1. What is Simple Harmonic Motion?

An oscillation occurs whenever an object moves repeatedly back and forth through an equilibrium position. The equilibrium position is the point at which the object would remain at rest if undisturbed. Not every oscillation qualifies as SHM; the motion must satisfy two strict conditions simultaneously.

These two conditions can be expressed compactly as:

The negative sign is essential: it encodes the requirement that the acceleration always acts back towards equilibrium, opposing the displacement. Without this restoring nature, the motion would not be oscillatory.

In this equation:

• $a$ is the acceleration of the object (m s$^{-2}$),
• $x$ is the displacement from the equilibrium position (m),
• $\omega$ is the angular frequency of the oscillation (rad s$^{-1}$), defined as $\omega = \frac{2\pi}{T} = 2\pi f$, where $T$ is the time period and $f$ is the frequency.

The constant of proportionality is $\omega^2$, which means that a shorter time period (faster oscillation) produces a larger acceleration for any given displacement.

Some key terms used throughout this lesson are worth defining precisely here.

The amplitude tells you how far the object travels from its rest position. Closely related are two quantities that describe how quickly the oscillations repeat.

The reciprocal of the time period gives the frequency of the oscillation.

The figure below shows the equilibrium position, an intermediate displacement, and both turning points; notice that amplitude is measured from the centre to an extreme while the restoring acceleration at the extremes points back towards equilibrium.

[DIAGRAM: asset_name: 6.1.2 - Simple Harmonic Motion - Diagram 1; asset_slug: 6.1.2 - Simple Harmonic Motion - Diagram 1; recommended_method: retained_png; description: An oscillating object shown on a horizontal line at three positions. The centre is labelled "equilibrium position, $x = 0$". The left turning point is labelled "$x = -A$" and the right turning point "$x = +A$". A shorter arrow from the centre to an intermediate point is labelled "displacement $x$". A full arrow from the centre to one turning point is labelled "amplitude $A$". Arrows at both turning points point back towards the centre to show the restoring acceleration.]

This diagram helps separate two ideas that students often merge. The displacement $x$ is the object's current signed distance from equilibrium, so it can be positive, negative, or zero. The amplitude $A$ is the largest value that $|x|$ ever reaches. Equilibrium is the midpoint of the motion, not one end of it.

Consider a simple pendulum: when displaced to one side and released, the restoring component of its weight always acts back towards the lowest point (the equilibrium). At maximum displacement (the amplitude), the acceleration is at its greatest magnitude because $|x|$ is at its maximum. At the equilibrium position ($x = 0$), the acceleration is zero.

2. Displacement as a Function of Time

The defining equation $a = -\omega^2 x$ is a second-order differential equation. Its solution gives displacement as a sinusoidal function of time. If we choose to start timing ($t = 0$) at the instant when the object is at maximum positive displacement ($x = +A$), the solution is:

Here $A$ is the amplitude and $\omega t$ must be calculated in radians (ensure your calculator is in radian mode).

This cosine form applies when the object starts at $x = +A$. If instead timing begins as the object passes through equilibrium moving in the positive direction, the displacement follows $x = A\sin(\omega t)$. Both are valid solutions; which one to use depends on the initial conditions of the problem.

The time period $T$ does not depend on the amplitude. Whether you pull a pendulum back by a small angle or a larger angle (within the SHM approximation), the period remains the same. This is a hallmark of SHM.

Worked Example: An object oscillates in SHM with amplitude 58 mm and time period 3.0 s. Find its displacement at $t = 0.50$ s, given that it starts at maximum positive displacement.

Solution:

At $t = 0.50$ s (which is $\frac{T}{6}$), the object has moved from 58 mm to 29 mm — it is returning towards equilibrium.

3. Velocity in SHM

Velocity in SHM can be expressed as a function of displacement using the equation:

The $\pm$ sign indicates that for any given displacement $x$ (other than $\pm A$), the object could be moving in either direction — towards or away from that point.

From this equation, several important results follow:

• At the equilibrium position ($x = 0$): $v = \pm\omega\sqrt{A^2 - 0} = \pm\omega A$. This is the maximum speed.
• At the extremes ($x = \pm A$): $v = \pm\omega\sqrt{A^2 - A^2} = 0$. The object is momentarily at rest before reversing direction.

This makes physical sense: the object is fastest as it passes through the centre and stationary at the turning points.

Worked Example: A mass on a spring oscillates with amplitude 0.040 m and frequency 1.5 Hz. Calculate the maximum speed of the mass.

Solution:

4. Acceleration in SHM

From the defining equation $a = -\omega^2 x$, the magnitude of the acceleration is greatest when $|x|$ is greatest, that is, when the object is at maximum displacement ($x = \pm A$).

