Lesson 6: Superposition of Waves | State Board · Class 12 · Physics

Progressive waves, wave equation and reflection

From daily life

A stone dropped in water sends ripples outward; the water only moves up and down while the wave travels on.

Simple definition

A progressive wave carries energy forward through a medium while the particles just vibrate in place.

In the figure : The blue wave moves right; A = amplitude, λ = length of one cycle.

Rule 1: Wave equation — A wave travelling in +x : y = A sin(ωt − kx); k = 2π ÷ λ, ω = 2π f, speed v = f λ = ω ÷ k. For −x use (ωt + kx).

Easy : y = 0.05 sin(40π t − 2π x) : ω = 40π ⇒ f = 20 Hz; k = 2π ⇒ λ = 1 m; v = 20 m/s.
Medium : f = 200 Hz, λ = 0.5 m ⇒ v = 100 m/s.
Exam (CET) : y = 2 sin(100π t − 0.5π x) cm : v = ω ÷ k = 100π ÷ 0.5π = 200 cm/s. (Options : 1) 200 cm/s 2) 50 3) 100 4) 20 → answer 1)

Rule 2: Speed of the wave — On a stretched string v = √(T ÷ μ) (T = tension, μ = mass per unit length). Sound in air ≈ 340 m/s; the speed depends on the medium and the frequency on the source.

Easy : T = 100 N, μ = 0.01 kg/m ⇒ v = 100 m/s.
Medium : T four times ⇒ v doubles.
Exam (CET) : The medium changes : f? (1) same (2) double (3) half (4) zero → answer 1). Trick : f belongs to the source.

Rule 3: Reflection — Reflection at a fixed end : phase change π (crest → trough). At a free end : no phase change. The reflected wave has the same A, f and v.

Easy : A rope tied to a wall : the reflected pulse is inverted.
Medium : A rope on a free ring : the pulse returns upright.
Exam (CET) : Phase change at a fixed end? (1) 0 (2) π/2 (3) π (4) 2π → answer 3)

Derivation : v = f λ

Step 1 : In one period T the wave advances one wavelength λ.
Step 2 : Speed = distance ÷ time = λ ÷ T.
Step 3 : 1 ÷ T = f ⇒ v = f λ.
Step 4 : ω = 2π f, k = 2π ÷ λ ⇒ v = ω ÷ k.

Remember

y = A sin(ωt − kx); v = fλ = ω/k; v = √(T/μ).

Common mistake / CET trap

When the medium changes f stays, v and λ change. (ωt − kx) and (ωt + kx) travel opposite ways. Fixed end ⇒ π, free end ⇒ 0.

Shortcut

Memory clue : "speed = frequency × length" (v = fλ).

Spot the difference

Quantity	Symbol	Unit / dimensions
Wavelength	λ	m; [L]
Frequency	f	Hz; [T⁻¹]
Wave number	k = 2π/λ	rad/m; [L⁻¹]
Speed	v = fλ	m/s; [L T⁻¹]
μ	M ÷ L	kg/m; [M L⁻¹]

Practice questions

Easy

1f = 50, λ = 4 : v?
2Formula for k?
3Phase change at a fixed end?

Medium

4y = sin(10t − 2x) : v?
5T = 36 N, μ = 4 g/m : v?
6f = 170 Hz, v = 340 : λ?

Challenging

7String tension ×9 ⇒ v?
8A wave passes from water to air : which of f, v, λ change?

Show answers (try them yourself first)

1v = 200 m/s
2k = 2π ÷ λ
3π
4v = ω ÷ k = 5 m/s
5μ = 0.004 ⇒ v = √9000 = 94.9 m/s
6λ = 2 m
7v triples
8v and λ change; f stays

Superposition principle and interference

From daily life

Ripples from two stones overlap; in some places the water is high, in others flat.

Simple definition

Interference is the redistribution of intensity when two waves superpose.

