Waves — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Waves" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Waves" — 8 important questions with detailed answers for CBSE board exam preparation.
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 11 Physics — Waves. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Waves — All 8 Questions Solved
Q1. A tuning fork vibrates at frequency 256 Hz. The speed of sound in air is 340 m/s. Calculate (a) the wavelength, (b) the period of the wave.
Step 1: Wavelength using wave equation:
v = fλ
λ = v/f
where v = speed of sound, f = frequency
λ = 340/256 = 1.328 m ≈ 1.33 m
Step 2: Period:
T = 1/f = 1/256 = 0.00391 s ≈ 3.91 × 10⁻³ s = 3.91 ms
Final Answer: (a) Wavelength = 1.33 m; (b) Period = 3.91 ms
Q2. Two coherent sound sources emit sound at 500 Hz. The speed of sound is 330 m/s. If the sources are 0.66 m apart, calculate the path difference for constructive and destructive interference.
Step 1: Calculate wavelength:
λ = v/f = 330/500 = 0.66 m
Step 2: For constructive interference:
Path difference = nλ where n = 0, 1, 2, 3, ...
For n = 0: Path difference = 0
For n = 1: Path difference = λ = 0.66 m
For n = 2: Path difference = 2λ = 1.32 m
Step 3: For destructive interference:
Path difference = (n + 1/2)λ where n = 0, 1, 2, 3, ...
For n = 0: Path difference = λ/2 = 0.33 m
For n = 1: Path difference = 3λ/2 = 0.99 m
For n = 2: Path difference = 5λ/2 = 1.65 m
Final Answer: For constructive interference: 0, 0.66 m, 1.32 m, ...; For destructive interference: 0.33 m, 0.99 m, 1.65 m, ...
Q3. A string of length 0.5 m and mass 10 g is fixed at both ends. When vibrating in its fundamental mode, the frequency is 100 Hz. Calculate the tension in the string.
Step 1: For a string fixed at both ends, fundamental frequency:
f₁ = (1/2L)√(T/μ)
where T = tension, μ = linear mass density = mass/length, L = length
Step 2: Rearrange for tension:
T = 4L²f₁²μ
Step 3: Calculate linear mass density:
m = 10 g = 0.01 kg
L = 0.5 m
μ = m/L = 0.01/0.5 = 0.02 kg/m
Step 4: Substitute values:
T = 4 × (0.5)² × (100)² × 0.02
T = 4 × 0.25 × 10,000 × 0.02
T = 4 × 0.25 × 200
T = 200 N
Final Answer: Tension in string = 200 N
Q4. A transverse wave on a string is represented by y = 0.02 sin(10x - 4πt), where x and y are in meters and t is in seconds. Calculate (a) amplitude, (b) wavelength, (c) frequency, (d) velocity of the wave.
Step 1: Standard wave equation: y = A sin(kx - ωt)
Given: y = 0.02 sin(10x - 4πt)
Amplitude A = 0.02 m = 2 cm
Step 2: Wave number:
k = 10 rad/m
Wavelength λ = 2π/k = 2π/10 = 0.628 m
Step 3: Angular frequency:
ω = 4π rad/s
Frequency f = ω/(2π) = 4π/(2π) = 2 Hz
Step 4: Wave velocity:
v = ω/k = 4π/10 = 1.256 m/s
Alternatively: v = fλ = 2 × 0.628 = 1.256 m/s
Final Answer: (a) Amplitude = 0.02 m or 2 cm; (b) Wavelength = 0.628 m; (c) Frequency = 2 Hz; (d) Wave velocity = 1.26 m/s
Q5. The intensity of sound is defined as power per unit area. If a sound has intensity 10⁻¹² W/m² (threshold of hearing), and frequency 1000 Hz, calculate the amplitude of oscillation of air particles. Speed of sound = 330 m/s, density of air = 1.2 kg/m³.
