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L-14: Wave Phenomena

Physics — Class 12 · NIOS Code 312 · Module 4 · Source: 312_Physics_Eng_Lesson14.pdf

Wave Phenomena — Energy Without Mass Transport

Waves carry energy, not matter — a straw on water bobs in place while ripples spread outward. Sound, light, and radio waves are essential to communication and life. This lesson covers progressive and stationary waves, superposition, musical instruments, electromagnetic spectrum, and the Doppler effect.

NIOS objectives: transverse and longitudinal propagation; v = fλ; Newton and Laplace formulas; stretched strings; SHM wave equation; beats and interference; stationary waves and organ pipes; Doppler effect; EM waves and spectrum.

14.1 Wave Propagation

On a slinky: sideways jerk → transverse pulses; push along length → longitudinal compressions and rarefactions (like sound in air).

  • Progressive waves — crests/troughs move forward.
  • Stationary waves — pattern fixed in space (discussed in 14.5).
  • Particles oscillate with same period T and amplitude A while the wave profile advances.
v = λ / T = f λ
v = wave speed (m·s⁻¹)
λ = wavelength (m)
f = frequency (Hz), T = period
Fundamental wave relation (Eq. 14.2).
k = 2π/λ , ω = 2πf , v = ω/k
k = propagation constant (rad·m⁻¹)
ω = angular frequency (rad·s⁻¹)
Phase change per unit distance/time.
Fig 14.1 — Transverse vs Longitudinal transverse (slinky sideways) compression rarefaction longitudinal (push along slinky) v →
Fig 14.1 — Transverse waves: displacement ⊥ propagation; longitudinal: compressions along direction

14.1.3 Equation of a Simple Harmonic Wave

For a transverse wave along +X with displacement along Y:

y(x,t) = a sin(ωt − kx + φ₀)
a = amplitude
φ₀ = initial phase at x = 0
Also: y = a sin 2π(vt − x)/λ
Example 14.1: y = 10⁻⁴ sin(100πt − 0.1πx) → f = 50 Hz, λ = 20 m, v = 1000 m/s.
Δφ = k Δx = (2π/λ)(x₂ − x₁)
Phase difference between two points.
Negative sign: point farther along +x acquires same phase later.

14.1.4 Transverse vs Longitudinal

  • Transverse: displacement ⊥ propagation; crests and troughs visible; solids and liquid surfaces.
  • Longitudinal: displacement ∥ propagation; compressions and rarefactions; solids, liquids, gases.
  • Mechanical waves need mass and elasticity (volume elasticity for longitudinal; rigidity for transverse).
  • EM waves are transverse but need no material medium.

14.2 Velocity of Waves in Elastic Media

14.2.1–14.2.2 Newton and Laplace

Newton assumed isothermal compression during sound propagation:

v = √(P/ρ) ≈ 280 m/s (air)
P = 1.01×10⁵ Pa, ρ = 1.29 kg·m⁻³ — 16% below measured 333 m/s.

Laplace corrected for adiabatic conditions (air is poor conductor; compressions are rapid):

v = √(γP/ρ) , γ = Cp/Cv
For air γ = 1.4 → v ≈ 333 m/s at 0°C — matches experiment.

14.2.3 Factors Affecting Velocity of Sound

  • Temperature: v ∝ √T → v ≈ 333 + 0.61t m/s (t in °C).
  • Pressure: no effect at constant T (P and ρ change proportionally).
  • Density: v ∝ 1/√ρ — sound faster in H₂ than O₂ (4× under same conditions).
  • Humidity: moist air less dense → speed increases slightly.
  • General: v_gas < v_liquid < v_solid.
v = √(T/m) (stretched string)
T = tension (N), m = mass per unit length (kg·m⁻¹).
Also v = √(E/ρ) for longitudinal waves in elastic media.

14.3 Principle of Superposition and Reflection

When two pulses overlap, resultant displacement = vector sum of individual displacements. After crossing, each pulse continues unchanged — enables tuning one radio station among many.

Reflection of Waves

  • Fixed end (denser medium): transverse pulse reflects inverted (phase π); crest → trough.
  • Free end (rarer medium): transverse pulse reflects same shape — crest → crest.
  • Longitudinal at denser boundary: reflected without change of type but sign reverses.
  • Longitudinal at rarer boundary: compression reflects as rarefaction and vice versa.

14.4 Interference and Beats

14.4.1 Interference

Two waves y₁ = a₁ sin(ωt − kx) and y₂ = a₂ sin(ωt − kx + φ) superpose to give amplitude:

A² = a₁² + a₂² + 2a₁a₂ cos φ
In phase (φ = 2mπ): A = a₁ + a₂ → I_max ∝ (a₁+a₂)².
Out of phase (φ = (2m+1)π): A = |a₁−a₂| → I_min ∝ (a₁−a₂)².
I_max/I_min = [(a₁+a₂)/(a₁−a₂)]².

14.4.2 Beats

Two tuning forks of frequencies f and f + Δf produce beats at frequency Δf per second. Audible as separate only if Δf < ~10 Hz.

Example: Unknown fork beats 5/s with 500 Hz fork → frequency is 495 Hz or 505 Hz.

Fig 14.12 — Standing Wave: Nodes & Antinodes N A N A λ/2 between nodes · λ/4 node to antinode
Fig 14.12 — Nodes (zero amplitude) and antinodes (maximum amplitude) in a stationary wave

14.5 Stationary (Standing) Waves

Two identical waves of same λ, same amplitude, same speed travelling in opposite directions produce stationary waves — nodes and antinodes do not travel.

y = −2a sin kx cos ωt
Stationary wave equation (Eq. 14.20).
Nodes: sin kx = 0 → spacing λ/2.
Antinodes: |sin kx| = 1 → spacing λ/2.
Node to antinode: λ/4.
         STANDING WAVE ON A STRING
         =========================
    N        A        N        A        N
    |   λ/4  |   λ/4  |   λ/4  |   λ/4  |
    zero     max      zero     max      zero
    amplitude          amplitude
    max strain         zero strain

Energy surges back and forth within segments — no net energy transport past a point.

14.6 Musical Sound and Organ Pipes

Pitch — subjective, related to frequency (high/sharp vs low/flat). Loudness — related to intensity I; β = 10 log(I/I₀) dB, I₀ = 10⁻¹² W·m⁻². Quality (timbre) — waveform shape; overtones 2n, 3n… distinguish instruments.