In the figure : Above constructive (double amplitude), below destructive (zero).

Live figure

Superposition principle and interference

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Rule 1: Principle of superposition — When two waves meet, the resultant displacement is the sum of the individual ones : y = y₁ + y₂. They then move on unchanged.

Easy : y₁ = +3 cm, y₂ = −1 cm ⇒ y = 2 cm.
Medium : Two equal pulses meeting give a doubled crest.
Exam (CET) : After meeting the waves? (1) vanish (2) continue unchanged (3) merge (4) stop → answer 2)

Rule 2: Constructive and destructive interference — Phase difference φ = 0, 2π… or path difference Δ = nλ ⇒ constructive (A = A₁ + A₂). φ = π, 3π… or Δ = (n + ½)λ ⇒ destructive (A = |A₁ − A₂|). Intensity I ∝ A².

Easy : A₁ = 3, A₂ = 4 : A_max = 7, A_min = 1; I_max ÷ I_min = 49 ÷ 1.
Medium : λ = 0.4 m, Δ = 1.0 m = 2.5 λ ⇒ destructive.
Exam (CET) : I₁ : I₂ = 9 : 1 ⇒ A₁ : A₂ = 3 : 1 ⇒ I_max : I_min = (4)² : (2)² = ? (1) 4 : 1 (2) 16 : 4 (3) 2 : 1 (4) 9 : 1 → answer 1)

Rule 3: Resultant amplitude — With phase difference φ : A = √(A₁² + A₂² + 2 A₁ A₂ cos φ). φ = 0 ⇒ sum; φ = π ⇒ difference; φ = π/2 ⇒ √(A₁² + A₂²).

Easy : A₁ = 3, A₂ = 4, φ = 90° ⇒ A = 5.
Medium : A₁ = A₂ = A, φ = 120° ⇒ A_res = A.
Exam (CET) : A₁ = A₂ = 2, φ = 60° ⇒ A = ? (1) 2√3 (2) 4 (3) 2 (4) 0 → answer 1). Trick : 2A cos(φ/2) = 4 cos 30° = 2√3.

Derivation : the resultant-amplitude formula

Step 1 : y₁ = A₁ sin ωt, y₂ = A₂ sin(ωt + φ); y = y₁ + y₂.
Step 2 : y = (A₁ + A₂ cos φ) sin ωt + (A₂ sin φ) cos ωt = R sin(ωt + δ).
Step 3 : R cos δ = A₁ + A₂ cos φ, R sin δ = A₂ sin φ.
Step 4 : Sum of squares : R² = A₁² + A₂² + 2 A₁ A₂ cos φ.

Remember

y = y₁ + y₂; Δ = nλ ⇒ constructive; (n+½)λ ⇒ destructive; I ∝ A².

Common mistake / CET trap

Take intensity from the square of the amplitude sum. Energy is not destroyed in destructive interference — it is redistributed.

Shortcut

Memory clue : "whole λ = add; half λ = cancel".

Spot the difference

Case	Phase difference φ	Path difference Δ	Amplitude
Constructive	0, 2π…	nλ	A₁ + A₂
Destructive	π, 3π…	(n+½)λ		A₁ − A₂

Practice questions

Easy

1Condition for constructive interference?
2A₁ = 2, A₂ = 5 : A_max?
3Their A_min?

Medium

4Δ = 1.5 λ : type?
5Intensity I is proportional to?
6A₁ = 6, A₂ = 8, φ = 90° : A?

Challenging

7A₁ : A₂ = 4 : 1 ⇒ I_max : I_min?
8φ = 120°, A₁ = A₂ = 5 : A?

Show answers (try them yourself first)

1Δ = nλ (φ = 2nπ)
27
33
4destructive
5A²
6A = 10
7(5)² : (3)² = 25 : 9
8A = 5

Stationary waves

From daily life

A plucked guitar string vibrates in place with some points staying still.

