Visual walkthrough — Velocity and acceleration in SHM — v = ω√(A² − x²)

We use one running character throughout: a dot that goes back and forth along a line, like a bead on a wire. Its distance from the middle we call $x$. That is the only idea we start with.

---

Step 1 — Draw the motion and fix a sign convention

WHAT. Picture a horizontal wire. A red dot slides left and right, never going past two walls. The middle of the wire is "home" (equilibrium). The farthest the dot ever reaches is a distance we name $==A==$ — the amplitude (in metres). We now make one firm choice and keep it forever: rightward is positive. So $x>0$ means the dot is right of home, $x<0$ means left; a velocity $v>0$ means moving rightward, $v<0$ means moving leftward.

WHY. Before any algebra we must pin down what each letter is on a picture, and which direction counts as "plus." $x$ = how far (and which side) the dot is from home right now. $A$ = the biggest that $|x|$ can ever be. The sign convention is what will later give the $\pm$ in our answer an honest meaning.

PICTURE. Look at figure s01: the dot sits at signed position $x$. The two dashed walls at $-A$ and $+A$ are the turning points — the dot touches them, freezes for an instant, and reverses. The green arrow shows which way is "positive."

For now we know only two facts, both readable from the picture:

• $x$ ranges between $-A$ and $+A$;
• at the walls the dot stops and turns.

No sine, no angle yet — we will build those in Step 2, not assume them.

---

Step 2 — Why a spinning dot? The reference circle

WHAT. Take a second dot walking around a circle of radius $A$ at steady speed. We measure its angle $==\theta ==$ from the upward vertical, turning anticlockwise. Now drop its shadow horizontally onto a vertical ruler (project sideways onto the up–down axis). That shadow slides up and down exactly like a bead — and we will let that up–down shadow be our $x$ laid on its side.

WHY. We choose a circle because uniform circular motion is the simplest thing whose shadow oscillates smoothly (the Reference circle (projection of uniform circular motion) trick). Measuring $\theta$ from the top and reading the shadow toward that same top axis is a deliberate choice: it makes the shadow equal to the sine of $\theta$. If we had measured from the side, we'd get a cosine. Same physics, different bookkeeping — we pick the sine convention so it matches the parent note's $x=A\sin (\omega t+\varphi )$.

PICTURE. In figure s02 the circle-dot is at angle $\theta$ from the top. Drop a horizontal dotted line to the vertical axis: it meets the axis at height $x$. The right triangle has hypotenuse $A$ (the radius) and the side along the direction we measure $\theta$ from has length $x$. Because $\theta$ is measured from the top, the side that grows as $\theta$ opens up is the one of length $x=A\sin \theta$.

$x=A\sin \theta ⟹\sin \theta =\frac{x}{A}$

• $\sin \theta$ = (side opposite the angle $\theta$) $÷$ (hypotenuse) — with $\theta$ read from the top, that opposite side is exactly the shadow $x$.
• $\frac{x}{A}$ = "how far along its full reach the dot is," a signed number from $-1$ to $+1$ (negative when the circle-dot is below the middle).
• The steady sweep means $\theta$ grows linearly in time: $\theta =\omega t+\varphi$, where $\omega$ is the spin rate and $\varphi$ is the starting angle. We never need $\varphi$ again — it just sets where the dot was at $t=0$.

---

Step 3 — Velocity is the speed of the shadow

WHAT. The circle-dot moves along the circle at a constant speed $\omega A$ (radius $\times$ spin rate). Its velocity arrow points along the circle (tangent). The shadow's velocity is just the projection of that arrow onto our measuring axis.

WHY. Velocity means "rate of change of position," written $v=\frac{dx}{dt}$. We use the derivative because it is the exact tool that converts a position rule into a speed rule. On the circle, differentiating $x=A\sin \theta$ pulls down a factor $\omega$ (the dial spins at rate $\omega$) and turns $\sin$ into $\cos$: this is where the perpendicular projection shows up.

PICTURE. In figure s03 the green tangent arrow has length $\omega A$. Its projection onto our $x$-axis (the part along the direction we measure $x$) is the dot's velocity $v$. Notice: when the circle-dot is level with the middle, the tangent arrow lies fully along the measuring axis, so the shadow's speed is largest.

