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.

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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…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

A mass of 500 g is attached to a spring of spring constant k = 100 N/m. The m…
A simple pendulum of length 1 m oscillates with amplitude 5 cm. If g = 10 m/s…
Two identical springs are connected in series with a mass m = 200 g attached.…
A particle undergoes SHM with equation x = 5 sin(2πt) cm, where t is in secon…
A uniform rod of mass M and length L is pivoted at one end. Find the period o…
A mass m is attached to a spring and oscillates in a viscous medium. The equa…
+ 2 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
A particle undergoes SHM with equation x = 5 sin(2πt) cm,…	✓ Solved
A uniform rod of mass M and length L is pivoted at one en…	✓ Solved
A mass m is attached to a spring and oscillates in a visc…	✓ Solved

Showing 6 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: Maximum velocity: v_max = ω × A where A = amplitude = 10 cm = 0.1 m v_max = 14.14 × 0.1 = 1.414 m/s ≈ 1.41 m/s Final Answer: (a) Angular frequency = 14.14 rad/s; (b) Period = 0.444 s; (c) Maximum velocit...

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 = ω² × A a_max = 10 × 0.05 = 0.5 m/s² Alternatively: a_max = (g/L) × A = (10/1) × 0.05 = 0.5 m/s² Final Answer: Maximum velocity = 0.16 m/s; Maximum acceleration = 0.5 m/s²

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: Period of oscillation = 0.2 s

Q4: A particle undergoes SHM with equation x = 5 sin(2πt) cm, where t is in seconds. Calculate (a) amplitude, (b) frequency, (c) period, (d) the velocity at t = 0.25 s.

Step 1: Standard SHM equation: x = A sin(ωt) Given: x = 5 sin(2πt) cm Amplitude A = 5 cm = 0.05 m Step 2: Angular frequency: ω = 2π rad/s Step 3: Frequency: f = ω/(2π) = 2π/(2π) = 1 Hz Step 4: Period: T = 1/f = 1/1 = 1 s Step 5: Velocity in SHM: v = dx/dt = d/dt[5 sin(2πt)] = 5 × 2π × cos(2πt) = 10π cos(2πt) cm/s At t = 0.25 s: v = 10π cos(2π × 0.25) = 10π cos(π/2) = 10π × 0 = 0 m/s Final Answer: (a) Amplitude = 5 cm; (b) Frequency = 1 Hz; (c) Period = 1 s; (d) Velocity at t = 0.25 s = 0 ...

Q5: A uniform rod of mass M and length L is pivoted at one end. Find the period of oscillation for small amplitude oscillations.

Step 1: For a physical pendulum: T = 2π√(I/(mgd)) where I = moment of inertia about pivot, d = distance from pivot to center of mass Step 2: For a uniform rod pivoted at one end: I = (1/3)ML² d = L/2 (center of mass is at midpoint) Step 3: Substitute: T = 2π√[(1/3)ML² / (Mg × L/2)] T = 2π√[(1/3)ML² / (MgL/2)] T = 2π√[(1/3)L / (g/2)] T = 2π√[(2L)/(3g)] T = 2π√(2L/3g) Final Answer: Period T = 2π√(2L/3g) ≈ 2.56√(L/g)

Q6: A mass m is attached to a spring and oscillates in a viscous medium. The equation of motion is m(d²x/dt²) + b(dx/dt) + kx = 0. If b = 0.1 kg/s, k = 100 N/m, m = 1 kg, determine whether the motion is underdamped, critically damped, or overdamped.

Step 1: For damped oscillations, compare b (damping coefficient) with critical damping b_c. Critical damping: b_c = 2√(km) Step 2: Calculate b_c: b_c = 2√(100 × 1) = 2√100 = 2 × 10 = 20 kg/s Step 3: Compare given b with b_c: Given: b = 0.1 kg/s b_c = 20 kg/s Since b < b_c (0.1 < 20), the motion is UNDERDAMPED Step 4: Verify using discriminant approach: Characteristic equation: mr² + br + k = 0 Δ = b² - 4mk = (0.1)² - 4(1)(100) = 0.01 - 400 = -399.99 < 0 Negative discriminant → comp...

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