Simple definition

A stationary wave forms when two equal waves travel in opposite directions; it has nodes and antinodes.

In the figure : N = node (zero amplitude), A = antinode (amplitude 2A); N–N = λ/2.

Rule 1: Equation — Two equal waves travelling oppositely : y = 2A sin(kx) cos(ωt). The amplitude depends on position as 2A |sin kx|; all particles are in step but with different amplitudes.

Easy : A = 1 cm ⇒ amplitude 2 cm at antinodes, 0 at nodes.
Medium : y = 4 sin(πx) cos(100πt) : nodes at x = 0, 1, 2 …; λ = 2 m.
Exam (CET) : Adjacent nodes are 25 cm apart ⇒ λ = ? (1) 50 cm (2) 25 cm (3) 12.5 cm (4) 100 cm → answer 1). Trick : λ = 2 × spacing.

Rule 2: Nodes and antinodes — Nodes : x = nλ/2 (zero amplitude). Antinodes : x = (2n + 1) λ/4 (amplitude 2A). Node–antinode spacing λ/4; node–node λ/2. No energy passes through a node.

Easy : λ = 0.8 m : nodes at 0, 0.4, 0.8 m; antinodes at 0.2, 0.6 m.
Medium : A closed end is a node; a free end is an antinode.
Exam (CET) : Antinode–node spacing 10 cm ⇒ λ? (1) 20 cm (2) 40 cm (3) 10 cm (4) 5 cm → answer 2)

Rule 3: Progressive vs stationary — Progressive : energy flows, all particles have the same amplitude, phase changes along x. Stationary : energy is trapped, amplitude varies, particles in a loop are in the same phase.

Easy : The vibrations of a guitar string are stationary.
Medium : Ripples on water are progressive.
Exam (CET) : Energy in a stationary wave? (1) flows (2) does not flow (3) grows (4) is destroyed → answer 2)

Derivation : y = 2A sin kx cos ωt

Step 1 : Forward : y₁ = A sin(ωt − kx); reflected at a fixed end (phase change π) : y₂ = −A sin(ωt + kx).
Step 2 : Sum : y = A [sin(ωt − kx) − sin(ωt + kx)].
Step 3 : Use sin P − sin Q = 2 cos((P+Q)/2) sin((P−Q)/2).
Step 4 : y = −2A sin(kx) cos(ωt) : amplitude 2A |sin kx|; nodes where sin kx = 0 ⇒ x = nλ/2.

Remember

y = 2A sin kx cos ωt; nodes nλ/2; antinodes (2n+1)λ/4; N–N = λ/2.

Common mistake / CET trap

N–N is λ/2, not λ. At a node displacement is zero but strain is maximum.

Shortcut

Memory clue : "node = still; antinode = double".

Spot the difference

Point	Progressive	Stationary
Energy	flows	trapped
Amplitude	same everywhere	varies with position
Nodes	none	present

Practice questions

Easy

1Amplitude at a node?
2Distance between two nodes?
3Is a closed end a node or an antinode?

Medium

4λ = 1.2 m : node–antinode distance?
5A = 3 cm : amplitude at an antinode?
6v = 300, f = 150 : node spacing?

Challenging

7y = 2 sin(0.5πx) cos(50t) : λ?
8Phase of particles within one loop of a stationary wave?

Show answers (try them yourself first)

1Zero
2λ/2
3Node
40.3 m
56 cm
6λ = 2 m ⇒ λ/2 = 1 m
70.5π = 2π ÷ λ ⇒ λ = 4 m
8All in the same phase

Air columns : harmonics, overtones and end correction

From daily life

Blowing across a bottle gives a note; filling it with water raises the pitch.

Simple definition

The air column resonates at definite frequencies (harmonics); the lowest is the fundamental, the others overtones.

In the figure : Closed pipe on the left (L = λ/4), open pipe on the right (L = λ/2).