$v=\frac{dx}{dt}=\underset{\text{full circle-speed}}{\underbrace{A\omega }}\underset{\text{how much lies along the axis}}{\underbrace{\cos \theta }}$

• $A\omega$ — the constant speed of the circle-dot.
• $\cos \theta$ — the fraction of that speed pointing along our $x$-axis; it is positive when $\theta$ is in the first/fourth part of the turn (dot moving toward $+x$) and negative on the way back. This sign is exactly the sign of $v$ from our convention in Step 1.
• Rearranged: $\cos \theta =\frac{v}{A\omega }$.

---

Step 4 — Two shadows, one circle: the key link

WHAT. We now have two readings off the same circle-dot: $\sin \theta =\frac{x}{A},\cos \theta =\frac{v}{A\omega }.$ One angle $\theta$ controls both. That single angle is the bridge that lets us throw time away.

WHY. We do not care when the dot is somewhere — we want a rule linking where it is ($x$) to how fast it moves ($v$). Since both come from the same $\theta$, we can eliminate $\theta$ using the one law every circle obeys: Pythagorean identity sin² + cos² = 1.

PICTURE. Figure s04 shows the right triangle inside the circle: one side $=A\sin \theta$, the perpendicular side $=A\cos \theta$, hypotenuse $=A$ (the radius). Pythagoras on this triangle is ${\sin }^2\theta +{\cos }^2\theta =1$.

$\underset{(x/A{)}^2}{\underbrace{{\sin }^2\theta }}+\underset{(v/A\omega {)}^2}{\underbrace{{\cos }^2\theta }}=1$

---

Step 5 — Kill the angle, keep $x$ and $v$

WHAT. Substitute the two shadow-readings into the identity and clear the fractions.

WHY. This is the payoff of Step 4: the identity is a constraint that ties $x$ and $v$ together with no $\theta$ left. One equation, two quantities we actually measure.

PICTURE. Figure s05 tracks the algebra as a shrinking of the triangle — each side scaled — until the radius-squared statement becomes a statement about $x$ and $v$ only.

${\left(\frac{x}{A}\right)}^2+{\left(\frac{v}{A\omega }\right)}^2=1$

Multiply every term by $A^2{\omega }^2$:

$\underset{\text{from the }x\text{ term}}{\underbrace{{\omega }^2{x}^2}}+\underset{\text{from the }v\text{ term}}{\underbrace{v^2}}=\underset{{\text{radius}}^2\text{ scaled}}{\underbrace{A^2{\omega }^2}}$

Solve for $v^2$:

$v^2={\omega }^2(A^2-x^2)$

---

Step 6 — Read the two extreme cases off the formula

WHAT. Test the two ends of the swing.

WHY. A formula you cannot check at its edges is a formula you do not trust. Plug in the two special positions.

PICTURE. Figure s06 overlays the speed on the wire: a fat arrow at the centre, no arrow at the walls.

• At home, $x=0$: $v=\omega \sqrt{A^2-0}=\omega A$. This is $==v_{\max }==$ — the fastest the dot ever moves. On the circle the dot is level with the middle, so the tangent arrow lies fully along our measuring axis and its projection is full-length.
• At a wall, $x=\pm A$: $v=\omega \sqrt{A^2-A^2}=0$. The dot freezes to reverse. On the circle the dot is at the very top or bottom, so the tangent arrow points straight across the measuring axis (perpendicular to it); its projection onto that axis is therefore zero. (This is the exact opposite of the common "fastest at the edge" mistake in the parent note.)

---

Step 7 — The degenerate cases: no swing, and forbidden positions

WHAT. Two boundary situations. First, amplitude zero, $A=0$. Second, asking for a position beyond the walls, $|x|>A$.

WHY. Every honest derivation must survive its own extreme inputs and clearly state where it is allowed to be used — its domain.

PICTURE. Figure s07 shows the circle shrunk to a single point at the origin (case $A=0$), and marks the allowed band $-A\le x\le A$ on the wire.

Case $A=0$. The circle has no radius; the dot cannot be anywhere but home, so the only legal position is $x=0$, and $v=\omega \sqrt{0-0}=0.$ No swing, no speed — consistent.

