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Waves Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Physics Waves exercise solutions — wave properties, transverse longitudinal, sound waves, Doppler effect, resonance, standing waves, beats.

Updated 15 June 202620 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

1. Solve first, check later — Attempt every exercise question on your own before reading the solutions below. 2. Follow the sequence — Work through sections I to VI in order; later sections build on earlier ones. 3. Understand the method — Focus on the reasoning and formula selection, not just the final answer. 4. Practise numericals — Re-work the worked examples with different values until you can set them up without hints. 5. Review mistakes — Read the "Common Mistakes to Avoid" section before every exam.

Key Concepts (Quick Recap)

Wave Basics

QuantitySymbolFormulaSI Unit
Wave speedvv = fλm/s
Frequencyff = 1/THz
Time periodTT = 1/fs
Wavelengthλλ = v/f = vTm
Angular frequencyωω = 2πf = 2π/Trad/s
Wave numberkk = 2π/λrad/m
Wave equationyy = A sin(kx − ωt)m

Types of Waves

TypeMotion of particlesExamplesCan polarise?
TransversePerpendicular to propagationLight, string waves, water surface wavesYes
LongitudinalParallel to propagationSound waves, spring compression wavesNo

Speed of Sound

MediumFormulaNotes
Generalv = √(E/ρ)E = elastic modulus, ρ = density
Solid (longitudinal)v = √(Y/ρ)Y = Young's modulus
Solid (transverse)v = √(η/ρ)η = shear modulus
Liquidv = √(B/ρ)B = bulk modulus
Gasv = √(γP/ρ) = √(γRT/M)γ = Cₚ/Cᵥ
Newton's formula (air)v = √(P/ρ)Incorrect (gives ~280 m/s)
Laplace correctionv = √(γP/ρ)Correct (gives ~331 m/s at 0°C)
Temperature dependencev ∝ √Tv_t = v₀√(1 + t/273) ≈ v₀ + 0.61t

Standing Waves

PropertyString (fixed both ends)Pipe (open both ends)Pipe (closed one end)
Fundamentalf₁ = v/(2L)f₁ = v/(2L)f₁ = v/(4L)
Overtonesfₙ = nv/(2L), n = 1,2,3…fₙ = nv/(2L), n = 1,2,3…fₙ = nv/(4L), n = 1,3,5…
HarmonicsAll harmonics presentAll harmonics presentOnly odd harmonics
Wavelengthλₙ = 2L/nλₙ = 2L/nλₙ = 4L/n

Doppler Effect

CaseFormula
Observer moving toward sourcef' = f(v + v₀)/v
Observer moving away from sourcef' = f(v − v₀)/v
Source moving toward observerf' = fv/(v − vₛ)
Source moving away from observerf' = fv/(v + vₛ)
Both moving (general)f' = f(v ± v₀)/(v ∓ vₛ)

Beats and Intensity

QuantityFormula
Beat frequencyf_beat = \f₁ − f₂\
IntensityI = P/A (power per unit area)
Intensity levelβ = 10 log₁₀(I/I₀) dB
Reference intensityI₀ = 10⁻¹² W/m² (threshold of hearing)

I. Choose the Best Answer (MCQs)

1. The speed of sound in air at 0°C is approximately:

  • (a) 280 m/s
  • (b) 331 m/s
  • (c) 340 m/s
  • (d) 380 m/s
Answer(b) 331 m/s
ExplanationThe speed of sound in air at 0°C is approximately 331 m/s. At room temperature (20°C), it is approximately 343 m/s. The speed increases with temperature approximately as v = 331 + 0.61t m/s, where t is in °C.

2. In a stationary wave, the distance between two consecutive nodes is:

  • (a) λ
  • (b) λ/2
  • (c) λ/4
  • (d) 2λ
Answer(b) λ/2
ExplanationIn a stationary wave, nodes occur at intervals of λ/2. The distance between any two consecutive nodes (or any two consecutive antinodes) is λ/2. The distance between a node and the next antinode is λ/4.

3. The Doppler effect is observed for:

  • (a) Sound waves only
  • (b) Light waves only
  • (c) Both sound and light waves
  • (d) Neither sound nor light waves
Answer(c) Both sound and light waves
ExplanationThe Doppler effect occurs for all wave types, including sound, light, and electromagnetic waves. For sound, the medium matters and the formula depends on whether the source or observer is moving. For light, only the relative velocity between source and observer matters (relativistic Doppler effect).