Live figure

Air columns : harmonics, overtones and end correction

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Rule 1: Closed pipe — One end closed : node at the closed end, antinode at the open end. f_n = (2n − 1) v ÷ 4L — only odd harmonics (f, 3f, 5f …).

Easy : v = 340 m/s, L = 0.5 m ⇒ f₁ = 340 ÷ 2 = 170 Hz; next 510 Hz.
Medium : L halved ⇒ f doubles.
Exam (CET) : The first two overtones of a closed pipe in terms of f₁? (1) 3f, 5f (2) 2f, 3f (3) 2f, 4f (4) 3f, 6f → answer 1). Trick : the even ones are missing.

Rule 2: Open pipe — Antinodes at both ends : f_n = n v ÷ 2L (n = 1, 2, 3 …) — all harmonics. The fundamental is twice that of a closed pipe of the same length.

Easy : L = 0.5 m, v = 340 ⇒ f₁ = 340 Hz; 680, 1020 …
Medium : A flute : an open pipe.
Exam (CET) : Ratio of f₁ of open to closed pipes of equal length? (1) 2 : 1 (2) 1 : 2 (3) 1 : 1 (4) 4 : 1 → answer 1)

Rule 3: End correction and resonance tube — The antinode forms slightly beyond the open end : e ≈ 0.3 d. From two resonance lengths L₁, L₂ : v = 2 f (L₂ − L₁) (e cancels).

Easy : f = 512 Hz, L₁ = 0.16 m, L₂ = 0.50 m ⇒ v = 2 × 512 × 0.34 = 348 m/s.
Medium : d = 4 cm ⇒ e = 1.2 cm.
Exam (CET) : L₂ − L₁ = λ ÷ 2. (Options : 1) λ 2) λ/2 3) λ/4 4) 2λ → answer 2)

Derivation : frequency of a closed pipe

Step 1 : Closed end = node, open end = antinode ⇒ length = odd multiples of λ/4.
Step 2 : L = (2n − 1) λ ÷ 4 ⇒ λ = 4L ÷ (2n − 1).
Step 3 : f = v ÷ λ.
Step 4 : f_n = (2n − 1) v ÷ 4L.

Remember

Closed : f = (2n−1)v/4L; open : f = nv/2L; v = 2f(L₂ − L₁).

Common mistake / CET trap

A closed pipe has no even harmonics. Do not swap the λ/4 and λ/2 relations. Do not forget e.

Shortcut

Memory clue : "closed = odd; open = all".

Spot the difference

Pipe	L	f_n	Harmonics
Closed	(2n−1)λ/4	(2n−1)v/4L	odd
Open	nλ/2	nv/2L	all

Practice questions

Easy

1f₁ of a closed pipe?
2f₁ of an open pipe?
3Formula for e?

Medium

4L = 0.85 m closed, v = 340 : f₁?
5L = 0.85 m open : f₁?
6L₁ = 20 cm, L₂ = 60 cm, f = 400 : v?

Challenging

7Third harmonic of a closed pipe as a multiple of f₁?
8Pipe length up by 1 % : f?

Show answers (try them yourself first)

1v ÷ 4L
2v ÷ 2L
30.3 d
4f₁ = 340 ÷ 3.4 = 100 Hz
5f₁ = 340 ÷ 1.7 = 200 Hz
6v = 2 × 400 × 0.4 = 320 m/s
75 times (5f₁)
8f falls by about 1 %

Vibrating strings, the sonometer and resonance

From daily life

A guitar note rises when the string is tightened, or shortened by pressing a finger.

Simple definition

The sonometer is an apparatus to test how a string's frequency depends on L, T and μ.

In the figure : A wire between two bridges, here the third harmonic (3 loops).

Rule 1: Frequency of a string — Both ends nodes : f_n = (n ÷ 2L) √(T ÷ μ), n = 1, 2, 3 … f ∝ 1 ÷ L, f ∝ √T, f ∝ 1 ÷ √μ.