Organ Pipe Harmonics

  • Open pipe (length l): n₁ = v/2l; harmonics n, 2n, 3n… — richer overtones.
  • Closed pipe: n₁ = v/4l; only odd harmonics 3n₁, 5n₁… — even harmonics missing.
  • Ratio: open fundamental = 2 × closed fundamental (same length).
Open: n = nv/(2l) | Closed: n = nv/(4l) (odd n only)
v = speed of sound in air inside pipe.
Closed end = displacement node; open end = antinode.
Fig 14.16 — EM Wave: E and B Fields k (propagation) E B E ⊥ B ⊥ direction of propagation · E = cB
Fig 14.16 — Electric and magnetic fields oscillate perpendicular to each other and to propagation

14.7 Electromagnetic Waves

  • Transverse oscillations of E and B fields perpendicular to each other and to propagation direction.
  • In vacuum: c = 1/√(μ₀ε₀) = 3×10⁸ m/s — independent of source motion.
  • In medium: v = c/√(μᵣεᵣ) < c.
  • Energy E ∝ frequency: E = hf = hc/λ.

14.7.2 Electromagnetic Spectrum

  • Radio waves (~10⁶–10⁹ Hz) — communications.
  • Microwaves (~10⁹–10¹¹ Hz) — radar, ovens, satellite links.
  • Infrared — heat radiation, thermography.
  • Visible light (~4×10¹⁴–7.5×10¹⁴ Hz) — violet to red.
  • Ultraviolet — sterilisation; absorbed by ozone layer.
  • X-rays — medical imaging, crystal structure.
  • Gamma rays — nuclear sources; most energetic and penetrating.
Fig 14.18 — Doppler Effect (approaching source) source crowded λ′ ear higher observed frequency when source approaches
Fig 14.18 — Moving source crowds wavelengths → higher pitch heard by observer

14.8 Doppler Effect

Apparent change in frequency when source and observer move relative to each other and the medium.

n′ = n(v − v₀)/(v − vₛ)
Observed frequency when source and observer move along line of sight.
v = sound speed in medium
vₛ, v₀ positive toward observer
Light: Δλ/λ = vₛ/c for recession (red shift).

Example 14.6: Red shift 0.032% → vₛ = c × 0.00032 ≈ 9.6×10⁴ m/s recession — evidence of expanding universe.

Quick Revision

  • v = fλ; y = a sin(ωt − kx).
  • Sound: Laplace v = √(γP/ρ); v ≈ 333 + 0.61t m/s.
  • String: v = √(T/m).
  • Superposition → interference, beats, standing waves.
  • Standing wave: nodes λ/2 apart; open pipe n₁ = v/2l; closed n₁ = v/4l.
  • EM spectrum — E and B ⊥ to propagation; c in vacuum.
  • Doppler: approach → higher pitch; recession → lower pitch / red shift.
23 cards · click any card to flip
Wave
Disturbance that transfers energy without net transport of matter. Types: progressive/stationary, mechanical/EM, transverse/longitudinal.
Wavelength λ
Distance between two nearest particles vibrating in the same phase.
Wave velocity
v = λ/T = fλ. Also v = ω/k where k = 2π/λ and ω = 2πf.
SHM wave equation
y(x,t) = a sin(ωt − kx + φ₀). Forward wave along +x.
Phase difference
Δφ = kΔx = (2π/λ)Δx for spatial separation. Δφ = 2πfΔt for time interval.
Transverse wave
Particle displacement ⊥ to propagation. Solids and liquid surfaces only (mechanical).
Longitudinal wave
Particle displacement ∥ to propagation. Solids, liquids, gases. Compressions and rarefactions.
Newton's formula for sound
v = √(P/ρ) assuming isothermal compression — gives ~280 m/s (16% low).
Laplace correction
v = √(γP/ρ) for adiabatic process. For air γ = 1.4 → v ≈ 333 m/s at STP.
Sound vs temperature
v ∝ √T. v ≈ 333 + 0.61t m/s for small t (°C). Pressure alone does not change v.
Stretched string
v = √(T/m) where T = tension, m = mass per unit length.
Principle of superposition
Resultant displacement = vector sum of individual displacements at a point.
Reflection — denser medium
Transverse: phase reversal (π). Longitudinal: same type, sign change.
Reflection — rarer medium
Transverse: no phase change. Longitudinal: type changes (compression ↔ rarefaction).
Interference
Imax ∝ (a₁+a₂)² when in phase; Imin ∝ (a₁−a₂)² when out of phase by odd π.
Beats
Superposition of nearly equal frequencies. Beat frequency = |f₁ − f₂|. Audible if < ~10 beats/s.
Stationary wave
y = −2a sin kx cos ωt. Nodes: zero amplitude, max strain. Antinodes: max amplitude, zero strain.
Node spacing
λ/2 between successive nodes or antinodes; λ/4 between node and nearest antinode.
Open organ pipe
n₁ = v/2l. Harmonics: n, 2n, 3n… (all harmonics present).
Closed organ pipe
n₁ = v/4l. Odd harmonics only: 3n₁, 5n₁… Open pipe fundamental = 2× closed pipe.
EM waves
Transverse E and B fields ⊥ to each other and to propagation. c = 1/√(μ₀ε₀) in vacuum.
Doppler effect
Apparent frequency change due to relative motion of source and observer. Applies to sound and light.
Carnot efficiency analogy
Red shift: Δλ/λ = vₛ/c for receding star — evidence of expanding universe.

Q1. The relation between wave velocity, frequency and wavelength is:

Q2. Equation of a progressive harmonic wave along +x is:

Q3. Laplace corrected velocity of sound in air at STP is approximately:

Q4. Velocity of a transverse wave on a stretched string is:

Q5. On reflection of a transverse wave from a fixed (denser) end:

Q6. Beat frequency equals:

Q7. Distance between two successive nodes in a stationary wave is:

Q8. Fundamental frequency of closed organ pipe of length l is:

Q9. Electromagnetic waves in vacuum travel with speed:

Q10. In an EM wave, E and B fields are oriented:

v = fλ = λ/T
k = 2π/λ, ω = 2πf, v = ω/k
y = a sin(ωt − kx + φ₀)
Δφ = kΔx = (2π/λ)Δx
v = √(P/ρ) (Newton)
v = √(γP/ρ), γ = Cp/Cv
v ≈ 333 + 0.61t m/s
v = √(T/m) (string)
A² = a₁² + a₂² + 2a₁a₂ cos φ
Beat frequency = Δf
y = −2a sin kx cos ωt
Open pipe: n = nv/(2l)
Closed pipe: n = nv/(4l) (odd n)
β = 10 log(I/I₀) dB
c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s
E = hf = hc/λ
n′ = n(v − v₀)/(v − vₛ)

1. Formulas & Definitions

Full Ch 14 study guide — progressive waves, sound, interference, standing waves, EM spectrum, and Doppler.

v = fλ = λ/T

Definition: Fundamental wave relation linking speed, frequency, and wavelength.

