NCERT Plus One (Class 11) Physics • Chapter 14

Waves

Q: Distinguish between transverse waves and longitudinal waves with one example each.

Transverse: Medium particles vibrate ⊥ to wave direction, forming crests & troughs (e.g. wave on stretched string). Longitudinal: Particles vibrate ∥ to wave direction, forming compressions & rarefactions (e.g. sound in air).

Q: A transverse wave on a string is given by y(x, t) = 3.0 sin(36 t + 0.018 x + π/4), where x, y are in cm and t in s. Find (a) direction & speed, (b) amplitude & frequency, and (c) wavelength.

(a) Same sign (+) between ωt and kx ⇒ travels along −x . Speed v = ω/k = 36 / 0.018 = 2000 cm/s = 20 m/s . (b) Amplitude A = 3.0 cm ; ν = ω/(2π) = 36/(2π) ≈ 5.73 Hz . (c) λ = 2π / k = 2π / 0.018 ≈ 349 cm = 3.49 m .

Q: Why did Newton's formula fail for the speed of sound in air? Explain Laplace's correction and write the corrected equation.

Newton assumed isothermal conditions ( B = P ), giving v = √(P/ρ) ≈ 280 m/s vs observed 331 m/s . Laplace's correction: Because air is a poor heat conductor and sound vibrations are rapid, compressions/rarefactions are adiabatic ( B = γP ). Thus v = √(γP / ρ) ≈ 331.3 m/s for γ = 1.4 .

Q: Explain why sound travels faster (a) on a hot summer day than a cold morning, and (b) in humid air than in dry air.

(a) v = √(γRT / M) ∝ √T : higher temperature raises sound speed. (b) Water vapour ( M = 18 ) is lighter than dry air ( M ≈ 29 ), lowering average molar mass M (and density ρ ) ⇒ v increases!

Q: Show that only odd harmonics are produced in an air column closed at one end (closed organ pipe).

Closed end ( x = 0 ) must be a Node ; open end ( x = L ) must be an Antinode . Pipe length fits an odd number of quarter-wavelengths: L = (2n − 1) λ/4 for n = 1, 2, 3... ⇒ λₙ = 4L / (2n − 1) ⇒ νₙ = v/λₙ = (2n − 1) v / (4L) = (2n − 1) ν₁ . Hence only odd harmonics 1 : 3 : 5... exist!

Q: A pipe 30 cm long is open at both ends. Find its fundamental frequency if speed of sound is 330 m/s. What would it be if one end is closed?

L = 0.30 m . • Open pipe: ν₁ = v / (2L) = 330 / (2 × 0.30) = 550 Hz . • Closed pipe: ν₁' = v / (4L) = 330 / (4 × 0.30) = 275 Hz (half!).

Q: A stretched string of length 1 m and mass 10 g is kept under a tension of 160 N. Find the speed of transverse waves and fundamental frequency.

μ = m/L = 0.010 / 1 = 0.01 kg/m . Wave speed v = √(T / μ) = √(160 / 0.01) = √16000 ≈ 126.5 m/s . Fundamental frequency ν₁ = v / (2L) = 126.5 / 2 ≈ 63.25 Hz .

Transverse vs longitudinal waves, travelling wave equations, string & sound speeds, superposition, standing waves & beats.

📄 Official NCERT Textbook PDF

Topic 14.2: Transverse vs Longitudinal Mechanical Waves

Picture cheering in a football stadium in Kozhikode doing a Mexican Wave! Each fan simply stands up and sits back down in their seat, YET the wave races all around the stadium! Waves carry energy and momentum without carrying matter from place to place. Flick a skipping rope up and down → Transverse. Push a Slinky coil forward and back → Longitudinal!

TRANSVERSE WAVESParticles vibrate ⊥ to wave travel (Crests & Troughs). Needs shear stiffness (solids & string)!

LONGITUDINAL WAVESParticles vibrate ∥ to wave travel (Compressions & Rarefactions). Travels in solids, liquids & gases!