Easy : L = 0.6 m, T = 100 N, μ = 0.01 kg/m : v = 100; f₁ = 100 ÷ 1.2 = 83.3 Hz.
Medium : Third harmonic = 3 × 83.3 = 250 Hz.
Exam (CET) : If L is halved, f? (1) doubles (2) halves (3) same (4) four times → answer 1)

Rule 2: Laws of the sonometer — (1) Law of length : f L = constant. (2) Law of tension : f ÷ √T = constant. (3) Law of mass : f √μ = constant.

Easy : f = 200 Hz at L = 0.5 m ⇒ at L = 0.4 m f = 250 Hz.
Medium : T four times ⇒ f doubles.
Exam (CET) : A wire twice as thick (μ four times) ⇒ f? (1) halves (2) doubles (3) four times (4) ¼ → answer 1)

Rule 3: Tuning a wire with a fork — At resonance of a tuning fork and the wire, f_fork = (1 ÷ 2L) √(T ÷ μ). Tuning is done by changing the tension.

Easy : L = 0.5 m, T = 64 N, μ = 4 × 10⁻³ ⇒ f = (1 ÷ 1) × √16000 = 126.5 Hz.
Medium : Tightening the peg raises the pitch.
Exam (CET) : Sign of resonance on a sonometer? (1) the paper rider is thrown off (2) the wire snaps (3) the rider stays still (4) the sound stops → answer 1)

Derivation : f = (n/2L)√(T/μ)

Step 1 : Both ends nodes ⇒ L = n λ ÷ 2 ⇒ λ = 2L ÷ n.
Step 2 : Speed on the string v = √(T ÷ μ).
Step 3 : f = v ÷ λ = (n ÷ 2L) v.
Step 4 : Substituting v : f_n = (n ÷ 2L) √(T ÷ μ).

Remember

f = (n/2L)√(T/μ); f ∝ 1/L; f ∝ √T; f ∝ 1/√μ.

Common mistake / CET trap

μ = mass ÷ length in kg/m, not g. L is the wire between the bridges. T is tension in N, not kg.

Shortcut

Memory clue : "short, tight, light = high pitch".

Spot the difference

Factor	Effect on f
Larger L	f decreases (1/L)
Larger T	f increases (√T)
Larger μ	f decreases (1/√μ)

Practice questions

Easy

1f₁ of a string?
2Raise T : f?
3Law of length?

Medium

4L = 1 m, v = 200 : f₁?
5f = 300 Hz at L = 0.6 : f at L = 0.4?
6T ×16 ⇒ f?

Challenging

7L = 0.5 m, T = 36 N, μ = 1 g/m : f₁ and f₂?
8In the 2nd harmonic what is at the middle of the string?

Show answers (try them yourself first)

1(1/2L)√(T/μ)
2f increases (√T)
3f L = constant
4f₁ = 100 Hz
5f = 450 Hz
6f ×4
7v = 189.7 ⇒ f₁ = 189.7 Hz; f₂ = 379.5 Hz
8A node

Beats

From daily life

Two nearly equal notes played together make the sound swell and fade, "wow-wow-wow".

Simple definition

Beats are the regular rise and fall of loudness when two sounds of nearly equal frequency superpose.

In the figure : The blue wave sits inside the red envelope.

Live figure

Beats

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Rule 1: Beat frequency — For two close frequencies f₁, f₂ the loudness rises and falls. Beats per second = |f₁ − f₂|; the resultant : 2A cos(π(f₁ − f₂)t) sin(π(f₁ + f₂)t).

Easy : f₁ = 256, f₂ = 260 ⇒ 4 beats/s.
Medium : Beat period = 1 ÷ 4 = 0.25 s.
Exam (CET) : Beats from 512 Hz and 508 Hz? (1) 4 (2) 2 (3) 8 (4) 1020 → answer 1)

Rule 2: Finding an unknown frequency — An unknown f gives b beats with a known f₀. So f = f₀ ± b. Decide the sign by loading the unknown fork with wax (f falls) or filing it (f rises) and watching how b changes.