Derivation

Eq. 14.2 — one wavelength passes a point in one period T.

Variables

v (m/s) · f (Hz) · λ (m) · T (s)

Why it works

Every wave problem starts here — find any one if you know two others.

Historical context

Unified description for sound, water ripples, light, and radio waves.

Deep understanding

v depends on medium (sound ~340 m/s in air, light 3×10⁸ m/s in vacuum). f and λ adjust together.

2. Diagrams & Visuals

λ v = fλ

Color-coded visual · step-by-step breakdown below

  1. Identify f or T
  2. Measure or find λ
  3. v = fλ
  4. Check units: m/s

3. Solved Examples

Basic

Q: f=50 Hz, λ=4 m.

Solution: v=200

Answer: 200 m/s

Intermediate

Q: Ex 14.1: f=50 Hz, λ=20 m.

Solution: v=1000

Answer: 1000 m/s

Advanced

Q: Double f, same medium?

Solution: λ halves

Answer: v unchanged

Exam

Q: Wave speed depends on?

Solution: Medium, not source f

Answer: Sec 14.1

k = 2π/λ, ω = 2πf, v = ω/k

Definition: Angular wave number and angular frequency forms of wave speed.

Derivation

k = phase change per metre; ω = phase change per second; v = ω/k.

Variables

k (rad/m) · ω (rad/s)

Why it works

Compact form for SHM wave equation y = a sin(ωt − kx).

Historical context

Standard in physics for progressive harmonic waves.

Deep understanding

ω = 2πf and T = 2π/ω. Phase velocity v = ω/k for single-frequency waves.

2. Diagrams & Visuals

ω/k = fλ = v k = 2π/λ

Color-coded visual · step-by-step breakdown below

  1. Find λ → k = 2π/λ
  2. Find f → ω = 2πf
  3. v = ω/k
  4. Cross-check v = fλ

3. Solved Examples

Basic

Q: λ=2 m, f=10 Hz.

Solution: k=π, ω=20π, v=20

Answer: 20 m/s

Intermediate

Q: ω=100π, k=0.1π.

Solution: v=1000

Answer: 1000 m/s

Advanced

Q: k doubles?

Solution: λ halves

Answer: v unchanged if ω fixed

Exam

Q: Units of k?

Solution: rad·m⁻¹

Answer: Sec 14.1

y = a sin(ωt − kx + φ₀)

Definition: Displacement equation for a progressive harmonic transverse wave along +x.

Derivation

SHM in time and space combined; −kx for wave travelling +x direction.

Variables

a = amplitude · φ₀ = initial phase

Why it works

Extract f, λ, v from given equation by comparing coefficients.

Historical context

Example 14.1: y = 10⁻⁴ sin(100πt − 0.1πx).

Deep understanding

+ sign in (ωt + kx) means wave travels −x. Compare with standard form.

2. Diagrams & Visuals

y = a sin(ωt−kx)

Color-coded visual · step-by-step breakdown below

  1. Match ω from t coefficient
  2. Match k from x coefficient
  3. f = ω/2π, λ = 2π/k
  4. v = ω/k

3. Solved Examples

Basic

Q: y = sin(20t − 4x).

Solution: ω=20, k=4

Answer: v=5 m/s

Intermediate

Q: Ex 14.1: 100πt, 0.1πx.

Solution: f=50 Hz, λ=20 m

Answer: v=1000 m/s

Advanced

Q: Wave travels −x?

Solution: Use ωt+kx

Answer: Sign of kx term

Exam

Q: Amplitude from equation?

Solution: Coefficient of sin

Answer: a in metres

Δφ = kΔx = (2π/λ)Δx

Definition: Phase difference between two points along a wave.

Derivation

Phase changes by k per unit distance along propagation direction.

Variables

Δφ (rad) · Δx = x₂ − x₁

Why it works

Determines constructive vs destructive interference at separated points.

Historical context

Path difference Δx links to phase difference for interference.

Deep understanding

Δφ = 2π if Δx = λ (in phase again). Farther along +x → lags in phase (−kx term).

2. Diagrams & Visuals

Δx → Δφ

Color-coded visual · step-by-step breakdown below

  1. Find separation Δx
  2. Δφ = kΔx
  3. Or Δφ = 2πΔx/λ
  4. Convert to degrees if needed

3. Solved Examples

Basic

Q: Δx = λ/2.

Solution: Δφ = π

Answer: 180° out of phase

Intermediate

Q: Δx = λ.

Solution: Δφ = 2π

Answer: In phase

Advanced

Q: Δx = λ/4?

Solution: Δφ = π/2

Answer: Quarter period apart

Exam

Q: Nodes separated by?

Solution: λ/2 in standing wave

Answer: Sec 14.5

v = √(P/ρ) (Newton)

Definition: Newton's isothermal formula for speed of sound in a gas.

Derivation

Assumes compressions are slow enough for heat exchange — isothermal bulk modulus = P.

Variables

P (Pa) · ρ (kg/m³)

Why it works

Historical first attempt — underestimates real sound speed in air.

Historical context

Isaac Newton derived ~280 m/s; 16% below measured 333 m/s.

Deep understanding

Real sound propagation in air is adiabatic (fast), not isothermal.

2. Diagrams & Visuals

v = √(P/ρ) ≈ 280 m/s too low for air

Color-coded visual · step-by-step breakdown below

  1. P ≈ 1.01×10⁵ Pa
  2. ρ ≈ 1.29 kg/m³ for air
  3. v = √(P/ρ)
  4. Compare with 333 m/s

3. Solved Examples

Basic

Q: P=10⁵, ρ=1.25.

Solution: v≈283

Answer: ~283 m/s

Intermediate

Q: Why too low?

Solution: Isothermal assumption wrong

Answer: Use Laplace

Advanced

Q: P doubles, T constant?

Solution: ρ doubles too

Answer: v unchanged

Exam

Q: Newton sound speed air?

