Reading the three curves of SHM

Position in simple harmonic motion is a cosine of time. Velocity is the slope of position, and acceleration is the slope of velocity, so both are cosines too: offset from $x(t)$ by fixed amounts. Stacked on a common time axis, the three curves tell one story: when velocity peaks, position and acceleration are zero; when position peaks, velocity is zero and acceleration is maximum in the opposite direction. The slope-vs-area rule swaps as you switch graph types, and that's where most slips happen.

§1

Position is a cosine.

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A block on an ideal spring oscillates back and forth in a sine wave. With no phase offset (the block released from rest at $x = +A$), the position as a function of time is

$$x(t) = A\cos(\omega t).$$

Here $A$ is the amplitude (the maximum distance from equilibrium), and $\omega = \sqrt{k/m}$ is the angular frequency, which sets the pace of the oscillation. The period is $T = 2\pi/\omega = 2\pi\sqrt{m/k}$, the time it takes to complete one full cycle.

Three things to notice. First, the cosine peaks at $+A$ and dips to $-A$, and the block visits both: the turning points sit equally far from equilibrium on each side. Second, the peaks happen at $t = 0, T, 2T, \ldots$, and the dips at $t = T/2, 3T/2, \ldots$ The zeros (where the block passes through equilibrium) happen at $t = T/4, 3T/4, 5T/4, \ldots$, halfway between a peak and the next dip. Third, the period is set entirely by $m$ and $k$, not by the amplitude: a bigger swing covers more ground but moves faster too, and those two effects cancel exactly.

Start the block somewhere else in its cycle and the cosine slides sideways. The general form is $x(t) = A\cos(\omega t + \varphi)$, where $\varphi$ is a phase constant that sets where you started. The shape never changes; only the starting position does.

§2

Velocity and acceleration are cosines too.

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Velocity is the slope of position; acceleration is the slope of velocity. For $x(t) = A\cos(\omega t)$, the velocity and acceleration work out to

$$v(t) = -A\omega\sin(\omega t), \qquad a(t) = -A\omega^{2}\cos(\omega t) = -\omega^{2}\,x(t).$$

Three relationships pin everything down:

$x$ and $v$ are a quarter cycle apart. When $x$ is at its peak (the block is at a turning point), $v = 0$. When $v$ is at its peak (the block is passing through equilibrium), $x = 0$. The two never peak together.
$a$ is the upside-down twin of $x$. The acceleration is $-\omega^{2}$ times the position, at every instant: same shape, opposite sign, scaled by $\omega^{2}$. Where $x$ peaks at $+A$, $a$ dips to $-\omega^{2}A$; where $x$ is zero, so is $a$.
$v$ and $a$ are a quarter cycle apart too. Each curve sits a quarter cycle ahead of the one below it: $x$, then $v$, then $a$. When $|v|$ peaks at equilibrium, $a$ is zero; when $|a|$ peaks at the turning points, $v$ is zero.

The two amplitudes follow from the same formulas:

$$v_{\max} = \omega A, \qquad a_{\max} = \omega^{2} A = \dfrac{kA}{m}.$$

So a heavier mass on the same spring (larger $m$, smaller $\omega$) gives a slower oscillation with smaller peak speed and smaller peak acceleration, even at the same amplitude. A stiffer spring (larger $k$) gives a faster oscillation with larger peak speed and larger peak acceleration. Amplitude scales the three curves linearly: doubling $A$ doubles $v_{\max}$ and $a_{\max}$, but leaves $T$ alone.

Fig. 7.3.1 The three curves of SHM, stacked. At $t = 0$ (left edge), $x$ is at its peak, $v$ is zero, and $a$ is at its negative peak. At $t = T/4$, the roles rotate: $x$ is zero, $v$ is at its negative peak, $a$ is zero. Each curve is a quarter cycle ahead of the previous one.

§3

Where students go wrong.

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Three pitfalls show up over and over on graph-reading items. Each one looks reasonable until you put it next to the kinematics.

Pitfall · 01

"When the block is at its peak displacement, it must also be at its peak speed."

The intuition: at the extreme, everything must be extreme. If the block is way out there, it has to be moving fast too. Picture a ball at the top of its arc, momentarily at rest after being thrown hard, and you can see why this is wrong: turning around takes a stop. But the language still trips people up.

Fix. $x$ and $v$ are a quarter cycle apart: extrema of one align with zeros of the other. When $x = \pm A$, the block is momentarily at rest, $v = 0$. When $|v|$ is largest, the block is at $x = 0$. The pattern repeats every quarter cycle.

Pitfall · 02

"At the turning point, the block is momentarily at rest, so its acceleration is zero."

The intuition: "no motion" reads as "no force" reads as "no acceleration." Each step is wrong in SHM, but the chain feels airtight because three slightly different ideas (instantaneous speed, force, and acceleration) collapse into one.

Fix. Acceleration depends on the force, not on the current speed. At the turning point the spring is stretched (or compressed) the most, so the restoring force is at its peak and so is the acceleration: $|a_{\max}| = kA/m = \omega^{2}A$. The block is at rest and being yanked hardest. That's exactly how it turns around.

Pitfall · 03

"The slope of $x$-vs-$t$ gives the acceleration, since slope means 'rate of change.'"

The intuition: "rate of change" feels like a slot any kinematic quantity could fill. If you don't ask which quantity is on which axis, the slope rule blurs across the three graph types and lands on the wrong one.

Fix. Two anchor rules carry across all of kinematics: slope of $x$-vs-$t$ is $v$; slope of $v$-vs-$t$ is $a$. And the partner pair: area under $v$-vs-$t$ is $\Delta x$; area under $a$-vs-$t$ is $\Delta v$. The rule depends on which graph you're reading, not on which quantity you want. When in doubt, name both axes first.

§4

Skill Check.

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Ten short scenarios that pin down the alignment rules and the slope-area rule. Pick a chip, hit Check; if you miss, the feedback names why that distractor is tempting, not just that it's wrong. Your progress saves as you go.