Easy : f₀ = 300, b = 5 : f = 295 or 305.
Medium : After waxing b decreases ⇒ f > f₀ (305).
Exam (CET) : After waxing b increases ⇒ ? (1) f > f₀ (2) f < f₀ (3) f = f₀ (4) undecided → answer 2)

Rule 3: Applications — Tuning a piano or a tanpura (until beats vanish); radio tuning; speed-measuring devices.

Easy : Zero beats ⇒ the two frequencies are equal.
Medium : Two 440 Hz forks, one waxed ⇒ beats are heard.
Exam (CET) : For audible beats the difference should be at most about? (1) 10 Hz (2) 100 Hz (3) 1000 Hz (4) 0 → answer 1)

Derivation : beat frequency |f₁ − f₂|

Step 1 : y₁ = A sin 2π f₁ t, y₂ = A sin 2π f₂ t; y = y₁ + y₂.
Step 2 : y = 2A cos[π(f₁ − f₂)t] sin[π(f₁ + f₂)t].
Step 3 : The amplitude is 2A |cos π(f₁ − f₂)t| ; intensity ∝ amplitude².
Step 4 : The amplitude peaks |f₁ − f₂| times per second ⇒ f_beat = |f₁ − f₂|.

Remember

f_beat = |f₁ − f₂|; f = f₀ ± b; amplitude 2A cos(π Δf t).

Common mistake / CET trap

The beat frequency is a difference, not a sum. Decide the ± by the wax/file test. Beats are not heard if the difference is large.

Shortcut

Memory clue : "difference = beats".

Spot the difference

Point	Formula	Unit
Beat frequency		f₁ − f₂	Hz
Beat period	1 ÷	f₁ − f₂	s
Mean frequency	(f₁ + f₂) ÷ 2	Hz

Practice questions

Easy

1What is a beat?
2f₁ = 400, f₂ = 404 : beats?
3Beat period?

Medium

4f₀ = 512, b = 3 : f?
5A and B give 6 beats; A = 300; B?
6Beats increase on waxing : what?

Challenging

7f₁ = 200, f₂ = 202, each of amplitude A : A_max?
8Frequencies differ by 100 Hz : are beats heard?

Show answers (try them yourself first)

1Periodic rise and fall of loudness
24
30.25 s
4f = 509 or 515
5B = 294 or 306
6unknown f < f₀
72A
8No (too rapid)

Doppler effect and characteristics of sound

From daily life

An ambulance siren sounds higher as it approaches and lower as it goes away.

Simple definition

The Doppler effect is the change in the heard frequency caused by relative motion of source and observer.

In the figure : For a moving source the wavefronts bunch up in front and spread out behind.

Live figure

Doppler effect and characteristics of sound

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Rule 1: Doppler effect — f' = f (v ± v₀) ÷ (v ∓ v_s). Observer towards source ⇒ + on top; source towards observer ⇒ − below. Approach ⇒ higher f', recede ⇒ lower. v = speed of sound.

Easy : Source approaching at 34 m/s, f = 500, v = 340 : f' = 500 × 340 ÷ 306 = 555.6 Hz.
Medium : Observer approaching at 17 m/s : f' = 500 × 357 ÷ 340 = 525 Hz.
Exam (CET) : Source receding at 34 m/s, f = 500 : f' = ? (1) 454.5 (2) 545 (3) 500 (4) 400 → answer 1). Trick : 340 ÷ 374 = 10/11.

Rule 2: Characteristics of sound — Pitch ↔ frequency; loudness ↔ intensity; quality (timbre) ↔ wave shape / overtones. Intensity level β = 10 log₁₀(I ÷ I₀) dB; I₀ = 10⁻¹² W/m².

