Oscillations — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Oscillations" — 8 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Oscillations" — 8 important questions with detailed answers for CBSE board exam prepar…

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Key Questions Covered:

  1. A mass of 500 g is attached to a spring of spring constant k = 100 N/m. The m…
  2. A simple pendulum of length 1 m oscillates with amplitude 5 cm. If g = 10 m/s…
  3. Two identical springs are connected in series with a mass m = 200 g attached.…
  4. + 5 more questions in the full chapter

Solutions Summary:

Question Status
A mass of 500 g is attached to a spring of spring constan… ✓ Solved
A simple pendulum of length 1 m oscillates with amplitude… ✓ Solved
Two identical springs are connected in series with a mass… ✓ Solved

Showing 3 of 8 questions

Q1: A mass of 500 g is attached to a spring of spring constant k = 100 N/m. The mass is displaced 10 cm from equilibrium and released. Calculate (a) the angular frequency, (b) the period of oscillation, (c) the maximum velocity.

Step 1: Angular frequency: ω = √(k/m) where k = spring constant, m = mass m = 500 g = 0.5 kg k = 100 N/m ω = √(100/0.5) = √200 = 14.14 rad/s Step 2: Period of oscillation: T = 2π/ω = 2π/14.14 = 0.444 s Alternatively: T = 2π√(m/k) = 2π√(0.5/100) = 2π√(0.005) = 2π × 0.0707 = 0.444 s Step 3: Maxim...

Q2: A simple pendulum of length 1 m oscillates with amplitude 5 cm. If g = 10 m/s², calculate the maximum velocity and maximum acceleration.

Step 1: For small oscillations (amplitude << length): Angular frequency ω = √(g/L) L = 1 m, g = 10 m/s² ω = √(10/1) = √10 = 3.16 rad/s Step 2: Maximum velocity: v_max = ω × A where A = amplitude = 5 cm = 0.05 m v_max = 3.16 × 0.05 = 0.158 m/s ≈ 0.16 m/s Step 3: Maximum acceleration: a_max ...

Q3: Two identical springs are connected in series with a mass m = 200 g attached. Each spring has k = 400 N/m. Find the period of oscillation.

Step 1: For springs in series, equivalent spring constant: 1/k_eq = 1/k₁ + 1/k₂ 1/k_eq = 1/400 + 1/400 = 2/400 = 1/200 k_eq = 200 N/m Step 2: Period of oscillation: T = 2π√(m/k_eq) m = 200 g = 0.2 kg k_eq = 200 N/m T = 2π√(0.2/200) T = 2π√(0.001) T = 2π × 0.0316 T = 0.199 s ≈ 0.2 s Final Answer:...

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