Step 1: Intensity relation with amplitude:
I = (1/2) × ρ × v × ω² × A²
where ρ = density, v = wave velocity, ω = angular frequency, A = amplitude
Step 2: Calculate angular frequency:
f = 1000 Hz
ω = 2πf = 2π × 1000 = 6283 rad/s
Step 3: Rearrange for amplitude:
A = √(2I/(ρ × v × ω²))
Step 4: Substitute values:
I = 10⁻¹² W/m²
ρ = 1.2 kg/m³
v = 330 m/s
ω = 6283 rad/s
A = √(2 × 10⁻¹² / (1.2 × 330 × 6283²))
A = √(2 × 10⁻¹² / (1.2 × 330 × 39,476,489))
A = √(2 × 10⁻¹² / 15.6 × 10⁹)
A = √(1.28 × 10⁻²²)
A = 1.13 × 10⁻¹¹ m ≈ 1.1 × 10⁻¹¹ m
Final Answer: Amplitude = 1.1 × 10⁻¹¹ m (approx)
Q6. A siren producing sound at 800 Hz is approaching a stationary observer at 20 m/s. The speed of sound is 330 m/s. Calculate the frequency heard by the observer.
Step 1: Doppler effect when source approaches stationary observer:
f' = f × (v/(v - v_s))
where f = frequency of source, v = speed of sound, v_s = velocity of source (toward observer)
Step 2: Substitute values:
f = 800 Hz
v = 330 m/s
v_s = 20 m/s
f' = 800 × (330/(330 - 20))
f' = 800 × (330/310)
f' = 800 × 1.0645
f' = 851.6 Hz ≈ 852 Hz
Final Answer: Frequency heard by observer = 852 Hz
Q7. In Youngs double slit experiment, the distance between slits is 1 mm, distance to screen is 1 m, and wavelength of light is 500 nm. Calculate the distance of 3rd bright fringe from the central maximum.
Step 1: In Young's double slit experiment, distance of nth bright fringe:
y_n = (n × λ × D)/d
where n = fringe order (1, 2, 3, ...), λ = wavelength, D = distance to screen, d = slit separation
Step 2: Substitute values:
n = 3 (for 3rd bright fringe)
λ = 500 nm = 500 × 10⁻⁹ m
D = 1 m
d = 1 mm = 1 × 10⁻³ m
y₃ = (3 × 500 × 10⁻⁹ × 1) / (1 × 10⁻³)
y₃ = (1500 × 10⁻⁹) / (10⁻³)
y₃ = 1500 × 10⁻⁶ m
y₃ = 1.5 × 10⁻³ m = 1.5 mm
Final Answer: Distance of 3rd bright fringe = 1.5 mm from central maximum
Q8. A sound wave travels through water and then enters air. The frequency is 1000 Hz. Speed of sound in water = 1500 m/s, speed in air = 330 m/s. Calculate the ratio of wavelengths in water and air.
Step 1: Wavelength in different media:
λ = v/f
Frequency remains constant when wave changes medium.
Step 2: Wavelength in water:
λ_w = v_w/f = 1500/1000 = 1.5 m
Step 3: Wavelength in air:
λ_a = v_a/f = 330/1000 = 0.33 m
Step 4: Ratio:
λ_w/λ_a = 1.5/0.33 = 4.55
Alternatively:
λ_w/λ_a = (v_w/f)/(v_a/f) = v_w/v_a = 1500/330 = 4.55
Final Answer: Ratio of wavelengths λ_w : λ_a = 1500 : 330 = 50 : 11 ≈ 4.55
Waves — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the SI unit of wave number (k)?
- A. m
- B. s
- C. m⁻¹
- D. Hz
Answer: C. m⁻¹ — Wave number (k) is defined as 2π/λ, where λ is the wavelength in meters. Therefore, its SI unit is m⁻¹.
MCQ 2. Which of the following types of waves can propagate through a vacuum?
- A. Sound waves
- B. Water waves
- C. Light waves
- D. Seismic S-waves
Answer: C. Light waves — Light waves are electromagnetic waves, which do not require a medium to propagate and can travel through a vacuum. All other options are mechanical waves.
