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State Board · Class 11 · Physics

Lesson 8: Sound

7 concepts · 56 practice questions with answers · Free to read

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Wave basics and the wave equation

From daily life

A stone dropped in a pond sends ripples out while the water only moves up and down. Link to earlier classes : sound, frequency and echo of classes 9-10 now extend to the wave equation and the Doppler effect.

Simple definition

A wave carries energy forward; particles only vibrate in place.

Wave : A, λ, vλAv = f λy = A sin(ωt − kx)

In the figure : A = amplitude, λ = wavelength.

Live figure

Wave basics and the wave equation

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Rule 1: Types of waves — Transverse : vibration perpendicular to the wave (string, light). Longitudinal : along the wave (sound; compressions and rarefactions).

  • Easy : Sound in air is longitudinal.
  • Medium : Pushing a spring to and fro ⇒ a longitudinal wave.
  • Exam (CET) : Sound wave is? (1) longitudinal (2) transverse (3) both (4) neither → answer 1)

Rule 2: v = f λ — v = f λ; ω = 2π f; k = 2π ÷ λ; v = ω ÷ k; T = 1 ÷ f.

  • Easy : f = 500 Hz, λ = 0.68 m ⇒ v = 340 m/s.
  • Medium : y = 0.01 sin(200πt − 1.85x) ⇒ f = 100 Hz, λ = 3.4 m, v = 340 m/s.
  • Exam (CET) : Medium changes, f? (1) same (2) changes (3) doubles (4) zero → answer 1) Trick : f belongs to the source.

Rule 3: Phase difference — Δφ = 2π Δx ÷ λ = 2π Δt ÷ T. Δx = λ ⇒ Δφ = 2π (same phase).

  • Easy : λ = 2 m, Δx = 0.5 m ⇒ Δφ = π/2.
  • Medium : Δx = λ/2 ⇒ Δφ = π (opposite phase).
  • Exam (CET) : Δx for Δφ = π? (1) λ/2 (2) λ (3) λ/4 (4) 2λ → answer 1)

Derivation : Wave equation y = A sin(ωt − kx)

  • Step 1 : Source at x = 0 : y₀ = A sin ωt.
  • Step 2 : At x the wave arrives a time x/v later.
  • Step 3 : y = A sin ω(t − x/v).
  • Step 4 : k = ω/v ⇒ y = A sin(ωt − kx).
Remember

v = fλ; ω = 2πf; k = 2π/λ; y = A sin(ωt − kx).

Common mistake / CET trap

f does not change with the medium; a wave does not carry matter.

Shortcut

Memory clue : "v = f lambda".

Spot the difference

TypeVibrationExample
Transverseperpendicularstring
Longitudinalparallelsound

Practice questions

Easy
  1. 1Relation among v, f and λ?
  2. 2Which kind of wave is sound?
  3. 3Formula for k?
Medium
  1. 4f = 256 Hz, v = 340 : λ?
  2. 5y = 0.02 sin(100t − 0.5x) : v?
  3. 6λ = 4 m, Δx = 1 m : Δφ?
Challenging
  1. 7Derive y = A sin(ωt − kx).
  2. 8When sound enters water, what happens to f and λ?
Show answers (try them yourself first)
  1. 1v = fλ
  2. 2Longitudinal
  3. 32π/λ
  4. 41.33 m
  5. 5200 m/s
  6. 6π/2
  7. 7Use the source delay x/v
  8. 8f unchanged; λ increases (v increases)

Speed of waves; Newton and Laplace

From daily life

You see lightning first and hear thunder later.

Simple definition

Speed of sound depends on the elasticity and density of the medium.

Speed of soundair 340 m/swater 1500steel 5000v = √(γP/ρ)v ∝ √T

In the figure : Solids > liquids > gases.

Rule 1: Newton and Laplace — Newton (isothermal) : v = √(P ÷ ρ) ≈ 280 m/s (wrong). Laplace (adiabatic) : v = √(γ P ÷ ρ) ≈ 332 m/s.