MECHANICAL VS EMSound/rope need a medium; Light (EM wave) needs NO medium & is transverse!

🧠 Memory Trick: Transverse = makes a T (perpendicular ⊥); Longitudinal = moves aLong the same line (parallel ∥)! Sound in air is ALWAYS longitudinal!

💡 Simple Kid-Check: Drop a stone in Vembanad lake and watch a floating dry leaf: the leaf just bobs up and down on the spot while ripples race outward! Medium particles only oscillate locally; the disturbance pattern travels!

Feature	Transverse Wave	Longitudinal Wave
Particle Motion	Perpendicular (⊥) to wave velocity	Parallel (∥) to wave velocity
Wave Parts	Crests (high) & Troughs (low)	Compressions (dense) & Rarefactions (thin)
Media Supported	Solids & liquid surfaces (needs shear rigidity)	Solids, liquids & gases (needs bulk elasticity)
Classic Examples	Guitar string wave, light (EM wave)	Sound in air, Slinky push-pull pulse

Exam Trap: Why can't transverse mechanical waves travel deep inside air or liquid interiors? Because fluids lack shear modulus (rigidity) to pull adjacent layers sideways! Only longitudinal sound waves pass through air.

Topic 14.3: Travelling Wave Equation (Amplitude, Phase, k & ω)

Want to write a GPS formula that predicts the exact height of a surfer's wave at ANY spot x on Kovalam beach and ANY second t on your wristwatch? Take SHM y = A sin(ωt) and replace time with (ωt − kx) — that sneaky minus sign tells the whole wave to slide steadily forward along +x!

TRAVELLING WAVEy(x,t) = A sin(kx − ωt + φ) (moves +x); (kx + ωt) moves −x!

WAVE NUMBER & OMEGAk = 2π / λ (rad/m) • ω = 2πν = 2π / T (rad/s)

WAVE SPEEDv = ω / k = ν λ = λ / T

🧠 Memory Trick: Check the sign between kx and ωt! Opposite signs (kx − ωt or ωt − kx) = moves RIGHT (+x)! Same signs (kx + ωt) = moves LEFT (−x)!

💡 Simple Kid-Check: Keep your eyes locked on ONE surfing crest — that means its phase kx − ωt = constant! As time t grows, x must grow at speed dx/dt = ω/k to keep the phase constant! That's why opposite signs move along +x!

y(x, t) = A sin(kx − ωt + φ)
Wavelength: λ = 2π / k | Period: T = 2π / ω | Wave Speed: v = ω / k = νλ

🧒 Quick Decoding Example: Given y = 0.05 sin(4x + 20t) (SI units):
• Same sign (+) ⇒ travels along −x! • Amplitude A = 0.05 m
• k = 4 rad/m ⇒ λ = 2π/4 = π/2 m • ω = 20 rad/s ⇒ v = ω/k = 20/4 = 5 m/s!

Giant Exam Trap: Do NOT confuse Wave Speed (v = ω/k, constant!) with Particle Speed (v_p = ∂y/∂t, oscillates like SHM with max Aω!). Wave speed moves the pattern horizontally; particle speed wiggles vertically!

Topic 14.4: Speed of Waves: Stretched String & Sound (Newton–Laplace)

Ever tuned a guitar or veena string in a music class? Tighten the peg (more tension T) and the sound snaps sharper and higher; pluck the thick bass wire (larger mass per length μ) and the wave travels slower! Even cooler: young Isaac Newton guessed sound travels isothermally at 280 m/s — until Pierre Laplace fixed it to 331 m/s using adiabatic compressions!

WAVE ON A STRINGv = √(T / μ) (T = tension in N, μ = m/L linear mass density in kg/m)

SOUND IN FLUIDS/SOLIDSFluid: v = √(B / ρ) • Solid rod: v = √(Y / ρ)

NEWTON–LAPLACE SOUND IN GASv = √(γ P / ρ) = √(γ R T / M) (Laplace adiabatic γ = 1.4!)

