Prerequisites
Before studying this topic, make sure you understand:
- Wave Motion - Wave equations and parameters
- Simple Harmonic Motion - For understanding wave composition
- Trigonometric Identities - For mathematical derivations
The Hook: When Waves Collide
Ever wondered how noise-canceling headphones work? They create a wave that’s exactly opposite to the noise, and when the two waves meet — silence! That’s wave interference in action.
Musical Connection: When you press a guitar string at different frets, you create standing waves of different wavelengths. The string vibrates, but certain points (nodes) never move at all!
The Marvel of Superposition: In Doctor Strange, when multiple portals open simultaneously, energy patterns overlap and combine. Similarly, when multiple waves meet, they don’t destroy each other — they simply add up, creating beautiful interference patterns.
The Question: What happens when two identical waves traveling in opposite directions meet? The answer leads to standing waves — one of the most important phenomena in physics!
Principle of Superposition
When two or more waves meet at a point, the resultant displacement is the algebraic sum of individual displacements.
In simple terms: Waves pass through each other without permanent disturbance. When they overlap, their effects simply add up (algebraically). After passing, each wave continues as if nothing happened.
Mathematical Statement: If $y_1$, $y_2$, $y_3$, … are displacements due to individual waves, then:
$$\boxed{y_{resultant} = y_1 + y_2 + y_3 + ...}$$This is called the Principle of Superposition.
Key Properties:
- Waves maintain their identity
- No permanent change after superposition
- Valid only for small amplitudes (linear systems)
- Basis for interference, beats, and standing waves
Interactive Demo
Watch how waves superpose to create interference patterns:
Interference of Waves
When two waves of same frequency superpose, we get interference.
Constructive Interference
Waves add up to give maximum amplitude.
Condition: Waves are in phase (phase difference = $0, 2\pi, 4\pi, ...$)
Path difference: $\Delta x = 0, \lambda, 2\lambda, 3\lambda, ...$ (integer multiples of $\lambda$)
$$\boxed{\Delta x = n\lambda, \quad n = 0, 1, 2, 3, ...}$$Resultant amplitude: $A_{max} = A_1 + A_2$
If $A_1 = A_2 = A$: $A_{max} = 2A$
Destructive Interference
Waves cancel out to give minimum amplitude.
Condition: Waves are out of phase (phase difference = $\pi, 3\pi, 5\pi, ...$)
Path difference: $\Delta x = \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, ...$ (odd multiples of $\frac{\lambda}{2}$)
$$\boxed{\Delta x = (2n + 1)\frac{\lambda}{2}, \quad n = 0, 1, 2, 3, ...}$$Resultant amplitude: $A_{min} = |A_1 - A_2|$
If $A_1 = A_2 = A$: $A_{min} = 0$ (complete cancellation!)
Noise-canceling headphones detect ambient sound (wave 1), then produce an identical wave with opposite phase (wave 2). When they superpose:
$y = y_1 + y_2 = A\sin(\omega t) + A\sin(\omega t + \pi) = A\sin(\omega t) - A\sin(\omega t) = 0$
Silence! That’s destructive interference.
General Case: Two Waves with Phase Difference
Two waves:
$$y_1 = A_1\sin(\omega t)$$ $$y_2 = A_2\sin(\omega t + \phi)$$Resultant:
$$y = A\sin(\omega t + \alpha)$$Resultant amplitude:
$$\boxed{A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2\cos\phi}}$$Phase of resultant:
$$\tan\alpha = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi}$$Special Cases:
| Phase Difference $\phi$ | Resultant Amplitude $A$ | Type |
|---|---|---|
| $0$ (in phase) | $A_1 + A_2$ | Constructive |
| $\pi$ (out of phase) | $ | A_1 - A_2 |
| $\frac{\pi}{2}$ | $\sqrt{A_1^2 + A_2^2}$ | Intermediate |
If $A_1 = A_2 = A_0$:
| $\phi$ | $A$ |
|---|---|
| $0$ | $2A_0$ |
| $\frac{\pi}{2}$ | $A_0\sqrt{2}$ |
| $\pi$ | $0$ |
Standing Waves (Stationary Waves)
When two identical waves travel in opposite directions, they create a standing wave.