  • Easy : P = 1.013 × 10⁵, ρ = 1.29 ⇒ Newton 280 m/s.
  • Medium : γ = 1.4 ⇒ Laplace 332 m/s.
  • Exam (CET) : Sound travels by a(n) ... process? (1) adiabatic (2) isothermal (3) isobaric (4) isochoric → answer 1) Trick : 'fast = adiabatic'.

Rule 2: Temperature and factors — v ∝ √T (K); v_t ≈ 331 + 0.61 t m/s. v is unchanged by pressure; more humidity ⇒ larger v.

  • Easy : 27 °C ⇒ v = 331 + 0.61 × 27 = 347.5 m/s.
  • Medium : T 300 → 1200 K ⇒ v doubles.
  • Exam (CET) : P doubled (constant T) ⇒ v? (1) same (2) double (3) half (4) four times → answer 1) Trick : P/ρ is unchanged.

Rule 3: Other media — String : v = √(T ÷ μ); solid : v = √(Y ÷ ρ); liquid : v = √(K ÷ ρ).

  • Easy : T = 100 N, μ = 0.01 kg/m ⇒ v = 100 m/s.
  • Medium : Steel Y = 2 × 10¹¹, ρ = 7800 ⇒ v = 5064 m/s.
  • Exam (CET) : Largest v in? (1) solids (2) liquids (3) gases (4) same → answer 1)

Derivation : Laplace : including γ

  • Step 1 : Compressions are fast so heat cannot escape : adiabatic, PVᵞ = constant.
  • Step 2 : Adiabatic bulk modulus K = γ P.
  • Step 3 : v = √(K ÷ ρ).
  • Step 4 : v = √(γ P ÷ ρ); isothermal K = P ⇒ Newton.
Remember

v = √(γP/ρ); v ∝ √T; v = √(T/μ) in a string.

Common mistake / CET trap

Pressure does not change v; T is in kelvin.

Shortcut

Memory clue : "Hotter = faster".

Spot the difference

Mediumv (m/s)
Air 0 °C331
Air 20 °C343
Water≈ 1500
Steel≈ 5000

Practice questions

Easy
  1. 1Laplace's formula?
  2. 2How does v vary with T?
  3. 3Newton's value?
Medium
  1. 4v = 331 at 0 °C; v at 40 °C?
  2. 5T = 400 N, μ = 0.04 : v?
  3. 6T from 27 °C → 327 °C : ratio of v?
Challenging
  1. 7Why do Newton and Laplace differ?
  2. 8v in a gas with γ = 1.67?
Show answers (try them yourself first)
  1. 1√(γP/ρ)
  2. 2v ∝ √T
  3. 3280 m/s
  4. 4355.4 m/s
  5. 5100 m/s
  6. 6√2 ≈ 1.41
  7. 7The process is adiabatic, not isothermal
  8. 8362 m/s (same P, ρ)

Superposition, echo, reverberation and acoustics

From daily life

Shout in the hills and the sound returns; in a big hall it booms.

Simple definition

Superposition: when two waves meet, the resultant displacement is their sum.

Superpositionconstructivedestructive

In the figure : Left : constructive, right : destructive.

Rule 1: Superposition — y = y₁ + y₂. Constructive : path difference = nλ; destructive : (n + ½)λ. Amplitude A₁ + A₂ or |A₁ − A₂|.

  • Easy : λ = 0.68 m, path difference 1.36 m ⇒ n = 2 ⇒ constructive.
  • Medium : Path difference 1.7 m ⇒ 2.5 λ ⇒ destructive.
  • Exam (CET) : Amplitudes 3 and 5 (destructive) ⇒ ? (1) 2 (2) 8 (3) 15 (4) 0 → answer 1)

Rule 2: Echo — d = v t ÷ 2; to be heard clearly ≥ 0.1 s ⇒ minimum distance ≈ 17 m (340 m/s).

  • Easy : t = 2 s, v = 340 ⇒ d = 340 m.
  • Medium : t = 0.1 s ⇒ d = 17 m.
  • Exam (CET) : In an echo d = ? (1) vt/2 (2) vt (3) 2vt (4) v/t → answer 1) Trick : to and fro ⇒ half.