At the equilibrium position ($x = 0$), the acceleration is zero. This is the exact opposite of velocity: velocity is maximum where acceleration is zero, and acceleration is maximum where velocity is zero.

Worked Example: An object on a spring oscillates with time period 0.48 s and maximum acceleration 9.8 m s$^{-2}$. Calculate (a) its frequency, and (b) its amplitude.

Solution:

(a) $f = \frac{1}{T} = \frac{1}{0.48} = 2.08$ Hz $\approx 2.1$ Hz

(b) From $a_{\max} = \omega^2 A$:

5. Graphical Representations: Linking $x$, $v$, and $a$ with Time

One of the most important skills in SHM is understanding and interpreting displacement–time, velocity–time, and acceleration–time graphs, and appreciating how they are connected through calculus (differentiation).

Displacement–time graph ($x$ vs $t$): Since $x = A\cos(\omega t)$, the graph is a cosine curve oscillating between $+A$ and $-A$ with period $T$.

The figure below stacks the three time graphs on one shared axis; notice how the dashed guide lines connect turning points, zero crossings, and the phase shifts between $x$, $v$, and $a$.

[DIAGRAM: asset_name: 6.1.2 - Simple Harmonic Motion - Diagram 2; asset_slug: 6.1.2 - Simple Harmonic Motion - Diagram 2; recommended_method: retained_png; description: Three vertically stacked graphs sharing the same time axis. Top graph: $x$ vs $t$, a cosine curve with amplitude $A$, labelled maxima at $+A$ and minima at $-A$. Middle graph: $v$ vs $t$, a negative sine curve with amplitude $\omega A$. Bottom graph: $a$ vs $t$, a negative cosine curve with amplitude $\omega^2 A$. Vertical dashed lines at $t = 0$, $T/4$, $T/2$, $3T/4$, and $T$ connect corresponding points across all three graphs.]

Velocity–time graph ($v$ vs $t$): Velocity is the gradient (derivative) of the displacement–time graph. If $x = A\cos(\omega t)$, then $v = \frac{dx}{dt} = -A\omega\sin(\omega t)$. This is a negative sine curve:

• When displacement is at a maximum ($x = +A$), the gradient of the $x$–$t$ graph is zero, so $v = 0$.
• When displacement is zero (object passing through equilibrium), the gradient is steepest, giving $v = \pm\omega A$.

The velocity–time graph leads the displacement–time graph by a quarter of a cycle ($\frac{\pi}{2}$ radians or 90°). More precisely, velocity reaches its maximum value a quarter period before displacement reaches its maximum.

Acceleration–time graph ($a$ vs $t$): Acceleration is the gradient of the velocity–time graph. Differentiating again: $a = \frac{dv}{dt} = -A\omega^2\cos(\omega t) = -\omega^2 x$, which confirms the defining equation. The acceleration–time graph is an inverted cosine curve:

• Maximum positive acceleration occurs at maximum negative displacement.
• Maximum negative acceleration occurs at maximum positive displacement.

The acceleration is exactly in antiphase with the displacement — a phase difference of $\pi$ radians (180°).

These relationships are summarised in the table below:

6. The Connection Between Circular Motion and SHM

The mathematics of SHM is intimately connected to uniform circular motion. Consider a point P moving at constant speed around a circle of radius $A$. If you project the position of P onto one axis (say the $x$-axis), the projection oscillates back and forth between $+A$ and $-A$ — this projected motion is SHM.

The figure below shows the circular-motion model; notice how the horizontal projection of point P gives the SHM displacement while the angle $\theta = \omega t$ controls where that projection falls.

[DIAGRAM: asset_name: 6.1.2 - Simple Harmonic Motion - Diagram 3; asset_slug: 6.1.2 - Simple Harmonic Motion - Diagram 3; recommended_method: retained_png; description: A circle of radius $A$ centred at the origin. A point P is shown on the circle at angle $\theta = \omega t$ from the positive $x$-axis. A vertical dashed line drops from P to the $x$-axis, showing the projection $x = A\cos(\omega t)$. The angular velocity $\omega$ is labelled along the arc.]

At time $t$, if P has moved through angle $\theta = \omega t$ from the positive $x$-axis, then the $x$-coordinate of P is:

The centripetal acceleration of P has magnitude $\omega^2 A$ and is directed towards the centre. Its component along the $x$-axis is $a_x = -\omega^2 A\cos(\omega t) = -\omega^2 x$, which is exactly the defining equation of SHM.

This is why $\omega$ — originally the angular velocity of circular motion — appears as the angular frequency in SHM, and why the constant of proportionality in $a = -\omega^2 x$ is specifically $\omega^2$.

By this stage, the separate pieces should fit into one model: define SHM with the restoring acceleration, describe the motion with the displacement equation, and then use gradients or phase differences to move between the graphs.