Formation
Two waves:
$$y_1 = A\sin(kx - \omega t)$$(traveling right)
$$y_2 = A\sin(kx + \omega t)$$(traveling left)
Resultant (using superposition):
$$y = y_1 + y_2 = A[\sin(kx - \omega t) + \sin(kx + \omega t)]$$Using $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$:
$$\boxed{y = 2A\sin(kx)\cos(\omega t)}$$This is the equation of a standing wave.
Characteristics
Equation: $y = 2A\sin(kx)\cos(\omega t)$
Can be written as: $y = [2A\sin(kx)]\cos(\omega t)$
- Amplitude varies with position: $A(x) = 2A\sin(kx)$ (not constant!)
- All particles oscillate in phase (or opposite phase)
- No net energy transfer (energy sloshes back and forth)
- Wave doesn’t propagate — it stands still!
Traveling wave: $y = A\sin(kx - \omega t)$
- Wave pattern moves
- All particles have same amplitude
- Energy propagates
Standing wave: $y = 2A\sin(kx)\cos(\omega t)$
- Wave pattern is stationary
- Amplitude varies with position
- No net energy transfer
Nodes and Antinodes
Nodes: Points of zero amplitude (no motion)
Condition: $\sin(kx) = 0$
$$kx = 0, \pi, 2\pi, 3\pi, ...$$ $$x = 0, \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2}, ...$$ $$\boxed{x_n = n\frac{\lambda}{2}, \quad n = 0, 1, 2, 3, ...}$$Distance between consecutive nodes: $\frac{\lambda}{2}$
Antinodes: Points of maximum amplitude (maximum motion)
Condition: $\sin(kx) = \pm 1$
$$kx = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...$$ $$x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, ...$$ $$\boxed{x_a = (2n + 1)\frac{\lambda}{4}, \quad n = 0, 1, 2, 3, ...}$$Distance between consecutive antinodes: $\frac{\lambda}{2}$
Distance between node and nearest antinode: $\frac{\lambda}{4}$
Node to Node = $\frac{\lambda}{2}$
Antinode to Antinode = $\frac{\lambda}{2}$
Node to Antinode = $\frac{\lambda}{4}$
Everything is either $\frac{\lambda}{2}$ or $\frac{\lambda}{4}$!
Energy in Standing Waves
- No energy transfer along the wave
- Energy oscillates between kinetic and potential
- At nodes: always zero displacement and velocity
- At antinodes: maximum displacement and velocity
Standing Waves on Strings
String Fixed at Both Ends
Boundary conditions: Nodes at both ends ($x = 0$ and $x = L$)
Allowed wavelengths:
$$L = n\frac{\lambda}{2}, \quad n = 1, 2, 3, ...$$ $$\boxed{\lambda_n = \frac{2L}{n}}$$Allowed frequencies (normal modes):
$$\boxed{f_n = \frac{nv}{2L} = n\frac{v}{2L}, \quad n = 1, 2, 3, ...}$$Where $v = \sqrt{\frac{T}{\mu}}$ is the wave speed on the string.
Fundamental frequency (first harmonic, $n = 1$):
$$\boxed{f_1 = \frac{v}{2L}}$$Overtones:
- Second harmonic: $f_2 = 2f_1$
- Third harmonic: $f_3 = 3f_1$
- $n$-th harmonic: $f_n = nf_1$
All harmonics are integer multiples of the fundamental.
When you pluck a guitar string:
- Fundamental ($n=1$): Lowest note, string vibrates as one segment
- Second harmonic ($n=2$): Octave higher, string vibrates in two segments
- Third harmonic ($n=3$): Even higher, three segments
The rich sound comes from superposition of all these harmonics!
String Fixed at One End (Open at Other)
Boundary conditions: Node at fixed end, antinode at free end
Allowed wavelengths:
$$L = (2n - 1)\frac{\lambda}{4}, \quad n = 1, 2, 3, ...$$ $$\boxed{\lambda_n = \frac{4L}{2n - 1}}$$Allowed frequencies:
$$\boxed{f_n = (2n - 1)\frac{v}{4L}, \quad n = 1, 2, 3, ...}$$Fundamental: $f_1 = \frac{v}{4L}$
Overtones: Only odd harmonics are present!
- $f_1, f_3, f_5, ...$ (fundamental, 3rd, 5th, …)
- $f_3 = 3f_1$, $f_5 = 5f_1$
Standing Waves in Air Columns (Organ Pipes)
Closed Organ Pipe (Closed at One End)
Boundary conditions:
- Closed end: Node (no displacement)
- Open end: Antinode (maximum displacement)
Same as string fixed at one end!
$$\boxed{f_n = (2n - 1)\frac{v}{4L}, \quad n = 1, 2, 3, ...}$$Where $v$ is the speed of sound in air.
