# Simple Harmonic Motion
*10 beats · narrated by Antoni · Canvas Tutor v0.5*
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## 1. Step 01 — Topic
> 🎙 *Here's a fascinating question: if you hang a mass on a spring and let it go, it bounces up and down forever in a perfectly repeating pattern. What determines how fast it oscillates? Today we'll figure that out.*
> "A mass bounces on a spring forever — what determines how fast it oscillates?"

## 2. Step 02 — Hooke's Law
> 🎙 *The key is Hooke's Law. A spring pushes or pulls with a restoring force proportional to how far you stretch it from its natural length. The constant k tells you how stiff the spring is — bigger k means a stiffer spring.*
Hooke's Law — the spring's restoring force:
$$F_{\text{spring}} = -k\,x$$
k = spring constant (N/m)x = displacement from equilibriumNegative sign → force opposes displacement
> ✓ The spring always pulls back toward equilibrium — that's what makes the motion repeat.
*[Diagram: 7 elements — text, dashedLine, text, block 'm', dashedLine, text, +1 more]*

## 3. Step 03 — Equation of Motion
> 🎙 *Now apply Newton's second law. The net force is negative k x, and force also equals m a. Setting them equal gives us a second-order differential equation — the heartbeat of SHM.*
Apply Newton's 2nd Law to the mass:
$$F_{\text{net}} = m\,a \quad \Rightarrow \quad -k\,x = m\ddot{x}$$
Rearrange:
$$\ddot{x} + \dfrac{k}{m}\,x = 0$$
> ↗ This says: acceleration is always proportional to — and opposite — displacement.

## 4. Step 04 — Solution: x(t)
> 🎙 *The solution to that equation is a cosine — or sine — function of time. Here x-naught is the amplitude, the maximum displacement, and omega is the angular frequency. This is what SHM looks like mathematically.*
The general solution is:
$$x(t) = x_0 \cos(\omega t)$$
x₀ = amplitude (maximum stretch)ω = angular frequency (rad/s)t = time
*[Diagram: 6 elements — axes, text, text, dashedLine 'x₀', dashedLine '−x₀', text]*

## 5. Step 05 — Finding ω
> 🎙 *Here's where it clicks. Plug the cosine solution back into the equation of motion and you get omega squared equals k over m. So the angular frequency is the square root of k over m. Stiffer spring means faster oscillation — makes physical sense!*
Substitute x = x₀ cos(ωt) back into the equation of motion:
$$-\omega^2 x_0\cos(\omega t) + \dfrac{k}{m}\,x_0\cos(\omega t) = 0$$
Cancel x₀ cos(ωt) — gives:
$$\omega^2 = \dfrac{k}{m}$$
> **Result:** $$\omega = \sqrt{\dfrac{k}{m}}$$

## 6. Step 06 — Period & Frequency
> 🎙 *From angular frequency we get two quantities students always need: the period T — the time for one full oscillation — and the frequency f, how many oscillations per second. Notice that a heavier mass gives a longer period.*
From ω = 2πf = 2π/T:
Period (time per cycle):
$$T = 2\pi\sqrt{\dfrac{m}{k}}$$
Frequency (cycles per second):
$$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$
> ✓ ↑ mass → ↑ T (slower)↑ spring stiffness → ↓ T (faster)

## 7. Step 07 — Worked Example
> 🎙 *Let's try a concrete example. A 0.5 kilogram mass hangs on a spring with k equals 50 newtons per metre. What is the period? Follow the recipe: plug into T equals two pi root m over k.*
Given: m = 0.5 kg, k = 50 N/mFind: period T
**Student work:**
- $$T = 2\pi\sqrt{\dfrac{m}{k}}$$
- $$T = 2\pi\sqrt{\dfrac{0.5}{50}}$$
- $$T = 2\pi\sqrt{0.01} = 2\pi \times 0.1$$
- $$T = 0.628 \text{ s}$$
> ✓ About 0.63 s per bounce — roughly one bounce per second. Feels right for a typical lab spring!

## 8. Step 08 — Trap: Amplitude
> 🎙 *Here's a trap that catches almost everyone. Students often think a bigger amplitude means a faster oscillation. It doesn't. The period T depends only on m and k — not on how far you pull the mass.*
Common mistake:"Pull the mass further → spring oscillates faster."
**Student work:**
- $$T = 2\pi\sqrt{\dfrac{m}{k}}$$
- $$\text{Amplitude } x_0 \text{ does NOT appear in } T$$
> ✗ ❌ Bigger pull ≠ faster oscillation.Period is isochronous — same regardless of amplitude.
> ✓ ✔ Amplitude only changes the maximum speed and maximum force — not T or f.

## 9. Step 09 — Verify: Limits
> 🎙 *Let's verify our formula makes sense at the limits. If the spring is infinitely stiff, k goes to infinity and the period goes to zero — the mass barely moves. If the mass is enormous, the period blows up — a very heavy mass barely accelerates. Both limits check out.*
Does T = 2π√(m/k) make physical sense?
**Verification:**
- $$k \to \infty \Rightarrow T \to 0$$ — Infinitely stiff spring: instant restoring force, no oscillation time — ✔
- $$m \to \infty \Rightarrow T \to \infty$$ — Infinite mass: force can't accelerate it, oscillation takes forever — ✔
- $$k \to 0 \Rightarrow T \to \infty$$ — No spring at all: zero restoring force, mass drifts — ✔

## 10. Step 10 — Summary
> 🎙 *Here's your complete picture of simple harmonic motion. The position follows a cosine, angular frequency is root k over m, and the period is two pi root m over k. These three results are the toolkit for every SHM problem you'll meet.*
Simple Harmonic Motion — the full picture:
Equation of motion:
$$\ddot{x} + \dfrac{k}{m}x = 0$$
Position:
$$x(t) = x_0\cos(\omega t)$$
Angular frequency:
$$\omega = \sqrt{k/m}$$
> **Result:** $$T = 2\pi\sqrt{\dfrac{m}{k}}$$