Visual walkthrough — Energy in SHM — KE + PE = ½kA² (constant)

Before any symbol, meet the cast in plain words.

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Step 1 — The stage: a block, a spring, a home position

WHAT. We set up the one physical picture everything else rests on: a block on a spring, sliding on a frictionless floor. "Home" ($x=0$) is where the spring is neither stretched nor squeezed.

WHY. Every energy formula below is a statement about this picture. If we don't nail down what $x$, $A$, and "home" mean geometrically, no later symbol has meaning.

PICTURE. Three snapshots: pulled left to $x=-A$, sitting at home $x=0$, pushed right to $x=+A$. The floor is marked so you can literally read off $x$.

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Step 2 — The spring's law: force grows with stretch

WHAT. When the block is at position $x$, the spring pushes it back toward home with a force $F=-kx$. This is Hooke's Law.

WHY. Energy will be defined as work done against this force, so we must see the force first. The minus sign is the whole personality of a spring: it always points back home.

PICTURE. Force arrows at several positions. Far from home = long arrow; near home = short arrow; at home = no arrow.

$F=\underset{\begin{array}{c}\text{points back}\\ \text{toward home}\end{array}}{\underbrace{-}}\underset{\begin{array}{c}\text{stiffness:}\\ \text{steeper}=\text{stronger}\end{array}}{\underbrace{k}}\underset{\begin{array}{c}\text{how far}\\ \text{from home}\end{array}}{\underbrace{x}}$

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Step 3 — Potential energy: the AREA under the force line

WHAT. Potential energy $U$ = the work you must do to drag the block from home out to $x$, fighting the spring the whole way. To fight the spring you apply $F_{\text{you}}=+kx$ (opposite to the spring). Work = force $\times$ distance, but the force keeps growing, so we add up thin slices.

WHY THIS TOOL — the integral. Work is force times distance only when the force is constant. Here the force grows as you go. So we chop the trip into slivers so thin that inside each sliver the force barely changes, compute force$\times$width for each, and stack them up. That "stack up infinitely many thin slices" operation is exactly what $\int$ means. We pick it because the force is not constant — nothing simpler would be honest.

PICTURE. The line $F_{\text{you}}=k{x}^{\prime }$; the shaded triangle under it from $0$ to $x$ is the stored energy. A triangle of base $x$ and height $kx$ has area $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}x\cdot kx=\frac{1}{2}k{x}^2$.

$U(x)={\int }_0^x\underset{\text{force at slice }{x}^{\prime }}{\underbrace{k{x}^{\prime }}}\underset{\text{width of slice}}{\underbrace{d{x}^{\prime }}}=\frac{1}{2}k{x}^2$

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Step 4 — Kinetic energy: the AREA that motion carries

WHAT. Kinetic energy $K$ = energy the block owns simply because it's moving: $K=\frac{1}{2}m{v}^2$.

WHY. Push a block, its speed $v$ builds; the energy invested becomes this $\frac{1}{2}m{v}^2$. It's the other piggy bank. We want it in the same language as $U$ so we can add them later.

PICTURE. A bar chart of the two banks at three moments — at the edge (all in $U$), halfway, and at home (all in $K$). Watch the money slosh from one bar to the other.

$K=\frac{1}{2}\underset{\text{mass}}{\underbrace{m}}\underset{\text{speed, squared}}{\underbrace{v^2}}$

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Step 5 — Put motion into words: $x(t)$ and $v(t)$

WHAT. The block's position over time (from SHM) is a cosine wave; its speed (from Velocity & Acceleration in SHM) is a sine wave.

$x(t)=A\cos (\omega t+\varphi ),v(t)=-A\omega \sin (\omega t+\varphi )$

WHY. To prove the total never changes, we watch it at every instant $t$. We need $x$ and $v$ as functions of time so we can drop them into $U$ and $K$.

PICTURE. Two stacked curves: $x(t)$ a cosine (starts high, at $A$), $v(t)$ a sine flipped (starts at zero). Where one is at its peak the other is at zero — they are "90° out of step."