Easy : I = 10⁻⁶ W/m² ⇒ β = 10 log(10⁶) = 60 dB.
Medium : I doubled ⇒ β + 3 dB (10 log 2).
Exam (CET) : I made ten times ⇒ β? (1) + 10 dB (2) × 10 (3) + 1 dB (4) + 100 dB → answer 1)

Rule 3: Speed and intensity of sound — In air v ≈ 331 + 0.6 t (°C) m/s; I = P ÷ A (power ÷ area); for a point source I ∝ 1 ÷ r².

Easy : t = 30 °C ⇒ v = 331 + 18 = 349 m/s.
Medium : Distance doubled ⇒ I becomes one quarter.
Exam (CET) : r made three times ⇒ I? (1) 1/9 (2) 1/3 (3) 9 (4) 3 → answer 1)

Derivation : f' for an approaching source

Step 1 : The source emits f; in one period T the sound moves v T and the source moves v_s T.
Step 2 : New wavelength λ' = v T − v_s T = (v − v_s) ÷ f.
Step 3 : The observer hears f' = v ÷ λ' = f v ÷ (v − v_s).
Step 4 : For a receding source the sign of v_s flips ⇒ f' = f v ÷ (v + v_s).

Remember

f' = f(v ± v₀)/(v ∓ v_s); β = 10 log(I/I₀); I ∝ 1/r².

Common mistake / CET trap

Signs : observer approaching ⇒ + on top; source approaching ⇒ − below. The real frequency does not change, only the heard one. dB is logarithmic, not additive.

Shortcut

Memory clue : "close = high, away = low".

Spot the difference

Feature	Related quantity	Unit
Pitch	frequency f	Hz
Loudness	intensity I	W/m²; dB
Quality	wave shape	—

Practice questions

Easy

1What is the Doppler effect?
2Pitch depends on?
3What is I₀?

Medium

4Source approaching at 20 m/s, f = 340, v = 340 : f'?
5I = 10⁻⁸ W/m² : β?
6t = 20 °C : v?

Challenging

7Observer receding at 34 m/s, f = 600, v = 340 : f'?
8A point source gives 60 dB at 2 m; β at 20 m? (I ∝ 1/r²)

Show answers (try them yourself first)

1Apparent change in frequency due to relative motion
2Frequency
310⁻¹² W/m²
4f' = 340 × 340 ÷ 320 = 361.25 Hz
5β = 40 dB
6v = 343 m/s
7f' = 600 × 306 ÷ 340 = 540 Hz
8I ÷ 100 ⇒ β = 60 − 20 = 40 dB

Quick revision of the whole lesson

Concept	Key formula / fact	Unit / dimensions
Progressive wave	y = A sin(ωt − kx); v = fλ; v = √(T/μ)	m/s; [L T⁻¹]
Interference	Δ = nλ constructive; (n+½)λ destructive; I ∝ A²	W/m²
Stationary wave	N–N = λ/2; y = 2A sin kx cos ωt	m
Air columns	closed (2n−1)v/4L; open nv/2L; e = 0.3d	Hz
String	f = (n/2L)√(T/μ)	Hz; μ : kg/m
Beats	f_b =	f₁ − f₂	Hz
Doppler	f' = f(v ± v₀)/(v ∓ v_s)	Hz
Intensity level	β = 10 log(I/I₀), I₀ = 10⁻¹²	dB

🎯 Most important points for the exam

v = fλ; on a string v = √(T/μ); f stays when the medium changes.
Interference : Δ = nλ constructive; I ∝ A².
Stationary : N–N = λ/2; amplitude zero at a node.
Closed pipe odd, open pipe all harmonics; v = 2f(L₂ − L₁).
String : f ∝ 1/L, √T, 1/√μ.
Beats = |f₁ − f₂|; wax test decides ±.
Doppler signs; β = 10 log(I/I₀).

Study mantra

Identify the wave type → boundary condition (node/antinode) → right formula → check units.