4. A pipe closed at one end has a fundamental frequency f. If the pipe is opened at both ends, the fundamental frequency becomes:

  • (a) f
  • (b) 2f
  • (c) f/2
  • (d) 4f
Answer(b) 2f
ExplanationClosed pipe: f_closed = v/(4L). Open pipe: f_open = v/(2L). Therefore f_open/f_closed = (v/2L)/(v/4L) = 4L/2L = 2. The fundamental frequency of an open pipe is twice that of a closed pipe of the same length.

5. Two sound waves of frequencies 256 Hz and 260 Hz produce beats. The beat frequency is:

  • (a) 258 Hz
  • (b) 4 Hz
  • (c) 516 Hz
  • (d) 2 Hz
Answer(b) 4 Hz
Explanation: Beat frequency =f₁ − f₂=260 − 256= 4 Hz. The beat frequency is simply the difference between the two frequencies. Four loudness maxima are heard per second.

6. Newton's formula for the speed of sound in air gave an incorrect value because:

  • (a) He used the wrong density of air
  • (b) He assumed isothermal conditions instead of adiabatic
  • (c) He did not account for humidity
  • (d) He used the wrong value of atmospheric pressure
Answer(b) He assumed isothermal conditions instead of adiabatic
ExplanationNewton assumed that the propagation of sound is an isothermal process (PV = constant), giving v = √(P/ρ) ≈ 280 m/s, which disagreed with experiment (~331 m/s). Laplace corrected this by showing that sound propagation is adiabatic (PVᵞ = constant), giving v = √(γP/ρ), which matches experimental values.

7. A transverse wave is described by y = 0.05 sin(2πx − 100πt). The wavelength of the wave is:

  • (a) 0.05 m
  • (b) 1 m
  • (c) 2π m
  • (d) 0.01 m
Answer(b) 1 m
ExplanationComparing with y = A sin(kx − ωt), we get k = 2π. Since k = 2π/λ, λ = 2π/k = 2π/(2π) = 1 m. Also, ω = 100π, so f = ω/(2π) = 50 Hz and v = fλ = 50 × 1 = 50 m/s.

II. Fill in the Blanks

1. The wave equation relating speed, frequency, and wavelength is v = fλ. 2. In a transverse wave, particles vibrate perpendicular to the direction of wave propagation. 3. The speed of sound in air increases by approximately 0.61 m/s for every 1°C rise in temperature. 4. In a closed pipe, only odd harmonics are present in the vibration pattern.

6. Newton's formula for the speed of sound is v = √(P/ρ), which was corrected by Laplace to v = √(γP/ρ). 7. The distance between a node and the nearest antinode in a standing wave is λ/4. 8. When a sound source moves away from a stationary observer, the observed frequency decreases (Doppler effect). 9. The intensity level of sound is measured in decibels (dB). 10. Sound waves are longitudinal waves that require a material medium for propagation. 11. In an open pipe of length L, the fundamental wavelength is 2L. 12. The phenomenon of a sound persisting after the source has stopped due to multiple reflections is called reverberation.

5. The beat frequency equals the difference of the two frequencies: f_beat =f₁ − f₂.

III. True or False (with corrections)

1.
AnswerFalse — Sound waves can travel through vacuum. Correction: Sound waves are mechanical (longitudinal) waves that require a material medium for propagation. They cannot travel through a vacuum. Light and electromagnetic waves can travel through vacuum, but sound cannot.
2.
AnswerTrue — The speed of sound in a solid is greater than the speed of sound in a gas.
3.
AnswerFalse — In a closed pipe, both even and odd harmonics are present. Correction: In a closed pipe (closed at one end), only odd harmonics are present: f₁, 3f₁, 5f₁, 7f₁, ... Even harmonics (2f₁, 4f₁, etc.) are absent. In an open pipe, all harmonics are present.
4.
AnswerFalse — The Doppler effect formula for sound is the same whether the source is moving or the observer is moving. Correction: The formulas are different. When the observer moves toward the source: f' = f(v + v₀)/v. When the source moves toward the observer: f' = fv/(v − vₛ). The results differ when v₀ = vₛ, especially at high speeds relative to the speed of sound.
5.
AnswerTrue — When two waves interfere constructively, the resultant amplitude equals the sum of the individual amplitudes.