$v=\underset{\begin{array}{c}\text{moving}\\ \text{left first}\end{array}}{\underbrace{-}}\underset{\text{amplitude}}{\underbrace{A}}\underset{\begin{array}{c}\text{angular}\\ \text{speed}\end{array}}{\underbrace{\omega }}\underset{\begin{array}{c}\text{zero at edges,}\\ \text{max at home}\end{array}}{\underbrace{\sin (\omega t+\varphi )}}$

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Step 6 — Substitute and reveal two mirror-image humps

WHAT. Plug the time-functions into each bank. Using $m{\omega }^2=k$ (because ${\omega }^2=k/m$):

$K=\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi )$ $U=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )$

WHY. Swapping $m{\omega }^2\to k$ makes both banks wear the same coat — every symbol out front is now $\frac{1}{2}k{A}^2$. Only the trig wiggle differs. That's the setup for the magic cancellation.

PICTURE. $K$ (a ${\sin }^2$ hump) and $U$ (a ${\cos }^2$ hump) plotted together. When one is up the other is down — perfect see-saw — and their heights always add to the same ceiling.

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Step 7 — The magic cancellation: add the humps

WHAT. Add the two banks:

$E=K+U=\frac{1}{2}k{A}^2\left[\underset{=1\text{always}}{\underbrace{{\sin }^2(\cdot )+{\cos }^2(\cdot )}}\right]=\frac{1}{2}k{A}^2$

WHY THIS TOOL — the Pythagorean identity. ${\sin }^2\theta +{\cos }^2\theta =1$ is not a coincidence: on a unit circle, $\cos \theta$ and $\sin \theta$ are the two legs of a right triangle whose hypotenuse is 1, and the legs² sum to hypotenuse² (Pythagoras). We use it because our two humps are exactly a ${\sin }^2$ and a ${\cos }^2$ with identical front factors — the one tool that collapses them to a constant.

PICTURE. The two humps stacked into one bar of total height $\frac{1}{2}k{A}^2$ — a flat ceiling across all time. The see-saw tips, but the roof never moves.

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Step 8 — The edge cases: check the corners

WHAT / WHY. A derivation isn't finished until we test the extreme moments — the reader must never meet a case we skipped.

PICTURE. Four labelled moments on the energy bar: the two turning points, home, and the equal-split point.

• At $x=+A$ (right edge): $v=0$ (block frozen). $K=0$, $U=\frac{1}{2}k{A}^2$. All in the stretch bank. ✓
• At $x=-A$ (left edge): identical, because $x^2=A^2$ regardless of sign. The mirror-symmetry of Step 1 pays off — both edges are the same energy. ✓
• At $x=0$ (home): $U=0$, so $K=\frac{1}{2}k{A}^2$ and $v=v_{\max }=\omega A$. All in the motion bank. ✓
• Degenerate $A=0$ (never pulled): $E=\frac{1}{2}k(0{)}^2=0$. No pull, no energy, block never moves. ✓
• Equal split $K=U$: each is half the ceiling, $\frac{1}{2}k{x}^2=\frac{1}{4}k{A}^2\Rightarrow x=\pm A/\sqrt{2}\approx 0.707A$. ✓

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The one-picture summary

Everything at once: the parabola bowl $U=\frac{1}{2}k{x}^2$, the upside-down $K=\frac{1}{2}k(A^2-x^2)$, and the flat ceiling $E=\frac{1}{2}k{A}^2$ they always sum to — with the block drawn as a marble rolling in the bowl, trading height for speed.

Recall Feynman retelling — the whole walkthrough in plain words

We put a block on a spring and marked "home." The spring always yanks it back, harder the further out it is — a straight-line law. To pull it out you fight that growing pull, and the work you spend piles up as a triangle of area, which is $\frac{1}{2}k{x}^2$ — that's the stretch bank. When the block moves it also owns $\frac{1}{2}m{v}^2$ — the motion bank. We wrote both over time: one turned out to be a ${\sin }^2$ hump, the other a ${\cos }^2$ hump, both wearing the same coat $\frac{1}{2}k{A}^2$. And here's the punchline: a ${\sin }^2$ and a ${\cos }^2$ always add to exactly 1 (that's just Pythagoras on a circle), so the two humps forever add up to one flat ceiling, $\frac{1}{2}k{A}^2$. The see-saw tips endlessly, the roof never moves. At the edges it's all stretch and the block is frozen; at home it's all motion and the block is fastest; and it never cares whether you're left or right of home, because energy only sees $x^2$. Pull twice as far and, since everything is squared, you get four times the energy.