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.
By Syllab.in · Updated
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 …
- + 5 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 |
Showing 3 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 ...
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
Ste...
Showing 3 of 8 questions. Visit the full page for complete solutions.