Only odd harmonics: $f_1, 3f_1, 5f_1, 7f_1, ...$
Open Organ Pipe (Open at Both Ends)
Boundary conditions: Antinodes at both ends
Same as string fixed at both ends!
$$\boxed{f_n = n\frac{v}{2L}, \quad n = 1, 2, 3, ...}$$All harmonics present: $f_1, 2f_1, 3f_1, 4f_1, ...$
Comparison:
| Pipe Type | Boundary Conditions | Harmonics | Fundamental |
|---|---|---|---|
| Closed (one end) | Node-Antinode | Odd only | $\frac{v}{4L}$ |
| Open (both ends) | Antinode-Antinode | All | $\frac{v}{2L}$ |
Closed pipe: Only odd harmonics → unique timbre (flute-like)
Open pipe: All harmonics → richer sound (clarinet-like)
This is why different musical instruments have distinct sounds even when playing the same note!
Beats
When two waves of slightly different frequencies $f_1$ and $f_2$ superpose, we hear periodic variations in loudness called beats.
Formation
Two waves:
$$y_1 = A\sin(2\pi f_1 t)$$ $$y_2 = A\sin(2\pi f_2 t)$$Resultant:
$$y = y_1 + y_2 = A[\sin(2\pi f_1 t) + \sin(2\pi f_2 t)]$$Using $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$:
$$y = 2A\cos\left[2\pi\frac{f_1 - f_2}{2}t\right]\sin\left[2\pi\frac{f_1 + f_2}{2}t\right]$$ $$\boxed{y = [2A\cos(2\pi f_{beat} t)]\sin(2\pi f_{avg} t)}$$Where:
- $f_{avg} = \frac{f_1 + f_2}{2}$ (average frequency)
- $f_{beat} = \frac{|f_1 - f_2|}{2}$ (beat modulation frequency)
Beat frequency (rate of loudness variation):
$$\boxed{f_{beats} = |f_1 - f_2|}$$Since intensity $\propto$ amplitude$^2$, and amplitude oscillates at frequency $f_{beat}$, the intensity (loudness) oscillates at twice that frequency, giving beat frequency $= |f_1 - f_2|$.
Characteristics
- Beat frequency: $f_{beats} = |f_1 - f_2|$
- Loudness maxima occur $|f_1 - f_2|$ times per second
- Useful for tuning instruments: Adjust until beats disappear ($f_1 = f_2$)
When tuning a guitar, you play a reference note and the string you’re tuning. If you hear beats (wah-wah-wah sound), the string is slightly out of tune. As you adjust the tension and frequencies get closer, the beats slow down. When beats disappear, the string is perfectly tuned!
Example: If reference is 440 Hz and you hear 4 beats per second, your string is at either 436 Hz or 444 Hz.
Limitations:
- Beat frequency should be less than ~10 Hz to be perceived clearly
- If $|f_1 - f_2|$ is too large, we hear two separate notes instead of beats
Common Mistakes to Avoid
Wrong: Distance between node and antinode = $\frac{\lambda}{2}$
Right: Distance = $\frac{\lambda}{4}$
- Node to node = $\frac{\lambda}{2}$
- Antinode to antinode = $\frac{\lambda}{2}$
- Node to nearest antinode = $\frac{\lambda}{4}$
Wrong: Closed pipe has all harmonics
Right: Closed pipe has only odd harmonics
$f_1, 3f_1, 5f_1, ...$ (not $2f_1, 4f_1, ...$)
Wrong: Beat frequency = $f_1 + f_2$
Right: Beat frequency = $|f_1 - f_2|$ (difference, not sum!)
You hear the difference in frequencies.
Wrong: All points in standing wave have same amplitude
Right: Amplitude varies with position!
$A(x) = 2A\sin(kx)$ — maximum at antinodes, zero at nodes
Wrong: Standing waves transfer energy like traveling waves
Right: Standing waves have no net energy transfer
Energy oscillates locally between KE and PE.