IV. Match the Following

Column AColumn B
1. Transverse wave(d) Particles vibrate perpendicular to wave direction
2. Longitudinal wave(f) Particles vibrate parallel to wave direction
3. Standing wave(a) Superposition of two identical waves travelling in opposite directions
4. Doppler effect(e) Apparent change in frequency due to relative motion
5. Beats(b) Periodic loudness variation from two close frequencies
6. Resonance(c) Maximum amplitude when driving frequency equals natural frequency

V. Answer Briefly (2-Mark Questions)

1. Distinguish between transverse and longitudinal waves with examples.
Answer

In transverse waves, particles vibrate perpendicular to the direction of wave propagation (e.g., waves on a string, electromagnetic waves, water surface waves). In longitudinal waves, particles vibrate parallel to the direction of wave propagation (e.g., sound waves in air, compression waves in a spring). Transverse waves exhibit polarisation; longitudinal waves do not. Transverse waves can propagate in solids and on surfaces; longitudinal waves can propagate in solids, liquids, and gases.

2. What is Laplace's correction to Newton's formula for the speed of sound?
Answer

Newton assumed isothermal conditions (PV = constant) and derived v = √(P/ρ), giving v ≈ 280 m/s at 0°C, which disagreed with experiment (~331 m/s). Laplace argued that sound propagation involves rapid compressions and rarefactions with no time for heat exchange — an adiabatic process. Replacing P with γP (since PVᵞ = constant gives the adiabatic bulk modulus as γP), the corrected formula is v = √(γP/ρ). With γ = 1.4 for air, this gives v ≈ 331 m/s, matching experiment.

3. Define beats and state the formula for beat frequency.
Answer
Beats are the periodic variations in loudness (intensity) produced when two sound waves of slightly different frequencies interfere at the same point. The resultant amplitude oscillates at a rate equal to the difference of the two frequencies. Beat frequency f_beat =f₁ − f₂. Beats are used to tune musical instruments — when the beat frequency becomes zero, the two frequencies are identical.
4. What is resonance? Give an example.
Answer

Resonance is the phenomenon where a system oscillates with maximum amplitude when the frequency of the external driving force equals the natural frequency of the system. Example: When a tuning fork of frequency 256 Hz is placed near an open pipe of appropriate length, the air column vibrates with maximum amplitude and a loud sound is heard — this is acoustic resonance. Another example: A swing pushed at its natural frequency goes higher with each push.

VI. Answer in Detail (5-Mark Questions)

1. Derive the expression for the speed of sound in a gas (Laplace correction) and solve numericals.

Newton assumed isothermal propagation of sound (slow compressions and rarefactions at constant temperature):

Isothermal bulk modulus: B_T = P Speed of sound: v = √(B_T/ρ) = √(P/ρ) At NTP: P = 1.013 × 10⁵ Pa, ρ = 1.293 kg/m³ v_Newton = √(1.013 × 10⁵/1.293) = √(78344) ≈ 280 m/s But the experimental value is ~331 m/s — a significant discrepancy.

Laplace's correction: Sound propagation involves rapid compressions and rarefactions. There is no time for heat exchange, so the process is adiabatic (PVᵞ = constant).

Adiabatic bulk modulus: B_S = γP (where γ = Cₚ/Cᵥ) Speed of sound: v = √(γP/ρ) At NTP: γ = 1.4 (for air), P = 1.013 × 10⁵ Pa, ρ = 1.293 kg/m³ v_Laplace = √(1.4 × 1.013 × 10⁵/1.293) = √(109681) ≈ 331 m/s This matches the experimental value.

Effect of temperature:

v = √(γP/ρ) = √(γRT/M) Since v ∝ √T (where T is in Kelvin): v₁/v₂ = √(T₁/T₂) In terms of Celsius: v_t = v₀√(1 + t/273) Approximately: v_t ≈ v₀ + 0.61t m/s

Worked Example 1: Calculate the speed of sound in air at 30°C. (Speed at 0°C = 331 m/s)

Given: v₀ = 331 m/s, t = 30°C Method 1 (approximate): v_t ≈ v₀ + 0.61t = 331 + 0.61 × 30 = 331 + 18.3 = 349.3 m/s Method 2 (exact): T₀ = 273 K, T = 273 + 30 = 303 K v = v₀ × √(T/T₀) = 331 × √(303/273) = 331 × √(1.1099) = 331 × 1.0535 = 348.7 m/s Answer: Speed of sound at 30°C ≈ 349 m/s

Worked Example 2: Find the speed of sound in helium gas at NTP. (γ = 5/3, M = 4 × 10⁻³ kg/mol, R = 8.314 J/(mol·K), T = 273 K)

Given: γ = 5/3 = 1.667, M = 4 × 10⁻³ kg/mol, R = 8.314 J/(mol·K), T = 273 K Step 1: Apply the formula v = √(γRT/M) Step 2: Substitute values v = √(1.667 × 8.314 × 273 / 4 × 10⁻³) = √(3781.5 / 0.004) = √(945375) = 972.3 m/s Answer: Speed of sound in helium ≈ 972 m/s (This is much higher than in air because helium is lighter — v ∝ 1/√M)

2. Derive expressions for the fundamental frequency and overtones of (a) an open pipe and (b) a closed pipe. Solve numericals.

