What Is the Defining Equation of SHM (a = -ω²x)?

For simple harmonic motion (SHM), the defining equation is $a=-\omega^2x$. It states that acceleration is proportional to displacement from equilibrium and always directed toward the equilibrium position.

This is syllabus point C.1.2 and applies to both SL and HL students.

Here, $a$ is the object's acceleration, $x$ is its displacement from equilibrium, and $\omega$ is its angular frequency in $\text{rad s}^{-1}$. Since $\omega^2$ is constant for a particular oscillator, the equation has the form

$$a \propto -x.$$

The negative sign describes direction, not merely a negative numerical value. Acceleration and displacement point in opposite directions, so the acceleration acts as a restoring acceleration.

The mechanism can also be seen from Newton's second law. If a restoring force is proportional to displacement, $F=-Kx$, then $ma=-Kx$, so $a=-(K/m)x$. Comparing this with the defining equation gives . For a massnegativespring oscillator, therefore, , although the defining equation applies to any system that satisfies the same accelerationnegativedisplacement relationship.

On an accelerationnegative displacement graph, SHM produces a straight line through the origin with negative gradient. Because the gradient is $-\omega^2$, its magnitude can be used to determine $\omega=\sqrt{|\text{gradient}|}$. A curved graph or one not passing through the origin does not satisfy ideal SHM over the full range.

For example, suppose $\omega=4.0\ \text{rad s}^{-1}$ and $x=0.050\ \text{m}$:

$$a=-(4.0)^2(0.050)=-0.80\ \text{m s}^{-2}.$$

The negative result means the acceleration is opposite to the chosen positive displacement direction. If the object were at $x=-0.050\ \text{m}$, its acceleration would be $+0.80\ \text{m s}^{-2}$.

A common misconception is that SHM has constant acceleration. It does not: acceleration changes continuously with displacement. At equilibrium, acceleration is zero but speed is maximum; at maximum displacement, acceleration magnitude is maximum but speed is zero.

Exam technique: When asked to show that motion is SHM, establish a relationship of the form $a=-kx$, where $k$ is a positive constant, and identify $k=\omega^2$. Always define $x$ relative to equilibrium and retain the minus sign.