CBSE Class 11 Physics Chapter 14 Notes: Oscillations

Chapter Introduction

A child on a swing, a plucked guitar string, a vibrating tuning fork, the pendulum of a clock — all exhibit oscillatory motion, a to-and-fro movement about an equilibrium position. Chapter 14 of NCERT Physics for Class 11 studies simple harmonic motion (SHM), the most fundamental type of oscillation, and its mathematical description. From the spring-mass system and the simple pendulum to damped and forced oscillations, this chapter provides the tools to analyse any vibrating system. SHM is the foundation for understanding waves (Chapter 15), ac circuits, and even quantum mechanics.

Key Definitions

Detailed Concepts

Periodic vs Oscillatory Motion

• Periodic motion: Repeats after a fixed time interval. Examples: Earth's revolution, rotation of fan blades.
• Oscillatory motion: Periodic + to-and-fro about an equilibrium position. Examples: Pendulum, spring-mass, vibrating string.
• SHM: A special case of oscillatory motion where F ∝ –x (restoring force proportional to displacement, opposite in direction).

Simple Harmonic Motion (SHM)

The defining condition for SHM: F = –kx (or equivalently, a = –ω²x)

where k = force constant, ω² = k/m, and x is displacement from equilibrium.

Displacement: x(t) = A cos(ωt + φ)

where A = amplitude, ω = angular frequency, φ = phase constant (determined by initial conditions)

Velocity: v(t) = dx/dt = –ωA sin(ωt + φ) = ω√(A² – x²)

Acceleration: a(t) = d²x/dt² = –ω²A cos(ωt + φ) = –ω²x

SHM Equations Summary

Spring-Mass System

For a mass m attached to a spring of force constant k:

Restoring force: F = –kx → a = –(k/m)x → ω² = k/m

T = 2π√(m/k), ν = (1/2π)√(k/m)

Key observations:

• T depends on m and k, not on amplitude or gravity
• A stiffer spring (larger k) gives shorter T
• A heavier mass gives longer T

Series and parallel springs:

• Springs in series: 1/k_eff = 1/k₁ + 1/k₂ → softer
• Springs in parallel: k_eff = k₁ + k₂ → stiffer

Simple Pendulum

For a pendulum of length l and bob mass m, displaced by angle θ:

Restoring force: F = –mg sin θ ≈ –mgθ (for small θ, sin θ ≈ θ)

Arc length: x = lθ → θ = x/l → F = –(mg/l)x

So ω² = g/l → T = 2π√(l/g)

Key observations:

• T is independent of mass (same as free fall: m cancels)
• T depends only on l and g
• At a place where g is less (e.g., at height or on the Moon), T increases
• The formula is valid only for small angles (θ < 15°)

Spring-mass vs Simple Pendulum:

Energy in SHM

Kinetic energy: KE = ½mv² = ½mω²(A² – x²)

Potential energy: PE = ½kx² = ½mω²x²

Total energy: E = KE + PE = ½mω²A² (constant — independent of x)

The total energy is conserved (no dissipation). Energy oscillates between KE and PE.

Energy graphs: KE and PE are sinusoidal functions of time, each oscillating at 2ω (twice the frequency of displacement). Total E is a horizontal line.

Damped Oscillations

In real oscillators, friction or viscosity dissipates energy. The equation of motion becomes:

m(d²x/dt²) + b(dx/dt) + kx = 0

where b is the damping constant.

Solution (underdamped case, b < 2√(mk)):

x(t) = A₀e⁻ᵇᵗ/²ᵐ cos(ω'd t + φ)

where ω'd = √(ω₀² – (b/2m)²) is the damped angular frequency.

• Amplitude decays exponentially: A(t) = A₀e⁻ᵇᵗ/²ᵐ
• Damped frequency ω'd < natural frequency ω₀
• If b is large enough (overdamped), oscillations stop entirely — the system returns to equilibrium without oscillating.

Forced Oscillations and Resonance

When an external periodic force F = F₀ cos ω_d t drives the oscillator:

m(d²x/dt²) + b(dx/dt) + kx = F₀ cos ω_d t

The steady-state solution gives oscillation at the driving frequency ω_d (not the natural frequency ω₀).

Resonance: When ω_d = ω₀, the amplitude becomes very large (theoretically infinite without damping). The system absorbs maximum energy from the driving force.

Resonance curve: Amplitude vs ω_d shows a peak at ω_d ≈ ω₀. The peak is sharper for lighter damping.