(a) Open pipe (open at both ends):

For a pipe open at both ends, both ends are antinodes.

Fundamental mode (first harmonic): Length L = λ₁/2 λ₁ = 2L f₁ = v/λ₁ = v/(2L) Second harmonic (first overtone): Length L = λ₂ λ₂ = L f₂ = v/λ₂ = v/L = 2f₁ Third harmonic (second overtone): Length L = 3λ₃/2 λ₃ = 2L/3 f₃ = v/λ₃ = 3v/(2L) = 3f₁ General: fₙ = nv/(2L) = nf₁, where n = 1, 2, 3, ... An open pipe has ALL harmonics: f₁, 2f₁, 3f₁, 4f₁, ...

(b) Closed pipe (closed at one end, open at the other):

For a pipe closed at one end, the closed end is a node and the open end is an antinode.

Fundamental mode: Length L = λ₁/4 λ₁ = 4L f₁ = v/λ₁ = v/(4L) First overtone (third harmonic): Length L = 3λ₃/4 λ₃ = 4L/3 f₃ = v/λ₃ = 3v/(4L) = 3f₁ Second overtone (fifth harmonic): Length L = 5λ₅/4 λ₅ = 4L/5 f₅ = v/λ₅ = 5v/(4L) = 5f₁ General: fₙ = nv/(4L), where n = 1, 3, 5, ... (odd only) A closed pipe has only ODD harmonics: f₁, 3f₁, 5f₁, 7f₁, ... Even harmonics (2f₁, 4f₁, ...) are absent.

Worked Example 1: An open pipe of length 0.5 m resonates at its fundamental frequency. Find the fundamental frequency and the first two overtone frequencies. (Speed of sound = 340 m/s)

Given: L = 0.5 m, v = 340 m/s, open pipe Step 1: Fundamental frequency f₁ = v/(2L) = 340/(2 × 0.5) = 340/1 = 340 Hz Step 2: First overtone (second harmonic) f₂ = 2f₁ = 2 × 340 = 680 Hz Step 3: Second overtone (third harmonic) f₃ = 3f₁ = 3 × 340 = 1020 Hz Answer: f₁ = 340 Hz, f₂ = 680 Hz, f₃ = 1020 Hz

Worked Example 2: A closed pipe of length 0.85 m has a fundamental frequency of 100 Hz. Find the speed of sound and the frequency of the next overtone.

Given: L = 0.85 m, f₁ = 100 Hz, closed pipe Step 1: Find speed of sound f₁ = v/(4L) v = f₁ × 4L = 100 × 4 × 0.85 = 340 m/s Step 2: Find next overtone (third harmonic, n = 3) f₃ = 3f₁ = 3 × 100 = 300 Hz (Note: The next overtone after fundamental in a closed pipe is the 3rd harmonic, not the 2nd, because even harmonics are absent) Answer: v = 340 m/s, next overtone = 300 Hz

3. State and derive the Doppler effect formula for sound and solve numericals.

The Doppler effect is the apparent change in frequency of a wave when there is relative motion between the source and the observer.

Case 1: Source stationary, observer moving toward source:

When the observer moves toward the source with speed v₀, the observer encounters more wavefronts per unit time. f' = f × (v + v₀)/v When the observer moves away: f' = f × (v − v₀)/v

Case 2: Observer stationary, source moving toward observer:

When the source moves toward the observer with speed vₛ, the wavefronts are compressed (shorter wavelength). The effective wavelength: λ' = (v − vₛ)/f The frequency: f' = v/λ' = v/((v − vₛ)/f) = fv/(v − vₛ) When the source moves away: f' = fv/(v + vₛ)

General formula (both moving):

f' = f(v ± v₀)/(v ∓ vₛ) Upper signs (+ in numerator, − in denominator): approach Lower signs (− in numerator, + in denominator): recede

Worked Example 1: A train whistle of frequency 600 Hz is heard by a stationary observer as the train approaches at 30 m/s. What frequency does the observer hear? (Speed of sound = 340 m/s)

Given: f = 600 Hz, vₛ = 30 m/s, v = 340 m/s, v₀ = 0 (observer stationary) Source moving toward observer Step 1: Apply Doppler formula f' = f × v/(v − vₛ) Step 2: Substitute values f' = 600 × 340/(340 − 30) = 600 × 340/310 = 600 × 1.0968 = 658.1 Hz Answer: Observed frequency ≈ 658 Hz (higher than original 600 Hz)

