§ 13.1NCERT Class 11 · Physics · Chapter 13

Periodic Motion, Oscillations & SHM Kinematics

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 4 min read

      Fundamental Definition:
      Periodic Motion: A motion that repeats itself at regular intervals of time (called Time Period T). Examples: Revolution of the Earth around the Sun, rotation of the hands of a clock.
Oscillatory (Vibratory) Motion: A periodic to-and-fro or back-and-forth motion of a body about a fixed central position (called Mean Position or Equilibrium Position). Examples: Swing of a pendulum, motion of a loaded spring.
Golden Rule: "Every oscillatory motion is periodic, but every periodic motion is NOT oscillatory." (Circular orbital motion is periodic, but not oscillatory).

    

1. Definition & Necessary Condition for Linear SHM

Simple Harmonic Motion (SHM) is the simplest and most fundamental form of oscillatory motion. A particle executes linear SHM if its acceleration is directly proportional to its displacement from the mean position and is always directed towards the mean position:

F = −k x ⇔ m (d²x / dt²) = −k x ⇔ d²x / dt² + ω²x = 0

Where:

k = Force constant or spring factor (N/m).
m = Inertia factor or mass of the oscillating particle (kg).
ω = √(k / m) = Angular frequency of the SHM (rad/s).

2. Kinematic Equations of Linear SHM

1. Displacement x(t)

The general solution to the differential equation:

x(t) = A sin(ωt + φ) or A cos(ωt + φ)

A: Amplitude (maximum displacement from mean position).
(ωt + φ): Phase of the motion at time t (determines state of motion).
φ: Initial phase or Epoch (phase at t = 0).

2. Velocity v(t)

Differentiating displacement with respect to time:

v(t) = dx/dt = Aω cos(ωt + φ) = ±ω √(A² − x²)

At Mean Position (x = 0): Velocity is maximum → v_max = ωA.
At Extreme Positions (x = ±A): Velocity is zero → v = 0.

3. Acceleration a(t)

Differentiating velocity with respect to time:

a(t) = dv/dt = −ω² A sin(ωt + φ) = −ω² x

At Mean Position (x = 0): Acceleration is zero → a = 0.
At Extreme Positions (x = ±A): Acceleration is maximum → a_max = ω²A (directed towards the mean position).

3. Phase Relationships Among x, v, and a

Variable	Mathematical Expression	Phase Relative to Displacement x	Lead / Lag
Displacement x	A sin(ωt)	0	Reference baseline
Velocity v	Aω cos(ωt) = Aω sin(ωt + π/2)	+π/2 rad (+90°)	Velocity leads displacement by π/2 rad
Acceleration a	−ω²A sin(ωt) = ω²A sin(ωt + π)	+π rad (+180°)	Acceleration leads displacement by π rad (opposite phase)

4. Reference Circle (Projection of Uniform Circular Motion)

      Theorem: Simple Harmonic Motion is the projection of Uniform Circular Motion (UCM) on any diameter of the circle of reference.
      Radius of the reference circle = Amplitude of SHM (A).
Uniform angular speed of revolving particle = Angular frequency of SHM (ω).
Projection on x-axis (diameter): x(t) = A cos(ωt + φ).
Projection on y-axis (diameter): y(t) = A sin(ωt + φ).

    

5. Solved Examples & Key JEE / NEET Traps

Example 1: A particle executes SHM with an amplitude of 10 cm and time period 4 s. Find: (i) the maximum velocity, (ii) the velocity when it is at a distance of 6 cm from the mean position, and (iii) the acceleration at x = 6 cm.

Solution:

Given: A = 10 cm = 0.1 m; T = 4 s ⇒ ω = 2π / T = 2π / 4 = π/2 rad/s ≈ 1.57 rad/s.

(i) Maximum velocity: v_max = ωA = (π/2) × 10 = 5π cm/s ≈ 15.7 cm/s.
(ii) Velocity at x = 6 cm: v = ω √(A² − x²) = (π/2) √(10² − 6²) = (π/2) × 8 = 4π cm/s ≈ 12.57 cm/s.
(iii) Acceleration at x = 6 cm: a = −ω² x = −(π/2)² × 6 = −(6π² / 4) = −1.5π² cm/s² ≈ −14.8 cm/s².

Superposition of two perpendicular SHMs (Lissajous Figures): If x = A₁ sin(ωt) and y = A₂ sin(ωt + Δφ):

If Δφ = 0 or π: Motion is a straight line (y = ±(A₂/A₁)x).
If Δφ = π/2: Motion is an ellipse (x²/A₁² + y²/A₂² = 1).
If Δφ = π/2 and A₁ = A₂: Motion is a circle!

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