Rule 3: Reverberation and acoustics — Sabine : T = 0.161 V ÷ A (V in m³, A = absorption in m²); suitable T ≈ 1–2 s in halls; curtains/carpets absorb.

  • Easy : V = 2000, A = 160 ⇒ T = 2.01 s.
  • Medium : Larger A ⇒ smaller T.
  • Exam (CET) : Reverberation means? (1) persistence of sound (2) echo (3) destruction (4) absorption → answer 1)

Derivation : Echo distance

  • Step 1 : The sound goes to the wall and back : total path 2d.
  • Step 2 : 2d = v t.
  • Step 3 : d = v t ÷ 2.
  • Step 4 : For t = 0.1 s, v = 340 ⇒ d = 17 m.
Remember

y = y₁ + y₂; d = vt/2; T = 0.161V/A.

Common mistake / CET trap

Divide by 2 in an echo; destructive path difference is (n + ½)λ.

Shortcut

Memory clue : "To and fro = half".

Spot the difference

EventConditionResult
ConstructiveΔx = nλmaximum amplitude
DestructiveΔx = (n+½)λminimum amplitude
Echot ≥ 0.1 sclear repetition

Practice questions

Easy
  1. 1What is superposition?
  2. 2Formula for echo distance?
  3. 3What is reverberation?
Medium
  1. 4t = 1.5 s : d?
  2. 5λ = 1 m, Δx = 2.5 m : type?
  3. 6V = 1000, A = 80 : T?
Challenging
  1. 7Amplitudes 4 and 4, constructive : intensity compared to one?
  2. 8Why 17 m minimum?
Show answers (try them yourself first)
  1. 1y = y₁ + y₂
  2. 2vt/2
  3. 3Persistence of sound
  4. 4255 m
  5. 5Destructive
  6. 62.01 s
  7. 7Amplitude 8 ⇒ intensity 4 times
  8. 8Persistence 0.1 s × 340 ÷ 2

Pitch, timbre, ultrasound and sonar

From daily life

A sitar and a flute play the same note, yet we tell them apart.

Simple definition

Pitch is how shrill a sound is; it depends on frequency.

Sonard = v t ÷ 2v ≈ 1500 m/sseabed

In the figure : Depth from an ultrasonic pulse.

Live figure

Pitch, timbre, ultrasound and sonar

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Rule 1: Audible range — Audible 20 Hz – 20 kHz; infrasonic < 20 Hz (earthquakes, elephants); ultrasonic > 20 kHz (bats, sonography).

  • Easy : 20 kHz ⇒ λ = 340 ÷ 20000 = 17 mm.
  • Medium : 256 Hz ⇒ λ = 1.33 m.
  • Exam (CET) : Ultrasonic f? (1) > 20 kHz (2) < 20 Hz (3) 100 Hz (4) 1 kHz → answer 1)

Rule 2: Pitch and timbre — Pitch ∝ frequency; loudness ∝ amplitude; timbre = waveform (tells instruments apart). Octave : f doubles.

  • Easy : The octave above 256 Hz = 512 Hz.
  • Medium : A woman's voice has a higher pitch.
  • Exam (CET) : Distinguishing instruments? (1) timbre (2) pitch (3) loudness (4) speed → answer 1)

Rule 3: Sonar and uses — d = v t ÷ 2, v ≈ 1500 m/s in water. Uses : depth, fish, sonography, cleaning, testing.

  • Easy : t = 0.4 s ⇒ d = 1500 × 0.4 ÷ 2 = 300 m.
  • Medium : A bat senses obstacles by echoes.
  • Exam (CET) : Sonar principle? (1) ultrasonic echo (2) light (3) radio (4) magnet → answer 1)

Derivation : Depth by sonar

  • Step 1 : An ultrasonic pulse goes to the bottom and back.
  • Step 2 : Total path 2d = v t.
  • Step 3 : d = v t ÷ 2.
  • Step 4 : v = 1500, t = 0.4 ⇒ d = 300 m.
Remember

Pitch ∝ f; loudness ∝ amplitude; d = vt/2.

Common mistake / CET trap

Pitch ≠ loudness; divide by 2 in sonar.