Memory Tricks & Patterns
Node-Antinode Spacing
“Half for same, Quarter for change”
- Same to same (N→N or A→A): $\frac{\lambda}{2}$
- Change type (N→A or A→N): $\frac{\lambda}{4}$
Harmonics in Pipes
“Closed is Odd, Open is All”
- Closed pipe: Odd harmonics only
- Open pipe: All harmonics
Beat Frequency
“Beats are the Difference”
$f_{beats} = |f_1 - f_2|$ (not sum!)
Fundamental Frequencies
“Open is Double”
For same length $L$:
- Closed: $f = \frac{v}{4L}$
- Open: $f = \frac{v}{2L} = 2 \times \frac{v}{4L}$
Open pipe fundamental is twice the closed pipe fundamental!
Practice Problems
Level 1: Foundation (NCERT Style)
Two waves $y_1 = 5\sin\omega t$ and $y_2 = 5\sin(\omega t + \frac{\pi}{3})$ superpose. Find the resultant amplitude.
Solution:
$$A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2\cos\phi}$$ $$A = \sqrt{25 + 25 + 2(5)(5)\cos\frac{\pi}{3}}$$ $$A = \sqrt{50 + 50 \times 0.5} = \sqrt{50 + 25} = \sqrt{75}$$ $$A = 5\sqrt{3}$$≈ 8.66
In a standing wave, two consecutive nodes are 20 cm apart. Find the wavelength.
Solution: Distance between consecutive nodes = $\frac{\lambda}{2}$
$$\frac{\lambda}{2} = 20 \text{ cm}$$ $$\lambda = 40$$cm = 0.4 m
Two tuning forks produce 6 beats per second. If one has frequency 256 Hz, what are the possible frequencies of the other?
Solution: Beat frequency: $f_{beats} = |f_1 - f_2| = 6$ Hz
Given $f_1 = 256$ Hz:
$$|256 - f_2| = 6$$So: $f_2 = 256 + 6 = 262$ Hz or $f_2 = 256 - 6 = 250$ Hz
Answer: 250 Hz or 262 Hz
Level 2: JEE Main
A string of length 1 m is fixed at both ends. The wave speed is 200 m/s. Find: (a) Fundamental frequency (b) Second harmonic frequency
Solution: (a) Fundamental: $f_1 = \frac{v}{2L} = \frac{200}{2 \times 1} = $ 100 Hz
(b) Second harmonic: $f_2 = 2f_1 = 2 \times 100 = $ 200 Hz
A closed organ pipe of length 50 cm produces its fundamental note. The speed of sound is 340 m/s. Find: (a) Fundamental frequency (b) Next allowed frequency
Solution: (a) Fundamental: $f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.5} = \frac{340}{2} = $ 170 Hz
(b) Closed pipe has only odd harmonics: Next frequency = $f_3 = 3f_1 = 3 \times 170 = $ 510 Hz
(Not $f_2$, because even harmonics don’t exist!)
An open organ pipe and a closed organ pipe have the same fundamental frequency. If the length of the closed pipe is 30 cm, what is the length of the open pipe?
Solution: Closed: $f_1 = \frac{v}{4L_c}$ Open: $f_1 = \frac{v}{2L_o}$
Since frequencies are equal:
$$\frac{v}{4L_c} = \frac{v}{2L_o}$$ $$\frac{1}{4L_c} = \frac{1}{2L_o}$$ $$2L_o = 4L_c$$ $$L_o = 2L_c = 2 \times 30 = $$60 cm
Open pipe must be twice as long to have same fundamental frequency!
Level 3: JEE Advanced
A string is fixed at both ends and vibrates in 5 loops (5th harmonic). If the length is 2 m and frequency is 500 Hz, find: (a) Wavelength (b) Wave speed on string
Solution: (a) For 5th harmonic ($n = 5$):
$$L = n\frac{\lambda}{2} = 5\frac{\lambda}{2}$$ $$2 = 5\frac{\lambda}{2}$$ $$\lambda = \frac{4}{5} = 0.8$$m = 80 cm
(b) $v = f\lambda = 500 \times 0.8 = $ 400 m/s
Two identical strings are in unison (same frequency). When tension in one is increased by 2%, how many beats per second will be heard if the original frequency was 200 Hz?