Solution: ~280 m/s

Answer: Sec 14.2.1

v = √(γP/ρ), γ = Cp/Cv

Definition: Laplace correction for adiabatic sound propagation in a gas.

Derivation

Adiabatic bulk modulus = γP; γ = 1.4 for diatomic air.

Variables

γ = ratio of specific heats

Why it works

Matches measured 333 m/s at 0°C — correct physics for sound.

Historical context

Pierre-Simon Laplace corrected Newton (~1816).

Deep understanding

γ = 1.4 for air → √(1.4) ≈ 1.18 factor over Newton formula.

2. Diagrams & Visuals

v = √(γP/ρ) γ=1.4 → 333 m/s air

Color-coded visual · step-by-step breakdown below

  1. γ for gas (1.4 air)
  2. P and ρ at temperature
  3. v = √(γP/ρ)
  4. Compare experiment

3. Solved Examples

Basic

Q: γ=1.4, P=10⁵, ρ=1.29.

Solution: v≈331

Answer: ~333 m/s

Intermediate

Q: Why γ appears?

Solution: Adiabatic compression

Answer: No heat escape

Advanced

Q: Monatomic gas γ?

Solution: 5/3 ≈ 1.67

Answer: Higher sound speed

Exam

Q: Laplace vs Newton?

Solution: Adiabatic vs isothermal

Answer: Sec 14.2.2

v ≈ 333 + 0.61t m/s

Definition: Empirical temperature dependence of sound speed in air.

Derivation

v ∝ √T (absolute); linearised for t in °C near 0°C.

Variables

t = temperature in °C

Why it works

Hot day → faster sound; explains thunder delay variation slightly.

Historical context

Practical rule from v ∝ √T with T = 273 + t.

Deep understanding

Pressure alone does not change v at fixed T (P and ρ scale together).

2. Diagrams & Visuals

v ≈ 333 + 0.61t (m/s)

Color-coded visual · step-by-step breakdown below

  1. Read t in °C
  2. v = 333 + 0.61t
  3. Use for air speed estimates

3. Solved Examples

Basic

Q: t = 0°C.

Solution: v=333

Answer: 333 m/s

Intermediate

Q: t = 30°C.

Solution: v=333+18.3

Answer: 351.3 m/s

Advanced

Q: Double pressure, same T?

Solution: v unchanged

Answer: ρ and P both scale

Exam

Q: Sound faster in H₂ or O₂?

Solution: H₂ (lower ρ)

Answer: v ∝ 1/√ρ

v = √(T/m) (string)

Definition: Speed of transverse waves on a stretched string.

Derivation

From wave equation on string; T = tension, m = mass per unit length.

Variables

T (N) · m (kg/m)

Why it works

Guitar pitch rises with tension; thicker string (larger m) → lower v.

Historical context

Foundation for vibrating strings in musical instruments.

Deep understanding

Also v = √(E/ρ) for longitudinal waves in elastic solids (mentioned in notes).

2. Diagrams & Visuals

v = √(T/m)

Color-coded visual · step-by-step breakdown below

  1. Tension T in N
  2. m = mass/length in kg/m
  3. v = √(T/m)
  4. Use with standing wave harmonics

3. Solved Examples

Basic

Q: T=100 N, m=0.01 kg/m.

Solution: v=100

Answer: 100 m/s

Intermediate

Q: 4× tension?

Solution: v doubles

Answer: v ∝ √T

Advanced

Q: Same T, 4× m?

Solution: v halves

Answer: v ∝ 1/√m

Exam

Q: String wave speed depends on?

Solution: T and m, not f

Answer: Sec 14.2.3

A² = a₁² + a₂² + 2a₁a₂ cos φ

Definition: Resultant amplitude when two harmonic waves of same ω, k superpose.

Derivation

Vector phasor addition; φ = phase difference between waves.

Variables

a₁, a₂ amplitudes · φ phase difference

Why it works

Explains interference maxima and minima.

Historical context

Principle of superposition — cornerstone of wave physics.

Deep understanding

φ=0: A=a₁+a₂. φ=π: A=|a₁−a₂|. I ∝ A².

2. Diagrams & Visuals

resultant A

Color-coded visual · step-by-step breakdown below

  1. Identify a₁, a₂
  2. Find phase difference φ
  3. A² = a₁²+a₂²+2a₁a₂ cosφ
  4. A = √(...)

3. Solved Examples

Basic

Q: a₁=a₂=2, φ=0.

Solution: A=4

Answer: Constructive

Intermediate

Q: a₁=a₂=2, φ=π.

Solution: A=0

Answer: Destructive

Advanced

Q: I_max/I_min ratio?

Solution: [(a₁+a₂)/(a₁−a₂)]²

Answer: Sec 14.4.1

Exam

Q: Superposition principle?

Solution: Sum displacements

Answer: Sec 14.3

Beat frequency = Δf

Definition: Number of loudness pulses per second from two close frequencies.

Derivation

Superposition of cos 2πft and cos 2π(f+Δf)t gives envelope at Δf.

Variables

Δf = |f₁ − f₂| · audible if Δf < ~10 Hz

Why it works

Tune musical instruments — count beats to match pitch.

Historical context

Example: 5 beats/s with 500 Hz fork → 495 or 505 Hz.

Deep understanding

Unknown frequency = f_known ± Δf. Two possibilities from beat count alone.

2. Diagrams & Visuals

beats = Δf

Color-coded visual · step-by-step breakdown below

  1. Two frequencies f₁, f₂
  2. Δf = |f₁−f₂|
  3. Count beats per second
  4. f_unknown = f_known ± Δf

3. Solved Examples

Basic

Q: f₁=256, f₂=260.

Solution: Δf=4

Answer: 4 beats/s

Intermediate

Q: 5 beats/s with 500 Hz.

Solution: 495 or 505 Hz

Answer: Two answers

Advanced

Q: Δf = 15 Hz audible?

Solution: Hard to distinguish

Answer: Limit ~10 Hz

Exam

Q: Beats need similar f?

Solution: Yes, small Δf

Answer: Sec 14.4.2

y = −2a sin kx cos ωt

Definition: Stationary (standing) wave from oppositely travelling waves of equal amplitude.

Derivation

Eq. 14.20 — sum of sin(ωt−kx) and sin(ωt+kx).

Variables

Nodes where sin kx=0 · Antinodes where |sin kx|=1

Why it works

Explains organ pipes, guitar strings, microwave cavities.