Domain of the formula. The dot physically lives only in the band $-A\le x\le A$. Inside this band $A^2-x^2\ge 0$, so the square root is a real, sensible speed. Outside it ($|x|>A$) we would be asking "how fast is the dot at a place it can never reach?" — and $A^2-x^2$ turns negative. A negative under a square root is the algebra's way of saying "impossible position, no real speed exists here." So the formula is valid exactly on $-A\le x\le A$, with $v=0$ at the two ends and $v=\pm \omega A$ in the middle.

---

Step 8 — Where the acceleration hides in the same picture

WHAT. The circle-dot, though it moves at constant speed, constantly changes direction, so it has an acceleration. For uniform circular motion this acceleration points straight to the centre and has magnitude ${\omega }^{2}A$. Its projection onto our $x$-axis is the sliding dot's acceleration.

WHY (deriving $a={\omega }^{2}R$). Why is the centre-pointing acceleration ${\omega }^{2}R$ with $R$ the radius? The velocity arrow has fixed length $v_c=\omega R$ but its direction turns at the same rate $\omega$ as the position. Acceleration is the rate of change of the velocity arrow; a vector of length $\omega R$ whose direction rotates at rate $\omega$ changes at rate $(\text{length})\times (\text{turn rate})=\omega R\times \omega ={\omega }^{2}R$, and the change points inward (toward the centre). With $R=A$ this is ${\omega }^{2}A$. (This is the same reasoning by which position of length $A$ turning at rate $\omega$ gave a velocity of length $\omega A$ in Step 3 — apply it once more to the velocity arrow.)

PICTURE. Figure s08 draws the inward arrow of length ${\omega }^{2}A$; its projection onto our $x$-axis is $-{\omega }^{2}x$ — pointing opposite to $x$ (always back toward home).

$a=-A{\omega }^2\sin \theta =-{\omega }^2\underset{=x}{\underbrace{(A\sin \theta )}}=-{\omega }^{2}x$

• The minus sign = the inward arrow's shadow always points toward home, opposite to displacement — the defining feature of SHM.
• Big when far out, zero at the centre — exactly the reverse of the speed pattern.

This same relation is energy book-keeping: $v^2+{\omega }^2{x}^2={\omega }^2{A}^2$ is $KE+PE=$ constant with $\frac{1}{2}m$ cancelled.

---

The one-picture summary

Figure s09 folds the whole story into one frame: the reference circle, the sliding shadow $x$, the tangent velocity whose projection is $v=\omega \sqrt{A^2-x^2}$, and the inward pull whose projection is $a=-{\omega }^{2}x$. Fast in the middle, frozen at the walls, always pulled home.

Recall Feynman retelling of the whole walkthrough

Put a dot on a wire; it slides between two walls a distance $A$ away, and we agree that rightward counts as positive. Now stand a friend walking round a circle of the same radius $A$ and read off their height as an angle $\theta$ measured from the top — their shadow is your sliding dot, and that shadow equals $A\sin \theta$. Your friend walks at a steady pace, so their speed arrow (pointing along the circle) has fixed length $\omega A$, and their "being-pulled-to-the-middle" arrow has fixed length ${\omega }^{2}A$ — each is a length turned at rate $\omega$, then turned once more. The part of the speed arrow along your axis is how fast your dot moves ($v$); the part of the pull arrow along your axis is your dot's push-back ($a$). When your friend is level with the middle, the speed arrow lies fully along your axis — shadow moves fastest — and that happens exactly when the shadow is at home. When the friend is at the very top or bottom, the speed arrow points straight across your axis — no motion along it — so the dot freezes at a wall. Pythagoras on the circle's right triangle (${\sin }^2+{\cos }^2=1$) ties the position and the speed together, and out pops $v=\omega \sqrt{A^2-x^2}$: the $\sqrt{A^2-x^2}$ is just "how much room is left before the wall," and the $\pm$ records which way the dot is going. The pull arrow always aims at the centre, which is why $a=-{\omega }^{2}x$ carries a minus — it's forever dragging the dot back home.

---

Connections

• Velocity and acceleration in SHM — v = ω√(A² − x²) (parent)
• Reference circle (projection of uniform circular motion)
• Pythagorean identity sin² + cos² = 1
• Angular frequency ω and time period T
• Phase and phase difference
• Simple Harmonic Motion — definition a = −ω²x
• Energy in SHM — kinetic and potential

Concept Map