Solution: Frequency $\propto \sqrt{T}$ (since $v = \sqrt{\frac{T}{\mu}}$ and $f = \frac{v}{2L}$)
Original: $f_1 = 200$ Hz
When tension increases by 2%: $T' = 1.02T$
$$f_2 = f_1\sqrt{\frac{T'}{T}} = 200\sqrt{1.02}$$Using $\sqrt{1 + x} \approx 1 + \frac{x}{2}$ for small $x$:
$$f_2 = 200(1 + \frac{0.02}{2}) = 200(1.01) = 202$$Hz
Beat frequency: $f_{beats} = |f_2 - f_1| = 202 - 200 = $ 2 Hz
A standing wave equation is $y = 0.04\sin(5\pi x)\cos(200\pi t)$ (SI units). Find: (a) Amplitude of component waves (b) Wavelength (c) Frequency (d) Speed of component waves
Solution: Comparing with $y = 2A\sin(kx)\cos(\omega t)$:
(a) $2A = 0.04$, so $A = 0.02$ m = 2 cm
(b) $k = 5\pi = \frac{2\pi}{\lambda}$
$$\lambda = \frac{2\pi}{5\pi} = \frac{2}{5} = 0.4$$m = 40 cm
(c) $\omega = 200\pi = 2\pi f$
$$f = \frac{200\pi}{2\pi} = 100$$Hz
(d) $v = f\lambda = 100 \times 0.4 = $ 40 m/s
Or: $v = \frac{\omega}{k} = \frac{200\pi}{5\pi} = $ 40 m/s ✓
Quick Revision Box
Interference
| Type | Phase Difference | Path Difference | Amplitude |
|---|---|---|---|
| Constructive | $0, 2\pi, 4\pi, ...$ | $n\lambda$ | $A_1 + A_2$ |
| Destructive | $\pi, 3\pi, 5\pi, ...$ | $(2n+1)\frac{\lambda}{2}$ | $ |
Standing Waves
| Feature | Formula |
|---|---|
| Equation | $y = 2A\sin(kx)\cos(\omega t)$ |
| Node positions | $x = n\frac{\lambda}{2}$ |
| Antinode positions | $x = (2n+1)\frac{\lambda}{4}$ |
| Node spacing | $\frac{\lambda}{2}$ |
| Node-Antinode spacing | $\frac{\lambda}{4}$ |
Strings and Pipes
| System | Boundary | Harmonics | Fundamental |
|---|---|---|---|
| String (both fixed) | Node-Node | All | $\frac{v}{2L}$ |
| String (one free) | Node-Antinode | Odd only | $\frac{v}{4L}$ |
| Closed pipe | Node-Antinode | Odd only | $\frac{v}{4L}$ |
| Open pipe | Antinode-Antinode | All | $\frac{v}{2L}$ |
Beats
$$f_{beats} = |f_1 - f_2|$$Related Topics
Within Oscillations & Waves
- Wave Motion — Foundation for superposition
- Simple Harmonic Motion — Individual wave components
- Doppler Effect — Frequency changes in moving sources
Connected Chapters
- Sound Waves — Applications of standing waves and beats
- Optics — Light interference and diffraction
- Electromagnetic Waves — EM wave superposition
Math Connections
- Trigonometric Identities — Sum-to-product formulas
- Complex Numbers — Alternative representation
- Fourier Series — Decomposition into harmonics
Teacher’s Summary
Superposition Principle: When waves meet, displacements add algebraically. Waves pass through each other unchanged.
Interference:
- Constructive: Path difference = $n\lambda$ → Maximum amplitude
- Destructive: Path difference = $(2n+1)\frac{\lambda}{2}$ → Minimum amplitude
Standing Waves: $y = 2A\sin(kx)\cos(\omega t)$
- Formed by two identical waves traveling in opposite directions
- Nodes (zero amplitude) at $x = n\frac{\lambda}{2}$
- Antinodes (max amplitude) at $x = (2n+1)\frac{\lambda}{4}$
- No energy transfer
Strings and Pipes:
- Both ends same (fixed-fixed or open-open): All harmonics, $f_n = n\frac{v}{2L}$
- One end different (fixed-free or closed-open): Odd harmonics only, $f_n = (2n-1)\frac{v}{4L}$
Beats: $f_{beats} = |f_1 - f_2|$
- Used for tuning instruments
- Audible for $|f_1 - f_2| < 10$ Hz
JEE Strategy:
- Memorize node-antinode spacing: $\frac{\lambda}{2}$ for same, $\frac{\lambda}{4}$ for different
- Know which systems have odd-only harmonics
- Beats = difference in frequencies
“When waves dance together, they create patterns more beautiful than either could alone — from music to light, superposition is nature’s symphony.”