Shortcut

Memory clue : "Pitch = f, loudness = A, timbre = shape".

Spot the difference

QualityDepends onExample
Pitchfrequencyshrill/deep
Loudnessamplitudeloud/soft
Timbrewaveforminstrument

Practice questions

Easy
  1. 1Audible range?
  2. 2What decides pitch?
  3. 3What is sonar?
Medium
  1. 4t = 0.2 s, v = 1500 : d?
  2. 530 kHz, v = 340 : λ?
  3. 6Octave of 320 Hz?
Challenging
  1. 7Why do two instruments differ on the same note?
  2. 8Why is ultrasound used in sonography?
Show answers (try them yourself first)
  1. 120 Hz – 20 kHz
  2. 2Frequency
  3. 3Distance from ultrasonic echoes
  4. 4150 m
  5. 511.3 mm
  6. 6640 Hz
  7. 7Different timbre (waveform)
  8. 8Short λ ⇒ clear detail, safe

Intensity, loudness and decibel

From daily life

A firecracker nearby hurts; a distant one does not.

Simple definition

Intensity is the sound energy passing per second through unit area.

Decibel scale0306090120whispertalktrafficpainβ = 10 log₁₀(I ÷ I₀)I₀ = 10⁻¹² W/m²+10 dB = intensity × 10+3 dB ≈ intensity × 2

In the figure : Decibel scale : every +10 = 10 times.

Live figure

Intensity, loudness and decibel

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Rule 1: Intensity — I = P ÷ A (W/m²); point source : I = P ÷ 4π r² ⇒ I ∝ 1/r²; also I ∝ (amplitude)².

  • Easy : P = 1 W, r = 10 m ⇒ I = 7.96 × 10⁻⁴ W/m².
  • Medium : r doubled ⇒ I ¼.
  • Exam (CET) : Amplitude doubled ⇒ I? (1) four times (2) double (3) half (4) same → answer 1)

Rule 2: Decibel — β = 10 log₁₀ (I ÷ I₀); I₀ = 10⁻¹² W/m² (hearing threshold = 0 dB). A log scale squeezes huge ratios.

  • Easy : I = 10⁻⁶ ⇒ β = 60 dB.
  • Medium : I = 7.96 × 10⁻⁴ ⇒ β = 89 dB.
  • Exam (CET) : I doubled ⇒ Δβ? (1) +3 dB (2) +6 dB (3) +10 dB (4) +2 dB → answer 1) Trick : 10 log 2 ≈ 3.

Rule 3: Distance and sources — β₂ − β₁ = 10 log₁₀ (I₂ ÷ I₁) = 20 log₁₀ (r₁ ÷ r₂). N equal sources : β + 10 log N.

  • Easy : r doubled ⇒ −6 dB.
  • Medium : Two 60 dB sources ⇒ 63 dB.
  • Exam (CET) : Intensity ratio of 90 dB and 60 dB? (1) 1000 (2) 30 (3) 90 (4) 100 → answer 1)

Derivation : dB change with distance

  • Step 1 : I ∝ 1/r² ⇒ I₂/I₁ = (r₁/r₂)².
  • Step 2 : Δβ = 10 log₁₀ (I₂/I₁).
  • Step 3 : = 10 log₁₀ (r₁/r₂)².
  • Step 4 : Δβ = 20 log₁₀ (r₁ ÷ r₂).
Remember

I = P/4πr²; β = 10 log(I/I₀); Δβ = 20 log(r₁/r₂).

Common mistake / CET trap

dB do not add directly; 60 + 60 ≠ 120.

Shortcut

Memory clue : "+10 dB = ten times; +3 dB = double".

Spot the difference

ChangeIntensity factordB change
I × 22+3
I × 1010+10
I × 100100+20
r × 2¼−6

Practice questions

Easy
  1. 1Formula for I?
  2. 2Formula for β?
  3. 3Value of I₀?
Medium
  1. 4I = 10⁻⁴ W/m² : β?
  2. 580 dB : I?
  3. 6r tripled : Δβ?
Challenging
  1. 7P = 4π W, r = 1 m : I and β?
  2. 8Ten 50 dB sources : β?
Show answers (try them yourself first)
  1. 1P/A
  2. 210 log (I/I₀)
  3. 310⁻¹² W/m²
  4. 480 dB
  5. 510⁻⁴ W/m²
  6. 6−9.5 dB
  7. 7I = 1 W/m², β = 120 dB
  8. 860 dB

Doppler effect: moving source

From daily life

An ambulance sounds shriller approaching and deeper receding.

