The reference circle gave the period of simple harmonic motion. It also gives the whole motion as a graph, and you can read velocity and acceleration from it just as you did with motion graphs along a line.
Why it works
Follow a point around a circle of radius $A$ at angular speed $\omega = \dfrac{2\pi}{T}$, starting at $x = A$. Its shadow is at $x = A\cos(\omega t)$. The point's velocity has size $A\omega$ along the circle, and the shadow gets the $x$-part of it: $v = -A\omega\sin(\omega t)$. Its acceleration has size $A\omega^2$ toward the center, and the shadow's part is $a = -A\omega^2\cos(\omega t) = -\omega^2 x$. Cosine and sine are the same wave shifted by a quarter turn, so the three graphs are a quarter cycle apart.
Key idea
Starting from $x = A$ at rest: $$x = A\cos\frac{2\pi t}{T}, \quad v = -\frac{2\pi A}{T}\sin\frac{2\pi t}{T}, \quad a = -\left(\frac{2\pi}{T}\right)^2 x.$$ The greatest speed is $v_{\max} = \dfrac{2\pi A}{T}$ and the greatest acceleration is $a_{\max} = \left(\dfrac{2\pi}{T}\right)^2 A$. Velocity peaks a quarter cycle after position; acceleration is always opposite to position.
| Moment | Position $x$ | Velocity $v$ | Acceleration $a$ |
|---|---|---|---|
| $t = 0$ | $+A$ | $0$ | $-a_{\max}$ |
| $t = T/4$ | $0$ | $-v_{\max}$ | $0$ |
| $t = T/2$ | $-A$ | $0$ | $+a_{\max}$ |
| $t = 3T/4$ | $0$ | $+v_{\max}$ | $0$ |
Worked examples
Worked example: Reading a graph
An $x$–$t$ graph of a mass on a spring has peaks at $x = 0.20\ \text{m}$, at $t = 0$ and $t = 2.0\ \text{s}$. Find $A$, $T$ and the greatest speed.
- Amplitude: from the middle line to a peak, $A = 0.20\ \text{m}$.
- Period: peak to peak, $T = 2.0\ \text{s}$.
- $v_{\max} = \dfrac{2\pi A}{T} = \dfrac{2\pi(0.20)}{2.0} = 0.63\ \text{m/s}$, when the graph crosses $x = 0$.
Answer: $A = 0.20\ \text{m}$, $T = 2.0\ \text{s}$, $v_{\max} = 0.63\ \text{m/s}$
Worked example: Position at a given time
For that motion, where is the mass at $t = 0.30\ \text{s}$?
- The angle is in radians: $\dfrac{2\pi(0.30)}{2.0} = 0.94\ \text{rad}$.
- $x = (0.20\ \text{m})\cos(0.94) = 0.12\ \text{m}$.
Answer: $0.12\ \text{m}$
Common mistake
Thinking position and velocity reach their peaks at the same moment. When $x$ is at its peak the mass is at a turning point, momentarily at rest: the graph is flat there, so $v = 0$. The velocity is largest a quarter cycle later, where the $x$–$t$ graph crosses zero at its steepest. Check by slopes: a peak of $x$ always lines up with a zero of $v$.