Historical context

No net energy transport — energy trapped between nodes.

Deep understanding

Node spacing λ/2; node to antinode λ/4. Fixed end = node for string.

2. Diagrams & Visuals

N A

Color-coded visual · step-by-step breakdown below

  1. Two opposite waves same a, λ
  2. Form standing pattern
  3. Nodes: sin kx = 0
  4. Antinodes: |sin kx| = 1

3. Solved Examples

Basic

Q: λ=4 m. Node spacing?

Solution: λ/2=2 m

Answer: 2 m

Intermediate

Q: Node to antinode?

Solution: λ/4

Answer: 1 m if λ=4

Advanced

Q: Energy transport?

Solution: Zero net past point

Answer: Sec 14.5

Exam

Q: Standing vs progressive?

Solution: Pattern fixed in space

Answer: Fig 14.12

Open pipe: n = nv/(2l)

Definition: Harmonic frequencies of an open organ pipe of length l.

Derivation

Both ends antinodes → l = nλ/2 → f_n = nv/(2l).

Variables

n = 1, 2, 3… all harmonics · v = sound speed in pipe

Why it works

Open pipe has all harmonics — richer timbre than closed pipe.

Historical context

Fundamental n₁ = v/(2l); overtones 2n₁, 3n₁…

Deep understanding

Open end = displacement antinode. Same length open pipe fundamental = 2× closed.

2. Diagrams & Visuals

open both ends

Color-coded visual · step-by-step breakdown below

  1. Length l, speed v
  2. n = harmonic number
  3. f_n = nv/(2l)
  4. All integer n allowed

3. Solved Examples

Basic

Q: l=0.5 m, v=340, n=1.

Solution: f=340

Answer: 340 Hz

Intermediate

Q: 2nd harmonic?

Solution: n=2, f=680

Answer: 680 Hz

Advanced

Q: vs closed same l?

Solution: Open f₁ = 2× closed f₁

Answer: Sec 14.6

Exam

Q: Open pipe harmonics?

Solution: n, 2n, 3n…

Answer: All present

Closed pipe: n = nv/(4l) (odd n)

Definition: Harmonic frequencies of a closed organ pipe — one closed, one open end.

Derivation

Closed end node, open end antinode → l = nλ/4, n odd only.

Variables

n = 1, 3, 5… only odd harmonics

Why it works

Even harmonics missing — characteristic hollow timbre of clarinet-like pipes.

Historical context

Fundamental n₁ = v/(4l) — half the open pipe for same length.

Deep understanding

Closed end = displacement node. 3rd harmonic = 3n₁, not 2n₁.

2. Diagrams & Visuals

closed | open

Color-coded visual · step-by-step breakdown below

  1. Length l, speed v
  2. n = 1, 3, 5… only
  3. f_n = nv/(4l)
  4. Skip even n

3. Solved Examples

Basic

Q: l=1 m, v=340, n=1.

Solution: f=85

Answer: 85 Hz

Intermediate

Q: Next harmonic?

Solution: n=3, f=255

Answer: 3rd harmonic

Advanced

Q: 2nd harmonic exists?

Solution: No — n must be odd

Answer: Missing even

Exam

Q: Closed vs open f₁ ratio?

Solution: 1:2 same length

Answer: Sec 14.6

β = 10 log(I/I₀) dB

Definition: Sound intensity level in decibels.

Derivation

Logarithmic scale; I₀ = 10⁻¹² W·m⁻² threshold of hearing.

Variables

β (dB) · I (W/m²) · I₀ = 10⁻¹²

Why it works

Human ear hears huge intensity range — log scale matches perception.

Historical context

Loudness subjective; β quantifies intensity logarithmically.

Deep understanding

10 dB increase ≈ perceived doubling of loudness. β = 0 at I = I₀.

2. Diagrams & Visuals

β = 10 log(I/I₀) dB

Color-coded visual · step-by-step breakdown below

  1. Intensity I in W/m²
  2. I₀ = 10⁻¹²
  3. β = 10 log₁₀(I/I₀)
  4. Result in dB

3. Solved Examples

Basic

Q: I = I₀.

Solution: β=0

Answer: 0 dB

Intermediate

Q: I = 100 I₀.

Solution: β=20

Answer: 20 dB

Advanced

Q: I = 10⁶ I₀.

Solution: β=60

Answer: 60 dB

Exam

Q: Loudness relates to?

Solution: Intensity I

Answer: Pitch to f

c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s

Definition: Speed of electromagnetic waves in vacuum.

Derivation

From Maxwell's equations; μ₀ = permeability, ε₀ = permittivity of free space.

Variables

c in m/s · independent of source motion

Why it works

All EM radiation (radio to gamma) travels at c in vacuum.

Historical context

Maxwell unified electricity, magnetism, and light (1860s).

Deep understanding

In medium v = c/√(μᵣεᵣ) < c. E ⊥ B ⊥ propagation.

2. Diagrams & Visuals

E ⊥ B

Color-coded visual · step-by-step breakdown below

  1. Vacuum: use c
  2. Medium: v = c/n or c/√(μᵣεᵣ)
  3. f and λ related by c = fλ

3. Solved Examples

Basic

Q: f=6×10¹⁴ Hz light.

Solution: λ=c/f≈500 nm

Answer: Visible green

Intermediate

Q: n=1.5 glass?

Solution: v=c/1.5

Answer: Slower in medium

Advanced

Q: EM needs medium?

Solution: No — unlike sound

Answer: Sec 14.7

Exam

Q: E and B in EM wave?

Solution: Mutually perpendicular

Answer: Fig 14.16

E = hf = hc/λ

Definition: Photon energy for electromagnetic radiation.

Derivation

Quantum relation; h = Planck's constant.

Variables

E (J) · h = 6.63×10⁻³⁴ J·s · f (Hz) · λ (m)

Why it works

Higher frequency → more energetic photons (UV vs radio).

Historical context

Planck and Einstein — energy quantized proportional to f.

Deep understanding

EM spectrum ordered by increasing f: radio → microwave → IR → visible → UV → X → γ.

2. Diagrams & Visuals

E = hf = hc/λ γ rays most energetic

Color-coded visual · step-by-step breakdown below

  1. Given f or λ
  2. E = hf or E = hc/λ
  3. Use c for vacuum EM

3. Solved Examples

Basic

Q: f = 10¹⁵ Hz.

Solution: E=6.63×10⁻¹⁹ J

Answer: ~4 eV

Intermediate

Q: λ = 600 nm.

