Simple Harmonic Motion and Uniform Circular Motion

The Connection Between Circular Motion and SHM

Simple harmonic motion and uniform circular motion might seem like completely different phenomena — one is a back-and-forth oscillation along a straight line, the other is steady rotation around a circle. Yet they share a deep geometric link. Understanding this connection gives you a powerful visual way to think about SHM, and it lets you derive the equations of SHM without solving a single differential equation.

Imagine a particle moving with constant angular speed $\omega$ around a circle of radius $A$. Now shine a light from the side, so that the particle casts a shadow on a screen placed behind the circle. As the particle goes around, the shadow moves back and forth along a straight line. That shadow's motion is simple harmonic.

This is the core idea of the section: the projection of uniform circular motion onto a diameter of the circle is simple harmonic motion.

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Setting Up the Geometry

Consider a particle $P$ moving on a circle of radius $A$ with constant angular speed $\omega$. At time $t = 0$, let $P$ be at the positive $x$-axis (angle zero). After time $t$, the radius vector $OP$ makes an angle $\theta = \omega t$ with the $x$-axis.

Now drop a perpendicular from $P$ onto the $x$-axis. The foot of this perpendicular is point $Q$. As $P$ moves around the circle, $Q$ slides back and forth along the $x$-axis. The position of $Q$ at any time $t$ is simply the $x$-coordinate of $P$:

This is exactly the equation of simple harmonic motion with amplitude $A$ and angular frequency $\omega$. The particle $P$ is called the reference particle, the circle is the reference circle, and $Q$ is the projection of $P$ onto the diameter.

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Velocity in SHM from Circular Motion

The reference particle $P$ moves with uniform circular motion, so its speed is constant: $v_P = \omega A$, directed tangentially to the circle. At any instant, this velocity vector makes an angle $\theta = \omega t$ with the vertical (or equivalently, an angle $90^\circ - \theta$ with the horizontal).

The $x$-component of $P$'s velocity is the velocity of the projection $Q$. From the geometry:

The negative sign appears because when $P$ is in the upper half of the circle ($\sin \omega t > 0$), the projection $Q$ is moving leftward (negative $x$-direction). This matches the SHM velocity formula $v = -\omega A \sin(\omega t)$.

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Acceleration in SHM from Circular Motion

The reference particle $P$ has centripetal acceleration $a_P = \omega^2 A$, directed radially inward toward the centre of the circle. This acceleration vector always points along the radius $OP$, making an angle $\theta = \omega t$ with the $x$-axis.

The $x$-component of $P$'s acceleration gives the acceleration of the projection $Q$:

But $A \cos(\omega t) = x$, so:

This is the hallmark of SHM: acceleration is proportional to displacement from the mean position and directed opposite to it. The negative sign confirms that acceleration always points toward the equilibrium position ($x = 0$).

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The Four Key Properties — Derived from the Circular Motion Picture

The textbook lists four properties that follow directly from this geometric interpretation. Each one is proved by considering the motion of the reference particle and its projection.

Property 1: The projection of uniform circular motion on any diameter is SHM

We have already shown this. For a particle moving with angular speed $\omega$ on a circle of radius $A$, the $x$-coordinate of its projection is $x = A \cos(\omega t + \phi)$, where $\phi$ is the initial phase. This satisfies the SHM equation $a = -\omega^2 x$.

›Proof

Formal proof

Let the reference particle have position vector $\mathbf{r} = A(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$ with $\theta = \omega t + \phi$. The projection onto the $x$-axis is $x = A\cos(\omega t + \phi)$. Differentiating twice:

$v_x = \frac{dx}{dt} = -\omega A \sin(\omega t + \phi)$

$a_x = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t + \phi) = -\omega^2 x$

Since $a_x \propto -x$, the motion of the projection is SHM.

Property 2: The amplitude of SHM equals the radius of the reference circle

The projection $Q$ oscillates between $x = -A$ and $x = +A$. These extremes occur when the reference particle $P$ is at the leftmost and rightmost points of the circle, respectively. The maximum displacement from the mean position is therefore $A$, which is exactly the radius of the reference circle.

Property 3: The time period of SHM equals the time period of the reference particle's circular motion

The reference particle takes time $T = \frac{2\pi}{\omega}$ to complete one full revolution. In that same time, the projection $Q$ goes from one extreme to the other and back — one complete oscillation. Hence:

This is the same time period we derived from the SHM differential equation.

Property 4: The phase of SHM equals the angular displacement of the reference particle

At any instant $t$, the reference particle's angular position is $\theta = \omega t + \phi$. The projection's position is $x = A \cos \theta$. The angle $\theta$ is precisely the phase of the SHM. When $\theta = 0$, the projection is at the positive extreme; when $\theta = \pi/2$, it passes through the mean position moving left; when $\theta = \pi$, it is at the negative extreme; and so on.

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A Useful Table: Correspondence Between Circular Motion and SHM