sqrt{\mu}} \)
Fundamental frequency is inversely proportional to radius of wire when \( L, T, \rho \) are constant.
\( f \propto \frac{1}{r} \)
Fundamental frequency is inversely proportional to square root of wire material density when \( L, T, r \) are constant.
\( f \propto \frac{1}{\sqrt{\rho}} \)
4. Sonometer Apparatus Experimental Verification
A Sonometer consists of a hollow rectangular wooden soundbox (to amplify sound by forced vibration) with a fixed pulley at one end. A uniform metallic wire fixed at one end passes over two movable knife-edge wooden bridges ($A$ and $B$) and a smooth pulley, carrying a hanger with weights ($M$) to provide tension $T = M g$.
Sonometer Experimental Setup
A light inverted 'V' shaped paper rider is placed at the center of wire segment $AB$. A vibrating tuning fork of known frequency $f$ is touched to the soundbox. The bridge $B$ is moved slowly until the paper rider violently flutter and falls off. This indicates maximum resonance amplitude, confirming that the fundamental frequency of string segment $AB$ matches the frequency of the tuning fork.
Waves in Pipes & Strings Laboratory
1. Real-Time Standing Wave Mode Visualizer
2. Organ Pipe Frequency & End Correction Calculator
3. Stretched String & Sonometer Calculator
Solved Numerical Examples
Numerical 1: Resonance Tube Speed of Sound & End Correction
"In a resonance tube experiment, a tuning fork of frequency $512\text{ Hz}$ produces first resonance at length $15.8\text{ cm}$ and second resonance at length $48.2\text{ cm}$. Calculate: (a) Speed of sound in air, (b) End correction of the tube, and (c) Internal radius of the tube."
Step-by-Step Solution:
Given Data:
- Frequency of tuning fork, $f = 512\text{ Hz}$
- First resonating length, $l_1 = 15.8\text{ cm} = 0.158\text{ m}$
- Second resonating length, $l_2 = 48.2\text{ cm} = 0.482\text{ m}$
(a) Speed of Sound ($v$):
\( v = 2 f (l_2 - l_1) = 2 \times 512 \times (0.482 - 0.158) \)
\( v = 1024 \times 0.324 = \mathbf{331.78\text{ m/s}} \)
(b) End Correction ($e$):
\( e = \frac{l_2 - 3 l_1}{2} = \frac{48.2 - 3(15.8)}{2} = \frac{48.2 - 47.4}{2} = \frac{0.8}{2} = \mathbf{0.4\text{ cm}} = 0.004\text{ m} \)
(c) Internal Radius ($r$):
Using Rayleigh formula \( e = 0.6 r \implies r = \frac{e}{0.6} = \frac{0.4}{0.6} = \mathbf{0.67\text{ cm}} \)
Numerical 2: Equal Frequency Pipe Length Relation
"An open organ pipe has a fundamental frequency equal to the third harmonic (1st overtone) of a closed organ pipe. If the length of the closed pipe is $60\text{ cm}$, calculate the length of the open pipe (neglecting end correction)."
Step-by-Step Solution:
Given Data:
- Length of closed pipe, $L_c = 60\text{ cm} = 0.60\text{ m}$
- Fundamental frequency of open pipe, $f_{1,open} = \frac{v}{2 L_o}$
- Third harmonic of closed pipe, $f_{3,closed} = \frac{3 v}{4 L_c}$
According to the problem condition: $f_{1,open} = f_{3,closed}$
\( \frac{v}{2 L_o} = \frac{3 v}{4 L_c} \)
Canceling $v$ from both sides: \( \frac{1}{2 L_o} = \frac{3}{4 L_c} \implies 3 \times 2 L_o = 4 L_c \implies L_o = \frac{2}{3} L_c \)
\( L_o = \frac{2}{3} \times 60\text{ cm} = \mathbf{40\text{ cm}} = 0.40\text{ m} \)
Numerical 3: Sonometer Wire Frequency & Mass
"A steel wire of length $1\text{ m}$ and mass $5\text{ g}$ is stretched with a tension of $400\text{ N}$. Find: (a) Speed of transverse wave on wire, (b) Fundamental frequency, and (c) Frequency of its 2nd overtone."
Step-by-Step Solution:
- Length $L = 1\text{ m}$, Mass $m = 5\text{ g} = 0.005\text{ kg}$
- Linear mass density \( \mu = \frac{m}{L} = \frac{0.005\text{ kg}}{1\text{ m}} = 0.005\text{ kg/m} \)
- Tension $T = 400\text{ N}$
(a) Wave Speed ($v$):
\( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.005}} = \sqrt{80000} = \mathbf{282.84\text{ m/s}} \)
(b) Fundamental Frequency ($f_1$):
\( f_1 = \frac{v}{2L} = \frac{282.84}{2 \times 1} = \mathbf{141.42\text{ Hz}} \)
(c) Frequency of 2nd Overtone ($f_3 = 3rd\text{ harmonic}$):
\( f_3 = 3 f_1 = 3 \times 141.42 = \mathbf{424.26\text{ Hz}} \)
Conceptual Questions & Practice Problems
NEB Conceptual Short Questions (With Answers)
Q1: Why is sound produced by an open organ pipe richer than that by a closed organ pipe? ↓
NEB Board Answer: In an open organ pipe, all harmonics (both even and odd: $1f_1, 2f_1, 3f_1, 4f_1 \dots$) are generated. In contrast, a closed organ pipe produces only odd harmonics ($1f_1, 3f_1, 5f_1 \dots$). Because the presence of a greater number of harmonic overtones enriches the quality and timbre of sound, an open organ pipe produces a richer and more pleasing musical sound.
Q2: What happens to the fundamental frequency of an organ pipe if temperature increases? ↓
NEB Board Answer: The velocity of sound in air increases with temperature ($v \propto \sqrt{T_K}$). Since fundamental frequency $f = \frac{v}{2L}$ (open) or $f = \frac{v}{4L}$ (closed) is directly proportional to speed of sound $v$, as temperature rises, speed of sound increases, and consequently the fundamental frequency of the organ pipe increases (pitch becomes higher).
Q3: Why are sound holes provided in the wooden soundbox of a sonometer? ↓
NEB Board Answer: Sound holes allow the air inside the hollow wooden soundbox to communicate with the outside atmosphere and set a larger volume of air into forced vibrations, significantly amplifying the loudness of the sound produced by the vibrating string.
Interactive Self-Assessment Test
1. In a closed organ pipe of length $L$, the ratio of frequencies of fundamental note, 1st overtone, and 2nd overtone is:
2. Rayleigh end correction formula for an organ pipe of internal radius $r$ is approximately:
3. In a resonance tube experiment, if $l_1$ and $l_2$ are first and second resonating lengths, speed of sound is:
4. If tension in a stretched sonometer wire is quadrupled (4x), keeping length and wire density constant, fundamental frequency will: