Simple harmonic motion
The same rhythm everywhere
- A pendulum, a mass on a spring, a child on a swing — all sway with the same rhythm.
- That shared rhythm is simple harmonic motion (SHM).
- One neat rule sits behind all of them.
The definition
- In SHM the acceleration is proportional to the displacement and points back to equilibrium.
- The defining equation: $a = -\omega^{2}x$ (the minus sign = "back toward the middle").
In simple harmonic motion, the acceleration is:
That is exactly $a = -\omega^{2}x$ — bigger displacement, bigger pull back; always toward the middle.
The key words
- amplitude $x_0$ — biggest displacement; period $T$; frequency $f = \dfrac{1}{T}$.
- angular frequency $\omega = \dfrac{2\pi}{T} = 2\pi f$ — find any of $\omega, f, T$ from another.
An oscillation has a period of $2.0\ \text{s}$. What is its angular frequency $\omega$?
$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.0} = \pi \approx 3.14\ \dfrac{\text{rad}}{\text{s}}$.
Displacement and speed
- Starting from the middle: $x = x_0\sin(\omega t)$.
- Speed without time: $v = \pm\omega\sqrt{x_0^{2} - x^{2}}$ — fastest at the middle ($\omega x_0$), zero at the ends.
In SHM the speed is greatest at the ____ position.
At the middle all the energy is kinetic, so the speed is maximum ($v = \omega x_0$).
At the extremes of the motion ($x = \pm x_0$), the speed is zero.
The oscillator stops for an instant at each end before reversing — speed zero, acceleration maximum.
An oscillator has amplitude $0.10\ \text{m}$ and $\omega = 10\ \dfrac{\text{rad}}{\text{s}}$. What is its maximum speed?
Maximum speed $= \omega x_0 = 10 \times 0.10 = 1.0\ \dfrac{\text{m}}{\text{s}}$.
Reading the a–x graph
- The acceleration–displacement graph is a straight line of gradient $-\omega^{2}$.
- So $\omega = \sqrt{|\text{gradient}|}$, and then $T = \dfrac{2\pi}{\omega}$.
On an acceleration-against-displacement graph for SHM, the gradient equals:
$a = -\omega^{2}x$ is a straight line of gradient $-\omega^{2}$, so $\omega = \sqrt{|\text{gradient}|}$.
You've got it
- SHM: $a = -\omega^{2}x$ (acceleration back toward equilibrium)
- $\omega = \dfrac{2\pi}{T} = 2\pi f$; fastest at the middle ($v = \omega x_0$), at rest at the ends
- the a–x graph is a line of gradient $-\omega^{2}$