Simple definition

The Doppler effect is the change in heard frequency due to relative motion.

Doppler : moving sourceλ' largerλ' shorterf' = f v ÷ (v ∓ v_s)approaching f' > f

In the figure : Waves bunch ahead and spread behind.

Live figure

Doppler effect: moving source

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Rule 1: Source approaching — f' = f v ÷ (v − v_s) (f' > f, λ' = (v − v_s) ÷ f).

  • Easy : f = 500, v_s = 20, v = 340 ⇒ f' = 531.2 Hz.
  • Medium : Whistle 400 Hz, 30 m/s ⇒ f' = 438.7 Hz.
  • Exam (CET) : v_s = v/10 ⇒ f'? (1) 10f/9 (2) 9f/10 (3) 11f/10 (4) f → answer 1)

Rule 2: Source receding — f' = f v ÷ (v + v_s) (f' < f).

  • Easy : 500 Hz, 20 m/s ⇒ f' = 472.2 Hz.
  • Medium : 400 Hz, 30 m/s ⇒ 367.6 Hz.
  • Exam (CET) : While receding f'? (1) lower (2) higher (3) same (4) zero → answer 1)

Rule 3: Shift and speed — Δf = f' − f; for small v_s, Δf ≈ ± f v_s ÷ v. Speed : v_s = v (f' − f)/f'.

  • Easy : f = 500, v_s = 20 ⇒ Δf ≈ 29 Hz (exact 31.25).
  • Medium : f = 500, f' = 531.25 ⇒ v_s = 340 × 31.25 ÷ 531.25 = 20 m/s.
  • Exam (CET) : Police radar principle? (1) Doppler (2) refraction (3) diffraction (4) polarisation → answer 1)

Derivation : f' = f v ÷ (v − v_s)

  • Step 1 : In one period T the source moves v_s T forward.
  • Step 2 : In front, λ' = vT − v_s T = (v − v_s) ÷ f.
  • Step 3 : f' = v ÷ λ'.
  • Step 4 : f' = f v ÷ (v − v_s).
Remember

Approaching : f v/(v − v_s); receding : f v/(v + v_s).

Common mistake / CET trap

Approaching = minus; receding = plus; v_s is relative to the medium.

Shortcut

Memory clue : "Close = minus, higher f".

Spot the difference

Sourcef'λ'
Approachingf v/(v − v_s) (higher)shorter
Recedingf v/(v + v_s) (lower)longer

Practice questions

Easy
  1. 1What is the Doppler effect?
  2. 2Why is f' higher on approach?
  3. 3Formula when receding?
Medium
  1. 4f = 600, v_s = 34 (approaching) : f'?
  2. 5f = 600, v_s = 34 (receding) : f'?
  3. 6v_s = v/5 : f'/f (approaching)?
Challenging
  1. 7f' = 1.1 f, approaching : v_s?
  2. 8Derive f'.
Show answers (try them yourself first)
  1. 1Apparent change in f due to relative motion
  2. 2Waves bunch up (shorter λ')
  3. 3f v/(v + v_s)
  4. 4667 Hz
  5. 5545 Hz
  6. 65/4
  7. 730.9 m/s
  8. 8Use λ' = (v − v_s)/f

Doppler effect: moving listener, both moving, reflection

From daily life

A speed gun measures a car's speed from the change in reflected frequency.

Simple definition

A moving listener meets waves at a different relative speed ⇒ f' changes.

Listener and source both moveSv_sLv_of' = f (v + v_o) ÷ (v − v_s)positive : from listener toward sourceapproach ⇒ f rises

In the figure : S → ← L : both approach.

Rule 1: Moving listener — Source at rest : f' = f (v ± v_o) ÷ v (listener toward : +, away : −).