🧠 Memory Trick: Every mechanical wave speed = √(Elasticity / Inertia)! String: √(T/μ); Air: √(γP/ρ)! Notice Pressure change at fixed T leaves sound speed UNCHANGED!

💡 Simple Kid-Check: Sound waves squeeze and stretch air SO fast (thousands of times per second!) that heat generated in a compression has ZERO time to leak away — making sound propagation Adiabatic (B = γP), NOT isothermal (B = P)!

1
Newton's Mistake (Isothermal): Assumed B = P, giving v = √(P/ρ) ≈ 280 m/s in air at STP — 15% too low compared to 331 m/s!
2
Laplace's Correction (Adiabatic): Rapid compressions don't exchange heat, so B_adia = γ P (γ = 1.4 for air), giving v = √(1.4 × P/ρ) = 331.3 m/s — bullseye match!
3
What Changes Sound Speed in Air? v = √(γRT/M) depends on √T (faster in warm air), 1/√M (faster in humid air since water vapour M=18 is lighter than dry air M≈29!), and is INDEPENDENT of Pressure P at constant T!

Top DHSE Trap: If atmospheric pressure doubles at constant temperature, what happens to the speed of sound in air? NO CHANGE! Because P/ρ = RT/M stays constant when T is fixed! Also remember: sound travels FASTER on a humid rainy day than on a dry day!

Topic 14.5: Principle of Superposition & Interference

Imagine two ripples from two rain drops colliding on a puddle! Do they smash together and shatter like toy cars? Not at all — they slide right through each other untouched! Right where both crests meet, the water jumps twice as high; where a crest meets a trough, the water magically flattens for a split second!

SUPERPOSITION PRINCIPLENet displacement = algebraic sum: y(x,t) = y₁(x,t) + y₂(x,t)

EQUAL-AMPLITUDE INTERFERENCEy_net = 2A cos(φ/2) sin(kx − ωt + φ/2) → Resultant Amp A_R = 2A cos(φ/2)

CONSTRUCTIVE VS DESTRUCTIVEφ = 0, 2π, 4π... ⇒ A_R = 2A (Max!) • φ = π, 3π... ⇒ A_R = 0 (Cancel!)

🧠 Memory Trick: When two equal waves meet with phase gap φ, resultant amplitude is 2A cos(φ/2), and Intensity is I = 4 I₀ cos²(φ/2)!

💡 Simple Kid-Check: Two friends whispering into the same room don't block each other's voices — sound waves cross right through one another independently! At any overlapping spot, nature simply adds their displacements like numbers (+3 cm + (−3 cm) = 0)!

General Resultant Amplitude: A_R = √(A₁² + A₂² + 2 A₁ A₂ cos φ)
If A₁ = A₂ = A: A_R = 2A |cos(φ/2)| | Constructive: φ = 2nπ | Destructive: φ = (2n+1)π

🧒 Quick Phase Check: Two identical waves of amplitude 5 mm meet with phase difference φ = 120°. Find resultant amplitude!
A_R = 2(5) cos(120°/2) = 10 cos(60°) = 10 × 0.5 = 5 mm!

Conceptual Trap: Does Destructive Interference destroy wave energy? NO! Energy is ALWAYS conserved — the energy missing from quiet dark spots simply shifts over into the louder constructive spots (average intensity stays I₁ + I₂)!

Topic 14.6: Reflection, Standing Waves & Harmonics in Strings/Organ Pipes

Pluck a guitar string or blow across a soda bottle! The wave shoots to the end, bounces back inverted (from a rigid wall!) or upright (from an open end!), and tangles with incoming waves to lock into frozen looping hills called Standing Waves! Points that never budge are Nodes (N); points screaming widest are Antinodes (A)!

REFLECTION PHASE SHIFTRigid boundary: π (180°) inversion! • Open/free boundary: 0° (no flip!)

