Lesson 10: Simple Harmonic Motion (SHM)

Week 10 — The physics of oscillation

🎯 Learning Objectives

Recognize and describe simple harmonic motion
Derive the period of a mass-spring system and pendulum
Relate SHM to uniform circular motion

🎯 Intuition First: The Back-and-Forth of Nature

Think of everything that bounces, swings, or vibrates: a guitar string, a pendulum clock, the suspension on your car, atoms in a crystal. They all share a common pattern — they oscillate.

💡 The Key Signature:

Simple harmonic motion occurs when the acceleration is always proportional to (and opposite to) the displacement. The further you pull, the harder it pulls back. That's the hallmark of SHM.

📋 Hooke's Law

F = −kx

The restoring force of a spring is proportional to displacement (k = spring constant, x = displacement from equilibrium). The negative sign means the force opposes the displacement.

📋 SHM Equations

SHM is mathematically identical to the projection of uniform circular motion onto a line:

x(t) = A cos(ωt + φ)

v(t) = −Aω sin(ωt + φ)

a(t) = −Aω² cos(ωt + φ) = −ω²x(t)

Where A = amplitude, ω = angular frequency, φ = phase constant.

📋 Period and Frequency

Mass on spring: T = 2π√(m/k), ω = √(k/m), f = 1/T

Simple pendulum (small angles): T = 2π√(L/g), f = 1/(2π)√(g/L)

📝 Key Insights:

For a mass-spring: period depends on mass and spring stiffness, but not on amplitude (for small oscillations)
For a pendulum: period depends only on length and gravity, not on mass or amplitude (small angles)
This is why grandfather clocks keep accurate time — the period is independent of amplitude!

📋 SHM Energy

KE = ½mv² = ½kA² sin²(ωt + φ)

PE = ½kx² = ½kA² cos²(ωt + φ)

E_total = ½kA² (constant!)

Energy continuously oscillates between kinetic and potential, but the total stays constant.

📝 Worked Example: Mass on a Spring

A 0.5-kg mass is attached to a spring (k = 200 N/m). It's pulled 0.1 m and released. Find the period, maximum speed, and maximum acceleration.

Step 1 — Angular frequency:

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

Step 2 — Period:

T = 2π/ω = 2π/20 = 0.314 s

Step 3 — Maximum speed:

v₍ = Aω = 0.1 × 20 = 2 m/s

Step 4 — Maximum acceleration:

a₍ = Aω² = 0.1 × 400 = 40 m/s²

✅ T = 0.314 s, v₍ = 2 m/s, a₍ = 40 m/s²

📝 Worked Example: Pendulum

A pendulum of length 1.5 m. What is its period? What is the period on the Moon (g = 1.6 m/s²)?

On Earth:

T = 2π√(L/g) = 2π√(1.5/9.8) = 2π√0.153 = 2π(0.391) = 2.46 s

On the Moon:

T = 2π√(1.5/1.6) = 2π√0.9375 = 2π(0.968) = 6.08 s

✅ Earth: 2.46 s, Moon: 6.08 s (slower oscillation due to weaker gravity)

📋 Connecting SHM to Circular Motion

💡 Visual Connection:

Imagine a ball moving in a circle at constant speed, projected onto a line (like a shadow on a wall). The shadow moves back and forth in exactly the same way as SHM. This is why the cosine function describes SHM — it's the x-coordinate of uniform circular motion!

✅ Check Your Understanding

Q1: Does a heavier pendulum swing with a different period than a lighter one (same length)?

Click to reveal answer

No. The period of a simple pendulum T = 2π√(L/g) doesn't depend on mass. Heavier and lighter pendulums of the same length swing at the same rate (in the small-angle approximation).

Q2 (mini-problem): A 0.2-kg mass on a spring oscillates with amplitude 0.08 m and period 0.5 s. What is the spring constant?