MCQ 3. If the frequency of a wave is doubled, what happens to its wavelength in a given medium, assuming wave speed remains constant?
- A. Halved
- B. Doubled
- C. Remains the same
- D. Quadrupled
Answer: A. Halved — The wave speed (v) is given by v = fλ. If v is constant and f is doubled, then λ must be halved to keep the product constant.
MCQ 4. A progressive wave is represented by y(x,t) = 0.05 sin(20πt - 4πx), where x and y are in meters and t is in seconds. What is the wavelength of the wave?
- A. 0.5 m
- B. 0.25 m
- C. 1 m
- D. 2 m
Answer: A. 0.5 m — Comparing with y(x,t) = A sin(ωt - kx), we have k = 4π. Since k = 2π/λ, then λ = 2π/k = 2π/(4π) = 0.5 m.
MCQ 5. A string of mass per unit length 0.1 kg/m is under a tension of 10 N. What is the speed of a transverse wave on this string?
- A. 1 m/s
- B. 10 m/s
- C. 100 m/s
- D. 0.1 m/s
Answer: B. 10 m/s — The speed of a transverse wave on a string is given by v = √(T/μ), where T is tension and μ is mass per unit length. So, v = √(10 N / 0.1 kg/m) = √(100) = 10 m/s.
MCQ 6. The phenomenon of interference of waves is based on which principle?
- A. Huygens' principle
- B. Doppler effect
- C. Principle of superposition
- D. Bernoulli's principle
Answer: C. Principle of superposition — Interference occurs when two or more waves overlap, and according to the principle of superposition, the resultant displacement at any point is the algebraic sum of the individual displacements.
MCQ 7. An open organ pipe produces a fundamental frequency of 300 Hz. What is the frequency of its first overtone?
- A. 300 Hz
- B. 600 Hz
- C. 900 Hz
- D. 1200 Hz
Answer: B. 600 Hz — For an open organ pipe, the frequencies of harmonics are f, 2f, 3f, ... The fundamental frequency is f. The first overtone is the second harmonic, which is 2f. So, 2 * 300 Hz = 600 Hz.
MCQ 8. A closed organ pipe has a fundamental frequency of 200 Hz. Which of the following frequencies can it produce?
- A. 400 Hz
- B. 600 Hz
- C. 800 Hz
- D. 1000 Hz
Answer: B. 600 Hz — A closed organ pipe produces only odd harmonics. If the fundamental frequency is f, the possible frequencies are f, 3f, 5f, ... So, 200 Hz, 600 Hz, 1000 Hz, etc.
MCQ 9. A train blowing its whistle moves towards a stationary observer with a speed of 30 m/s. The frequency of the whistle emitted is 500 Hz. If the speed of sound in air is 330 m/s, what is the apparent frequency heard by the observer?
- A. 458 Hz
- B. 500 Hz
- C. 550 Hz
- D. 590 Hz
Answer: C. 550 Hz — Using the Doppler effect formula for a source moving towards a stationary observer: f' = f (v / (v - vs)). So, f' = 500 * (330 / (330 - 30)) = 500 * (330 / 300) = 550 Hz.
MCQ 10. Beats are formed due to the superposition of two waves having:
- A. Same frequency
- B. Same amplitude
- C. Slightly different frequencies
- D. Different directions of propagation
Answer: C. Slightly different frequencies — Beats are formed when two sound waves of slightly different frequencies and similar amplitudes travel simultaneously in the same direction.
MCQ 11. For a wave described by y(x,t) = A sin(kx - ωt), the maximum particle velocity is:
- A. Aω
- B. Aω²
- C. Ak
- D. Aω/k
Answer: A. Aω — The particle velocity is found by differentiating y(x,t) with respect to t: v_p = ∂y/∂t = -Aω cos(kx - ωt). The maximum value of cos(kx - ωt) is 1, so the maximum particle velocity is Aω.