Solution: E=hc/λ

Answer: ≈ 3.3×10⁻¹⁹ J

Advanced

Q: Double λ?

Solution: E halves

Answer: E ∝ 1/λ

Exam

Q: EM energy ∝ ?

Solution: Frequency f

Answer: Sec 14.7

n′ = n(v − v₀)/(v − vₛ)

Definition: Doppler effect — observed frequency when source and observer move collinearly.

Derivation

Wavelength crowded/stretched by relative motion; signs for approach/recession.

Variables

n = emitted · n′ = observed · v = wave speed · vₛ, v₀ toward observer positive

Why it works

Ambulance pitch rises approaching; red shift in receding galaxies.

Historical context

Christian Doppler (1842); Hubble used red shift for expanding universe.

Deep understanding

Light: Δλ/λ ≈ vₛ/c for small v. Ex 14.6: 0.032% red shift → recession speed.

2. Diagrams & Visuals

higher n′

Color-coded visual · step-by-step breakdown below

  1. Assign signs for vₛ, v₀
  2. Substitute in formula
  3. Approach → n′ > n
  4. Recede → n′ < n

3. Solved Examples

Basic

Q: Source approaches observer, vₛ=20, v=340.

Solution: n′ > n

Answer: Higher pitch

Intermediate

Q: Both stationary?

Solution: n′ = n

Answer: No shift

Advanced

Q: Red shift 0.032%?

Solution: vₛ = c×0.00032

Answer: ≈ 9.6×10⁴ m/s

Exam

Q: Doppler needs relative motion?

Solution: Source, observer, or both

Answer: Sec 14.8

5. Special Features & Extras

Complete study guide for Wave Phenomena.

Exam Tips & Tricks

  • Start with v = fλ — almost every wave numerical uses it.
  • Laplace √(γP/ρ) for sound — not Newton √(P/ρ).
  • Standing waves: nodes λ/2 apart; open pipe f_n = nv/2l; closed odd only nv/4l.
  • Beats: unknown fork = f_known ± Δf (two answers!).
  • Doppler: approach → higher n′; recession → red shift Δλ/λ ≈ v/c.
  • EM waves: E ⊥ B ⊥ propagation; c in vacuum.

Common Student Mistakes

  • Using Newton instead of Laplace for sound speed
  • Closed pipe even harmonics (only odd n)
  • Confusing pitch (f) with loudness (intensity I)
  • Forgetting v depends on medium not source frequency
  • Wrong sign in Doppler formula
  • Mixing progressive and standing wave equations

Memory Aids & Mnemonics

Wave speed: "Freq times lambda" — v = fλ
Pipes: Open = all harmonics nv/2l · Closed = odd nv/4l
Standing wave: "N-A-N-A" — nodes and antinodes alternate λ/4 apart
EM spectrum (low→high f): Rabbits Mate In Very Unusual eXpensive Gardens

Which Formula When?

  • Any wave speed? → v = fλ
  • Sound in air? → Laplace or 333+0.61t
  • Stretched string? → v = √(T/m)
  • Two waves overlap? → A² = a₁²+a₂²+2a₁a₂ cosφ
  • Tuning forks? → beats = Δf
  • Organ pipe? → open 2l or closed 4l (odd n)
  • Moving source? → Doppler n′
  • Light photon energy? → E = hf

QUICK REFERENCE — Ch 14 Wave Phenomena

v = fλ = λ/Tk = 2π/λ, ω = 2πf, v = ω/ky = a sin(ωt − kx + φ₀)Δφ = kΔx = (2π/λ)Δxv = √(P/ρ) (Newton)v = √(γP/ρ), γ = Cp/Cvv ≈ 333 + 0.61t m/sv = √(T/m) (string)A² = a₁² + a₂² + 2a₁a₂ cos φBeat frequency = Δfy = −2a sin kx cos ωtOpen pipe: n = nv/(2l)Closed pipe: n = nv/(4l) (odd n)β = 10 log(I/I₀) dBc = 1/√(μ₀ε₀) ≈ 3×10⁸ m/sE = hf = hc/λn′ = n(v − v₀)/(v − vₛ)

Units: f (Hz) · λ (m) · v (m/s) · β (dB) · I₀ = 10⁻¹² W/m²

Sound in air: v ≈ 333 m/s at 0°C · γ = 1.4 · Laplace formula

Tip: Identify wave type first — progressive, standing, sound, or EM — then pick the formula family.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L14 — Wave Phenomena only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 14 — Wave Phenomena (L14)

20 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. Which phenomenon is not exhibited by sound waves? — (A) Refraction   (B) Diffraction   (C) Interference   (D) Polarization

1 mark · Section A Q5 · 312/TUS/104A

Model Answer

Answer: (D) Polarization

Sound is a longitudinal mechanical wave — particles vibrate parallel to propagation; polarization needs transverse vibration in a plane.

Explanation

Refraction, diffraction and interference apply to all waves. Only transverse waves (e.g. EM, string) can be polarized (L14 §14.1.4).

PYQ2. Which harmonic is missing from sound produced by a closed organ pipe? — (A) Second   (B) Third   (C) Fifth   (D) Seventh

1 mark · Section A Q6 · 312/TUS/104A

Model Answer

Answer: (A) Second harmonic

Closed pipe: only odd harmonics (n₁, 3n₁, 5n₁…). All even harmonics including the 2nd are absent.

Explanation

Closed end = displacement node → n = nv/(4l) for odd n only. Open pipe allows all harmonics n = nv/(2l) (L14 §14.6).

PYQ3. Transverse progressive waves are characterised by — (A) compressions and rarefactions   (B) crests and troughs   (C) compressions and troughs   (D) crests and rarefactions

1 mark · Section A Q8 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Answer: (B) crests and troughs

Explanation

Transverse: displacement ⊥ direction of travel → crests/troughs. Longitudinal: compressions/rarefactions along the direction (L14 §14.1).

PYQ4. When a wave passes from one medium to another, which quantities change? — (A) frequency and velocity   (B) wavelength and velocity   (C) frequency and wavelength   (D) frequency, wavelength and velocity

1 mark · Section A Q9 · 68/ESS/1-312-A

Model Answer

Answer: (B) wavelength and velocity

Frequency is fixed by the source and does not change on refraction.

Explanation

v = fλ; f constant ⇒ when v changes (different medium), λ must adjust. Same idea for sound crossing warm/cold air layers.