STANDING WAVE EQUATIONy = [2A sin(kx)] cos(ωt) • Adjacent N–N or A–A = λ/2; N–A = λ/4

HARMONICS CHEAT SHEETString / Open Pipe: νₙ = n(v/2L) (All 1:2:3) • Closed Pipe: ν = (2k+1)v/4L (Odd only 1:3:5!)

🧠 Memory Trick: Closed Pipe = Odd Only! (Closed end forces a Node, open end forces an Antinode → fits λ/4, 3λ/4, 5λ/4 → harmonics 1 : 3 : 5!). Open pipe & string have ALL harmonics 1 : 2 : 3!

💡 Simple Kid-Check: In a Standing (Stationary) Wave, energy stays trapped locally between Nodes — it does NOT travel down the string! Look at 2A sin(kx) cos(ωt): space and time sit in SEPARATE trig functions (never kx − ωt together!).

System of Length L	Boundary Rules	Fundamental Freq (ν₁)	Allowed Harmonics
Stretched String	Nodes at BOTH ends	`v / (2L)`	All: `1, 2, 3, 4...`
Open Organ Pipe (both open)	Antinodes at BOTH ends	`v / (2L)`	All: `1, 2, 3, 4...`
Closed Organ Pipe (1 closed)	Node at closed, Antinode at open	`v / (4L)`	Odd ONLY: `1, 3, 5...`

Watch the Overtone vs Harmonic naming trap in Closed Organ Pipes! Since even harmonics are missing, the 1st Overtone of a closed pipe is the 3rd Harmonic (3ν₁), and the 2nd Overtone is the 5th Harmonic (5ν₁)! Meanwhile in an open pipe, 1st Overtone = 2nd Harmonic.

Topic 14.7: Beats & Beat Frequency Calculation

Hit two guitar strings that are almost — but NOT quite — in tune! Instead of a smooth hum, you hear a rhythmic pulsing "WAH-wah-WAH-wah-WAH!" Piano tuners in Kochi use this exact throbbing loudness pulse, called Beats, tuning the string until the "wah-wah" slowing completely disappears!

BEAT FREQUENCYν_beat = |ν₁ − ν₂| (Beats heard per second!)

WAXING & WANING1 Beat = 1 Loud (wax) + 1 Quiet (wane). Beat period T_b = 1 / |ν₁ − ν₂|

WAXING / FILING TUNING FORKLoading wax: ↑m ⇒ ↓ν • Filing prong: ↓m ⇒ ↑ν!

🧠 Memory Trick: Think human ears! Our ear can only separate beats clearly when |ν₁ − ν₂| ≤ 10 Hz! And Waxing a tuning fork makes it heavy (↓ν); Filing makes it light (↑ν)!

💡 Simple Kid-Check: Imagine two drummers tapping at 10 beats/s and 12 beats/s. Sometimes their taps line up (BOOM — loud!), then drift out of step (quiet!), then line up again exactly 12 − 10 = 2 times every second!

y = y₁ + y₂ = 2A cos[2π ((ν₁ − ν₂)/2) t] · sin[2π ((ν₁ + ν₂)/2) t]
Amplitude hits BOTH +2A and −2A (loud!) twice per cosine cycle ⇒ ν_beat = |ν₁ − ν₂|!

🧒 Classic Wax-Loading Numerical: Fork A (unknown ν_A) gives 4 beats/s with fork B (256 Hz). So ν_A = 256 ± 4 = 260 or 252 Hz.
Loading wax on A lowers ν_A, and beats drop to 2/s! Since lowering 260 → 258 brings it closer to 256 (shrinking gap to 2), the original frequency MUST be 260 Hz!

Watch the sneaky factor-of-2 derivation trap! The cosine envelope oscillates at frequency (ν₁ − ν₂)/2, YET loudness depends on amplitude squared (|cos|), which peaks TWICE per cycle — giving beat frequency 2 × (ν₁ − ν₂)/2 = |ν₁ − ν₂|!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Distinguish between transverse waves and longitudinal waves with one example each.