Practical examples of resonance:

• Tuning a radio: circuit resonates at the broadcast frequency
• Musical instruments: air column resonates at specific frequencies
• Swing: pushing at the natural frequency gives maximum amplitude
• Tacoma Narrows Bridge: wind-induced resonance caused collapse

Combination of SHMs

When two SHMs at right angles are combined:

x = A₁ cos(ω₁t + φ₁) and y = A₂ cos(ω₂t + φ₂)

The resulting path (Lissajous figure) depends on the frequency ratio and phase difference:

Important Derivations

1. Time Period of a Spring-Mass System

For mass m on a spring of constant k:

F = –kx → ma = –kx → d²x/dt² = –(k/m)x

This is the SHM equation with ω² = k/m.

2. Time Period of a Simple Pendulum

For a pendulum of length l, displaced by small angle θ:

Torque: τ = –mgl sin θ ≈ –mglθ (small angle)

Moment of inertia: I = ml²

Angular equation: Iα = τ → ml²(d²θ/dt²) = –mglθ

d²θ/dt² = –(g/l)θ → SHM with ω² = g/l

T = 2π/ω = 2π√(l/g)

3. Total Energy in SHM is Constant

E = KE + PE = ½mv² + ½kx²

v = dx/dt = –Aω sin(ωt + φ) and x = A cos(ωt + φ)

E = ½mω²A²sin²(ωt + φ) + ½kA²cos²(ωt + φ)

Since k = mω²: E = ½mω²A²[sin²(ωt + φ) + cos²(ωt + φ)] = ½mω²A² = constant

Common Mistakes

1. Confusing periodic and oscillatory motion: All SHM is oscillatory, all oscillatory is periodic, but not all periodic is oscillatory. Earth's rotation is periodic but not oscillatory. 2. Using the pendulum formula for large angles: T = 2π√(l/g) is only valid for small angles (θ < 15°). For large angles, the period increases. 3. Forgetting the minus sign in F = –kx: The restoring force is always directed opposite to displacement. Without the minus sign, the equation does not represent SHM. 4. Mixing up angular frequency (ω) and frequency (ν): ω = 2πν. They differ by a factor of 2π. Always check which one the question asks for. 5. Assuming damping increases the time period: Damping decreases the frequency (increases T slightly) but the primary effect is amplitude decay. For light damping, T ≈ T₀(1 + small correction). 6. Ignoring phase constant: If initial conditions are given (e.g., released from rest at x = A), you must determine φ. For release from x = A at t = 0: φ = 0, so x = A cos ωt.

Numerical Practice

Example 1: Spring-Mass Frequency

A body of mass 0.5 kg is attached to a spring with k = 200 N/m. Find the frequency and time period.

ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s

ν = ω/2π = 20/(2π) ≈ 3.18 Hz

T = 1/ν ≈ 0.314 s

Example 2: Simple Pendulum on the Moon

A pendulum has T = 2 s on Earth (g = 9.8 m/s²). What is its period on the Moon (g = 1.6 m/s²)?

T ∝ 1/√g → T_Moon = T_Earth × √(g_Earth/g_Moon) = 2 × √(9.8/1.6) = 2 × √6.125 = 2 × 2.47 ≈ 4.95 s

Example 3: Energy in SHM

A particle of mass 0.2 kg executes SHM with A = 0.1 m and ω = 10 rad/s. Find the total energy and the KE when x = 0.06 m.

Total energy: E = ½mω²A² = ½ × 0.2 × 100 × 0.01 = 0.1 J

KE at x = 0.06 m: KE = ½mω²(A² – x²) = ½ × 0.2 × 100 × (0.01 – 0.0036) = 10 × 0.0064 = 0.064 J

Exam Tips

1. SHM equation: Be ready to derive x = A cos(ωt + φ) from F = –kx. This is a 3-mark derivation. 2. Spring-mass and pendulum: Both derivations are standard 3–5 mark questions. Practise with correct diagrams. 3. Energy in SHM: Draw KE, PE, and total E graphs on the same axes. The 2ω frequency of KE and PE vs the ω frequency of x is a common MCQ point. 4. Damped oscillations: Usually 1–2 marks. Know that amplitude decays exponentially and frequency slightly decreases. 5. Resonance: Be ready to define it, draw the resonance curve, and give one practical example (positive or negative). 6. Numericals: Always identify ω first, then use it to find T, ν, v, a, or E. Watch for the difference between ω and ν. 7. Phase constant: If initial conditions are given, determine φ before writing the equation.

Quick Revision Notes

SHM key: F ∝ –x, a ∝ –x, sinusoidal displacement, energy conserved

Spring-mass: T depends on m and k (not g, not A)

Pendulum: T depends on l and g (not m, not A, only small θ)