Simple Harmonic Motion

Topics covered: Simple Harmonic Motion (SHM), kinematics, energy, spring–mass systems, pendulums and physical pendulum (JEE Main focus).

1. Introduction to SHM

• Motion is oscillatory when it repeats periodically about a stable equilibrium position.
• In SHM, restoring force is directly proportional to displacement and directed towards mean position.

F ∝ x

F = −kx

Using Newton’s second law:

d²x/dt² = −(k/m)x = −ω²x

Angular frequency:

ω = √(k/m)

2. Kinematics of SHM

Displacement:

x = A sin(ωt + φ)

• A = amplitude
• φ = initial phase
• (ωt + φ) = phase

Time period:

T = 2π / ω

Frequency:

ν = 1 / T

Angular frequency:

ω = 2πν

3. Velocity in SHM

v = dx/dt = ωA cos(ωt + φ)

v = ω √(A² − x²)

• Maximum velocity: v_max = ωA (at x = 0)
• Velocity is zero at x = ±A

4. Acceleration in SHM

a = dv/dt = −ω²A sin(ωt + φ)

a = −ω²x

• Acceleration is maximum at x = ±A
• Acceleration is zero at x = 0

5. Velocity–Displacement Relation

v² = ω² (A² − x²)

• Graph of v vs x is an ellipse
• Graph of a vs x is a straight line with slope −ω²

6. Energy in SHM

Kinetic Energy (K)

K = ½ mv² = ½ mω² (A² − x²)

• K_max = ½ mω²A² (at x = 0)
• K = 0 at x = ±A

Potential Energy (U)

U = ½ kx² = ½ mω²x²

• U_max = ½ mω²A² (at x = ±A)
• U = 0 at x = 0

Total Energy (E)

E = K + U = ½ mω²A² (constant)

7. Time Period of SHM

Linear SHM

T = 2π √(m / k)

Angular SHM

T = 2π √(I / k)

General form:

T = 2π √(Inertia factor / Force factor)

8. Spring–Mass System

(a) Series Combination

1 / k_eq = 1 / k₁ + 1 / k₂

T = 2π √(m / k_eq)

(b) Parallel Combination

k_eq = k₁ + k₂

T = 2π √(m / k_eq)

9. Relation Between Spring Length and Stiffness

k ∝ 1 / l

If a spring of length l is cut into parts l₁ and l₂:

k₁ = K (l / l₁)

k₂ = K (l / l₂)

10. Simple Pendulum

Time period (small oscillations):

T = 2π √(l / g)

• Independent of mass
• Seconds pendulum: T = 2 s

Effective gravity cases

• Lift accelerating upwards: g_eff = g + a
• Lift accelerating downwards: g_eff = g − a
• Horizontal acceleration: g_eff = √(g² + a²)
• Free-falling lift: g_eff = 0 → no oscillation

T = 2π √(l / g_eff)

11. Physical Pendulum

Restoring torque:

τ = −mgdθ

Equation of motion:

I d²θ/dt² = −mgdθ

Time period:

T = 2π √(I / mgd)

Special cases

• Rod of length l about one end:
  T = 2π √(2l / 3g)
• Ring of radius R pivoted at rim:
  T = 2π √(2R / g)