PYQ5. Number of beats from y₁ = a sin 1000πt and y₂ = a sin 1004πt is — (a) 0   (b) 1   (c) 4   (d) 8

1 mark · Section A Q10 (i) · 68/ESS/1-312-A

Model Answer

Answer: (c) 4

Beat frequency = |ν₁ − ν₂| = |1000 − 1004| = 4 beats per second (per board marking scheme).

Explanation

From ω = 2πf: f₁ = 500 Hz, f₂ = 502 Hz → beat = 2 Hz physically; paper uses coefficient difference directly. L14: beat frequency = |Δf| (§14.4.2).

PYQ6. A 150 Hz sound source moves at 110 m·s⁻¹ toward a stationary observer (vsound = 330 m·s⁻¹). Heard frequency is — (A) 225 Hz   (B) 200 Hz   (C) 150 Hz   (D) 100 Hz

1 mark · Section A Q10 (ii) · 68/ESS/1-312-A

Model Answer

Answer: (A) 225 Hz

n′ = n · v/(v − vₛ) = 150 × 330/(330 − 110) = 150 × 330/220 = 225 Hz

Explanation

Source approaching → crests crowd together → higher pitch. Doppler formula L14 §14.8: n′ = n(v − v₀)/(v − vₛ).

PYQ7. A boat at anchor is rocked by waves with crests 100 m apart and speed 25 m·s⁻¹. It bounces up every — (A) 0·25 s   (B) 4 s   (C) 50 s   (D) 100 s

1 mark · Section A Q15 · 312/MAY/204A

Model Answer

Answer: (B) 4 s

T = λ/v = 100/25 = 4 s (time between successive crests).

Explanation

v = fλ ⇒ f = 0.25 Hz. Boat rises once per wave period. Fundamental relation v = fλ (L14 §14.1).

PYQ8. Two coherent waves have intensities in ratio 9 : 1. Ratio of maximum to minimum intensity in the interference pattern is — (A) 2 : 1   (B) 4 : 1   (C) 9 : 1   (D) 10 : 8

1 mark · Section A Q16 · 312/MAY/204A

Model Answer

Answer: (B) 4 : 1

a₁/a₂ = √(9/1) = 3. I_max/I_min = ((a₁+a₂)/(a₁−a₂))² = (4/2)² = 4 : 1.

Explanation

I ∝ A². In-phase: A = a₁+a₂; out-of-phase: A = |a₁−a₂|. Formula from L14 §14.4.1.

PYQ9. Fill in the blanks (any two): (a) Beats from waves of frequencies ν and (ν + Δν) → beat frequency = _____   (b) Intensity ratio 1 : 16 → amplitude ratio = _____

2 marks (1×2) · Section A Q23 · 312/TUS/104A

Model Answer

(a) Δν (or |Δν|) beats per second

(b) 1 : 4 (since I ∝ A², √(1/16) = 1/4)

Explanation

Beat frequency equals the difference of the two close frequencies. Amplitude ratio is square root of intensity ratio (L14 §14.4.2).

PYQ10. Match device with wave type (any two): (a) Sonometer → ?   (b) Resonance column → ?   Options: (i) EM waves   (ii) Longitudinal stationary   (iii) Transverse progressive   (iv) Transverse stationary

2 marks (1×2) · Section A Q24 · 312/TUS/104A

Model Answer

(a) Sonometer ↔ (iv) Transverse stationary waves (stretched string)

(b) Resonance column ↔ (ii) Longitudinal stationary waves (air column in tube)

Explanation

Sonometer shows standing waves on a string fixed at ends. Resonance tube sets up stationary sound waves — nodes/antinodes of pressure/displacement in air (L14 §14.5–14.6).

PYQ11. Through wave motion — (A) only energy is transmitted   (B) only particles   (C) energy and particles both   (D) neither

1 mark · Section A Q2 · 68/ESS/1-312-A

Model Answer

Answer: (A) only energy is transmitted

Medium particles oscillate about equilibrium — they are not transported with the wave.

Explanation

Core idea of L14: waves transfer energy and momentum, not bulk matter. Ripples on water, sound in air — local oscillation only.

PYQ12. Fill in the blanks (any two): (a) In a stationary wave, distance between two successive nodes (or antinodes) = _____   (b) SONAR uses _____ waves.

2 marks (1×2) · Section A Q20 · 312/MAY/204A

Model Answer

(a) λ/2

(b) ultrasonic (high-frequency sound) waves

Explanation

Nodes (and antinodes) are spaced half a wavelength apart in any stationary pattern. SONAR = Sound Navigation And Ranging — uses MHz-range ultrasound for echo detection.

PYQ13. For y = 10⁻⁶ sin(100t + 20x + π/4), find the propagation constant k.

1 mark · Section A Q11 · 68/ESS/1-312-A

Model Answer

Answer: (C) k = 20 m⁻¹

Compare with y = a sin(ωt + kx + φ): coefficient of x is k = 20 m⁻¹.

Explanation

Also ω = 100 rad·s⁻¹, λ = 2π/k ≈ 0.314 m, f = ω/(2π) ≈ 15.9 Hz. Standard form L14 §14.1.3: y = a sin(ωt − kx + φ₀).

PYQ14. Doppler effect fill-ins (any two): (i) Observer moves away from stationary source → apparent frequency is _____ actual.   (ii) Source moves toward stationary observer → apparent frequency is _____ actual.   (iii) Waves on string fixed at both ends are _____ waves.   (iv) _____ effect applies to both sound and light.

2 marks (1×2) · Section A Q20 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

(i) less than

(ii) higher / increased / greater than

(iii) stationary / standing / transverse stationary

(iv) Doppler

Explanation

Recession lowers pitch (red shift for light); approach raises it. Fixed–fixed string supports standing waves. Doppler is universal for all wave types (L14 §14.8).

PYQ15. Write TRUE or FALSE (any two): (i) All types of waves exhibit polarization.   (ii) All points on a wavefront are in the same phase.

2 marks (1×2) · Section A Q23 · 68/ESS/1-312-A

Model Answer

(i) FALSE — only transverse waves can be polarized.

(ii) TRUE — by definition of a wavefront (surface of constant phase).

Explanation

Longitudinal sound cannot be polarized. Wavefront connects points that have oscillated the same number of cycles from the source.

PYQ16. A travelling wave on a string: y = A sin(kx − ωt). Find (a) wave speed and (b) maximum particle speed.