  • Easy : f = 500, v_o = 20 (toward) ⇒ 529.4 Hz.
  • Medium : away ⇒ 470.6 Hz.
  • Exam (CET) : Listener toward source ⇒ f'? (1) higher (2) lower (3) same (4) zero → answer 1)

Rule 2: Both move — f' = f (v + v_o) ÷ (v − v_s); approaching each other : upper signs; same direction : f (v − v_o) ÷ (v − v_s) (source behind).

  • Easy : f = 500, v_s = v_o = 20 (toward each other) ⇒ 562.5 Hz.
  • Medium : Source 20 behind, listener 10 ahead (same direction) ⇒ 515.6 Hz.
  • Exam (CET) : Equal speeds, same direction ⇒ f'? (1) f (2) 2f (3) f/2 (4) 0 → answer 1) Trick : no relative motion.

Rule 3: Reflection — A source moving toward a wall : the wall is first a listener, then a source ⇒ f'' = f (v + v_s) ÷ (v − v_s).

  • Easy : Car 30 m/s, f = 1000 ⇒ echo 1193.5 Hz.
  • Medium : Beat = f'' − f = 193.5 Hz.
  • Exam (CET) : Reflected f from a wall? (1) f(v+v_s)/(v−v_s) (2) f (3) f/2 (4) 2f → answer 1)

Derivation : f' = f (v + v_o) ÷ (v − v_s)

  • Step 1 : Source motion gives wavelength λ' = (v − v_s) ÷ f.
  • Step 2 : The waves' speed relative to the listener = v + v_o.
  • Step 3 : f' = (relative speed) ÷ λ'.
  • Step 4 : f' = f (v + v_o) ÷ (v − v_s).
Remember

f' = f (v + v_o)/(v − v_s); wall : f (v + v_s)/(v − v_s).

Common mistake / CET trap

Listener sign : toward +; source sign : toward −; v relative to the medium.

Shortcut

Memory clue : "Listener toward + on top, source toward − below".

Spot the difference

CaseFormula
Moving listenerf (v ± v_o)/v
Moving sourcef v/(v ∓ v_s)
Bothf (v ± v_o)/(v ∓ v_s)
Wallf (v + v_s)/(v − v_s)

Practice questions

Easy
  1. 1Formula for a moving listener?
  2. 2Formula for both moving?
  3. 3Echo frequency from a wall?
Medium
  1. 4f = 400, v_o = 17 (away) : f'?
  2. 5f = 800, v_s = v_o = 34 (approaching) : f'?
  3. 6Car 20 m/s, f = 600 : echo?
Challenging
  1. 7Source 30, listener 10, same direction (source behind), f = 600 : f'?
  2. 8How are beats formed (wall)?
Show answers (try them yourself first)
  1. 1f (v ± v_o)/v
  2. 2f (v + v_o)/(v − v_s)
  3. 3f (v + v_s)/(v − v_s)
  4. 4380 Hz
  5. 5978 Hz
  6. 6675.0 Hz
  7. 7638.7 Hz
  8. 8From mixing f and f''

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Wavev = fλ; y = A sin(ωt − kx)m/s; L T⁻¹
Speed (air)v = √(γP/ρ) ∝ √Tm/s
Stringv = √(T/μ)m/s
Echod = vt/2m
ReverberationT = 0.161V/As
IntensityI = P/4πr²W/m²; M T⁻³
Decibelβ = 10 log(I/I₀)dB
Dopplerf' = f (v + v_o)/(v − v_s)Hz; T⁻¹

🎯 Most important points for the exam

  • Sound is longitudinal; v = fλ; f does not depend on the medium.
  • v ∝ √T; pressure does not change v; Laplace includes γ.
  • Constructive nλ; destructive (n + ½)λ.
  • Audible 20 Hz – 20 kHz; sonar d = vt/2.
  • I ∝ 1/r²; +10 dB = 10 times.
  • Approach raises f.
  • In the formula listener +, source − (when approaching).
Study mantra

v = fλ → v = √(γP/ρ) → d = vt/2 → β = 10 log(I/I₀) → Doppler.