Transverse: Medium particles vibrate ⊥ to wave direction, forming crests & troughs (e.g. wave on stretched string).
Longitudinal: Particles vibrate ∥ to wave direction, forming compressions & rarefactions (e.g. sound in air).

Q2 (Kerala HSE Say • 3 Marks): A transverse wave on a string is given by y(x, t) = 3.0 sin(36 t + 0.018 x + π/4), where x, y are in cm and t in s. Find (a) direction & speed, (b) amplitude & frequency, and (c) wavelength.

(a) Same sign (+) between ωt and kx ⇒ travels along −x. Speed v = ω/k = 36 / 0.018 = 2000 cm/s = 20 m/s.
(b) Amplitude A = 3.0 cm; ν = ω/(2π) = 36/(2π) ≈ 5.73 Hz.
(c) λ = 2π / k = 2π / 0.018 ≈ 349 cm = 3.49 m.

Q3 (Kerala HSE March • 3 Marks): Why did Newton's formula fail for the speed of sound in air? Explain Laplace's correction and write the corrected equation.

Newton assumed isothermal conditions (B = P), giving v = √(P/ρ) ≈ 280 m/s vs observed 331 m/s.
Laplace's correction: Because air is a poor heat conductor and sound vibrations are rapid, compressions/rarefactions are adiabatic (B = γP). Thus v = √(γP / ρ) ≈ 331.3 m/s for γ = 1.4.

Q4 (Kerala HSE Model • 2 Marks): Explain why sound travels faster (a) on a hot summer day than a cold morning, and (b) in humid air than in dry air.

(a) v = √(γRT / M) ∝ √T: higher temperature raises sound speed.
(b) Water vapour (M = 18) is lighter than dry air (M ≈ 29), lowering average molar mass M (and density ρ) ⇒ v increases!

Q5 (Kerala HSE March • 3 Marks): Show that only odd harmonics are produced in an air column closed at one end (closed organ pipe).

Closed end (x = 0) must be a Node; open end (x = L) must be an Antinode.
Pipe length fits an odd number of quarter-wavelengths: L = (2n − 1) λ/4 for n = 1, 2, 3...
⇒ λₙ = 4L / (2n − 1) ⇒ νₙ = v/λₙ = (2n − 1) v / (4L) = (2n − 1) ν₁.
Hence only odd harmonics 1 : 3 : 5... exist!

Q6 (Kerala HSE Say • 2 Marks): A pipe 30 cm long is open at both ends. Find its fundamental frequency if speed of sound is 330 m/s. What would it be if one end is closed?

L = 0.30 m.
• Open pipe: ν₁ = v / (2L) = 330 / (2 × 0.30) = 550 Hz.
• Closed pipe: ν₁' = v / (4L) = 330 / (4 × 0.30) = 275 Hz (half!).

Q7 (NCERT / DHSE Model • 3 Marks): What are beats? Show analytically that the beat frequency equals the difference between the frequencies of the two superposing waves.

Beats are periodic waxing and waning of sound intensity when two waves of nearly equal frequencies ν₁, ν₂ superpose.
y = A sin(2πν₁t) + A sin(2πν₂t) = 2A cos[π(ν₁ − ν₂)t] sin[π(ν₁ + ν₂)t].
Amplitude envelope |2A cos[π(ν₁ − ν₂)t]| peaks twice per cosine period, at interval 1 / |ν₁ − ν₂| ⇒ ν_beat = |ν₁ − ν₂|.

Q8 (Kerala HSE March • 2 Marks): A stretched string of length 1 m and mass 10 g is kept under a tension of 160 N. Find the speed of transverse waves and fundamental frequency.

μ = m/L = 0.010 / 1 = 0.01 kg/m.
Wave speed v = √(T / μ) = √(160 / 0.01) = √16000 ≈ 126.5 m/s.
Fundamental frequency ν₁ = v / (2L) = 126.5 / 2 ≈ 63.25 Hz.