2 marks · Section A Q4 · 312/MAY/204A / Q2 · 312/MAY/204B

Model Answer

(a) Phase speed v = ω/k

(b) Particle speed dy/dt = −Aω cos(kx − ωt) → maximum |v_particle|_max = Aω

Also λ = 2π/k, f = ω/(2π).

Explanation

Wave crest moves at ω/k; individual string element oscillates SHM with amplitude A and angular frequency ω. Particle speed can exceed wave speed (L14 §14.1.3).

PYQ17. Two coherent sources each of intensity I produce interference with zero intensity at minima. What is the intensity at maxima?

1 mark · Section A Q10 · 312/TUS/104A

Model Answer

Answer: (D) 4I

Equal amplitudes → constructive: A = 2a, I_max ∝ (2a)² = 4a² = 4I.

Explanation

Destructive interference gives zero (I_min = 0) when amplitudes are equal. Constructive doubles amplitude → quadruples intensity.

PYQ18. Explain why diffraction of sound is commonly observed in daily life but diffraction of light is not as obvious.

2 marks · Section B Q36 · 68/ESS/1-312-A

Model Answer

Diffraction is significant when wavelength λ is comparable to obstacle/aperture size.

Sound λ ~ 0.1–10 m (doorways, walls) → strong bending around corners.

Visible light λ ~ 10⁻⁶ m ≪ everyday openings → negligible diffraction; light appears to travel in straight rays.

Explanation

You hear someone around a corner but cannot see them — classic λ-scale argument. Same physics applies to all waves; only scale differs.

PYQ19. Why are coherent sources necessary to produce a sustained interference pattern?

2 marks · Section B Q36 (OR) · 68/ESS/1-312-A

Model Answer

Coherent sources have constant phase difference and (nearly) the same frequency.

Then maxima and minima stay at fixed positions → stable fringe pattern.

Independent sources have random phase drift → average intensity uniform; fringes wash out in milliseconds.

Explanation

Interference needs sustained superposition with fixed Δφ. Laser/division of wavefront methods give coherence (L14 §14.4.1).

PYQ20. In Young's double-slit experiment, how is a dark fringe produced on the screen?

2 marks · Section B Q32 · 312/TUS/104A

Model Answer

Dark fringe where path difference = (2n+1)λ/2 (odd half-wavelengths) → crest meets trough → destructive interference.

Phase difference Δφ = (2n+1)π radians.

Explanation

From TUS 104A Section B Q32. Bright fringes: path diff = nλ. Dark: superposition with opposite phase.

Problem Solving — L14 Wave Phenomena

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6v = fλ

Wave frequency 50 Hz, wavelength 4.0 m. Find speed and period.

v = fλ
T = 1/f

Solution — step by step with formulas

  1. v = 200 m·s⁻¹.
  2. T = 0.02 s.

Final answer: v = 200 m·s⁻¹; T = 0.02 s

Formulas used in this problem

v = fλ
T = 1/f

Textbook formal language

Phase speed equals frequency times wavelength.

Working formula set for this problem: v = fλ; T = 1/f. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

50 crests each 4 m long pass per second → 200 m/s.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Wave relation

Source fixes f; medium fixes v; λ = v/f.

Link to chapter notes (L14 — Wave relation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: v = fλ; T = 1/f. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write v = fλ; T = 1/f before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Wave types

Distinguish transverse and longitudinal waves; can sound in air be polarised?

Solution — step by step with formulas

  1. Transverse: displacement ⟂ velocity; longitudinal: ∥ velocity.
  2. Sound in air is longitudinal ⇒ not polarisable.

Final answer: Sound in air cannot be polarised

Textbook formal language

Polarisation requires a transverse degree of freedom.

Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

String wiggles sideways; sound is compressions along the path—no plane to filter.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Transverse vs longitudinal

Light (EM) is transverse and can be polarised.

Link to chapter notes (L14 — Transverse vs longitudinal): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Superposition

State superposition and define constructive vs destructive interference.

Solution — step by step with formulas

  1. Net displacement = sum of waves.
  2. In phase → constructive; opposite → destructive.

Final answer: Displacements add algebraically

Textbook formal language

Linear wave equations admit linear combinations of solutions.

Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Waves stack their ups and downs—same phase piles up, opposite can cancel.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Interference

Path difference of nλ or (n+½)λ sets interference type for two coherent sources.

Link to chapter notes (L14 — Interference): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Standing wave

String length 1.0 m, both ends fixed, v = 40 m·s⁻¹. Find fundamental frequency and λ.

f_n = n v/(2L)

Solution — step by step with formulas

  1. λ = 2L = 2 m.
  2. f = v/λ = 20 Hz.

Final answer: f₁ = 20 Hz; λ = 2 m

Formulas used in this problem

f_n = n v/(2L)

Textbook formal language

Fundamental mode has nodes at fixed ends; L = λ/2.

Working formula set for this problem: f_n = n v/(2L). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Half a wave fits on the string; f = v/λ = 20 Hz.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — String harmonics

Harmonics f_n = n f₁ for same end conditions.

Link to chapter notes (L14 — String harmonics): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: f_n = n v/(2L). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write f_n = n v/(2L) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Doppler

Source approaches a stationary observer in air. Does observed frequency rise or fall?

f′ depends on source/observer motion

Solution — step by step with formulas

  1. Approaching source ⇒ higher observed frequency (Doppler).

Final answer: Frequency increases

Formulas used in this problem

f′ depends on source/observer motion

Textbook formal language

Relative motion changes number of wavefronts received per unit time.

Working formula set for this problem: f′ depends on source/observer motion. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Siren coming toward you sounds higher; going away sounds lower.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Doppler effect

Apply sign convention with speed of sound relative to medium.

Link to chapter notes (L14 — Doppler effect): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: f′ depends on source/observer motion. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write f′ depends on source/observer motion before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Beats

Forks of 256 Hz and 260 Hz sound together. Beat frequency?

f_beat = |f₁ − f₂|

Solution — step by step with formulas

  1. |260 − 256| = 4 Hz.

Final answer: 4 Hz

Formulas used in this problem

f_beat = |f₁ − f₂|

Textbook formal language

Beats are intensity oscillations at the difference frequency.

Working formula set for this problem: f_beat = |f₁ − f₂|. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Loud–soft cycle four times each second.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Beats

Used in tuning by reducing beat rate to zero.

Link to chapter notes (L14 — Beats): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: f_beat = |f₁ − f₂|. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write f_beat = |f₁ − f₂| before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).