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.

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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 Jul 23, 2026

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Key Questions Covered:

A tuning fork vibrates at frequency 256 Hz. The speed of sound in air is 340 …
Two coherent sound sources emit sound at 500 Hz. The speed of sound is 330 m/…
A string of length 0.5 m and mass 10 g is fixed at both ends. When vibrating …
A transverse wave on a string is represented by y = 0.02 sin(10x - 4πt), wher…
The intensity of sound is defined as power per unit area. If a sound has inte…
A siren producing sound at 800 Hz is approaching a stationary observer at 20 …
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
A tuning fork vibrates at frequency 256 Hz. The speed of …	✓ Solved
Two coherent sound sources emit sound at 500 Hz. The spee…	✓ Solved
A string of length 0.5 m and mass 10 g is fixed at both e…	✓ Solved
A transverse wave on a string is represented by y = 0.02 …	✓ Solved
The intensity of sound is defined as power per unit area.…	✓ Solved
A siren producing sound at 800 Hz is approaching a statio…	✓ Solved

Showing 6 of 8 questions

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 con...

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...

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

Showing 6 of 8 questions. Visit the full page for complete solutions.