Worked Example 2: An ambulance with a siren of frequency 800 Hz approaches and then passes a stationary observer at 25 m/s. Find the apparent frequencies (a) while approaching, and (b) while receding. (v = 340 m/s)

Given: f = 800 Hz, vₛ = 25 m/s, v = 340 m/s, v₀ = 0 (a) Approaching: f₁ = f × v/(v − vₛ) = 800 × 340/(340 − 25) = 800 × 340/315 = 800 × 1.0794 = 863.5 Hz ≈ 864 Hz (b) Receding: f₂ = f × v/(v + vₛ) = 800 × 340/(340 + 25) = 800 × 340/365 = 800 × 0.9315 = 745.2 Hz ≈ 745 Hz Apparent frequency drop: Δf = f₁ − f₂ = 864 − 745 = 119 Hz Answer: Approaching ≈ 864 Hz, Receding ≈ 745 Hz

Worked Example 3: Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. Find the beat frequency. How many beats are heard in 5 seconds?

Given: f₁ = 256 Hz, f₂ = 260 Hz Step 1: Calculate beat frequency f_beat = |f₂ − f₁| = |260 − 256| = 4 Hz Step 2: Number of beats in 5 seconds Number of beats = f_beat × time = 4 × 5 = 20 beats Answer: Beat frequency = 4 Hz, 20 beats heard in 5 seconds

Common Mistakes to Avoid

1. Confusing transverse and longitudinal waves — In transverse waves, particles move perpendicular to wave direction (e.g., light, string vibrations). In longitudinal waves, particles move parallel to wave direction (e.g., sound in air). Sound cannot be a transverse wave in air because air cannot sustain shear forces.

2. Using v = fλ with inconsistent units — Ensure that v is in m/s, f in Hz, and λ in metres. A common error is mixing cm/s with m, or kHz with km. Always convert to SI units before substituting.

3. Mixing up open and closed pipe harmonics — Open pipe: all harmonics present (f₁, 2f₁, 3f₁, ...). Closed pipe: only odd harmonics present (f₁, 3f₁, 5f₁, ...). The fundamental of a closed pipe is v/(4L), half that of an open pipe of the same length.

4. Wrong Doppler formula sign — The Doppler formula f' = f(v ± v₀)/(v ∓ vₛ) uses upper signs for approach and lower signs for recede. A mnemonic: "approaching increases frequency" — so use the sign that makes f' > f when source and observer approach each other.

5. Forgetting Laplace correction — Newton's formula v = √(P/ρ) gives an incorrect speed of sound (~280 m/s). Always use Laplace's corrected formula v = √(γP/ρ) for adiabatic propagation. For air, γ = 1.4, which gives ~331 m/s at 0°C.

6. Confusing node/antinode positions — In a standing wave on a string fixed at both ends: both ends are nodes. In an open pipe: both ends are antinodes. In a closed pipe: closed end is a node, open end is an antinode. Getting these wrong leads to incorrect wavelength calculations.

7. Applying Doppler effect incorrectly for light — The formulas given above are for sound waves in a medium. For light, the relativistic Doppler formula must be used, and only relative velocity matters (the medium-based formulas do not apply).

8. Confusing beat frequency with resultant frequency — Beat frequency isf₁ − f₂(how many times the loudness varies per second). The resultant sound has an average frequency of (f₁ + f₂)/2 with an amplitude modulated at the beat frequency. These are different quantities.

Exam Checklist

  • [ ] I can write and use the wave equation v = fλ and its variations
  • [ ] I can distinguish between transverse and longitudinal waves with examples
  • [ ] I can derive and apply Laplace's corrected formula for the speed of sound: v = √(γP/ρ)
  • [ ] I can calculate the speed of sound at different temperatures using v = v₀ + 0.61t
  • [ ] I can find fundamental and overtone frequencies for open and closed pipes
  • [ ] I can explain why only odd harmonics are present in a closed pipe
  • [ ] I can apply the Doppler effect formula for various source/observer motion combinations
  • [ ] I can calculate beat frequency and explain how beats are used for tuning
  • [ ] I can define and identify nodes, antinodes, and wavelengths in standing waves
  • [ ] I can solve numericals involving wave speed, frequency, wavelength, Doppler effect, and resonance

Frequently Asked Questions

v = fλ, where v is the wave speed (m/s), f is the frequency (Hz), and λ is the wavelength (m). This is the fundamental wave equation applicable to all types of waves.