MCQ 12. When a transverse wave is reflected from a denser medium, what happens to its phase?
- A. Increases by π/2
- B. Decreases by π/2
- C. Changes by π (180°)
- D. Remains unchanged
Answer: C. Changes by π (180°) — When a transverse wave is reflected from a denser medium (fixed end), the reflected wave undergoes a phase change of π (or 180°). A crest is reflected as a trough.
MCQ 13. Sound waves are an example of:
- A. Transverse electromagnetic waves
- B. Longitudinal electromagnetic waves
- C. Transverse mechanical waves
- D. Longitudinal mechanical waves
Answer: D. Longitudinal mechanical waves — Sound waves require a medium to propagate and involve oscillations parallel to the direction of wave propagation, making them longitudinal mechanical waves.
MCQ 14. A string fixed at both ends vibrates in its third harmonic. How many nodes are present along the string?
- A. 2
- B. 3
- C. 4
- D. 5
Answer: C. 4 — For a string fixed at both ends, the nth harmonic has (n+1) nodes and n antinodes. For the third harmonic (n=3), there will be 3+1 = 4 nodes.
MCQ 15. Two tuning forks A and B produce 4 beats/s. Tuning fork A is in resonance with a string of length 30 cm. Tuning fork B is in resonance with the same string of length 30.5 cm. The frequency of A is:
- A. 240 Hz
- B. 244 Hz
- C. 236 Hz
- D. 250 Hz
Answer: B. 244 Hz — Frequency is inversely proportional to length (f ∝ 1/L). So, f_A / f_B = L_B / L_A = 30.5 / 30. Since L_B > L_A, f_A > f_B. The beat frequency is f_A - f_B = 4 Hz. Solving f_A / (f_A - 4) = 30.5/30 gives f_A = 244 Hz.
MCQ 16. The maximum displacement of a particle from its mean position during wave propagation is called:
- A. Wavelength
- B. Frequency
- C. Amplitude
- D. Period
Answer: C. Amplitude — Amplitude is defined as the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position.
MCQ 17. The speed of sound in a gas is proportional to:
- A. Square root of its absolute temperature
- B. Absolute temperature
- C. Square of its absolute temperature
- D. Reciprocal of its absolute temperature
Answer: A. Square root of its absolute temperature — The speed of sound in a gas is given by v = √(γRT/M), where T is the absolute temperature. Thus, v is proportional to √T.
MCQ 18. Two waves are given by y₁ = a sin(ωt - kx) and y₂ = a cos(ωt - kx). The phase difference between the two waves is:
- A. 0
- B. π/4
- C. π/2
- D. π
Answer: C. π/2 — We can write y₂ as y₂ = a sin(ωt - kx + π/2). Comparing this with y₁ = a sin(ωt - kx), the phase difference is π/2.
MCQ 19. A standing wave is described by y(x,t) = 0.02 sin(5πx) cos(100πt). What is the distance between two consecutive nodes?
- A. 0.1 m
- B. 0.2 m
- C. 0.4 m
- D. 0.5 m
Answer: B. 0.2 m — For a standing wave y = 2A sin(kx) cos(ωt), k = 5π. The wavelength λ = 2π/k = 2π/(5π) = 0.4 m. The distance between two consecutive nodes is λ/2. So, 0.4 m / 2 = 0.2 m.
MCQ 20. An observer is moving away from a stationary source of sound with a speed of 20 m/s. The source emits sound of frequency 400 Hz. If the speed of sound in air is 340 m/s, what is the apparent frequency heard by the observer?
- A. 376.5 Hz
- B. 388.2 Hz
- C. 400 Hz
- D. 424.2 Hz
Answer: A. 376.5 Hz — Using the Doppler effect formula for an observer moving away from a stationary source: f' = f ((v - v_o) / v). So, f' = 400 * ((340 - 20) / 340) = 400 * (320 / 340) ≈ 376.5 Hz.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.