Show that the projection of uniform circular motion is SHM

Question

Show that the projection of uniform circular motion on any diameter is simple harmonic motion (SHM).

Solution — Step by Step

Consider a particle P moving in a circle of radius $A$ with uniform angular velocity $\omega$ (rad/s). Let the centre of the circle be at the origin O. At time $t = 0$ , let the particle be at position $(A, 0)$ on the positive x-axis.

At time $t$ , the particle has moved through angle $\omega t$ from the starting position. The position of P is:

x_P = A\cos(\omega t), \quad y_P = A\sin(\omega t)

Let Q be the foot of the perpendicular from P onto the y-axis (the vertical diameter). The y-coordinate of Q equals the y-coordinate of P:

y_Q = y_P = A\sin(\omega t)

This gives the displacement of Q from the centre O as a function of time.

Differentiating displacement with respect to time:

v_Q = \frac{dy_Q}{dt} = A\omega\cos(\omega t)

Differentiating velocity:

a_Q = \frac{d^2y_Q}{dt^2} = -A\omega^2\sin(\omega t)

Now observe: since $y_Q = A\sin(\omega t)$ :

a_Q = -\omega^2 y_Q

The acceleration of Q is:

a_Q = -\omega^2 y_Q

This is exactly the defining condition of SHM: acceleration is proportional to displacement and directed towards the mean position (negative sign).

Therefore, the projection Q executes SHM with:

Amplitude $= A$ (radius of circle)
Angular frequency $= \omega$
Period $= T = 2\pi/\omega$

Why This Works

Simple harmonic motion is, by definition, motion where restoring acceleration is proportional to displacement from equilibrium. The projection of uniform circular motion automatically satisfies this condition because the circular motion’s centripetal acceleration has components along and perpendicular to any diameter — and the component along the diameter varies sinusoidally with displacement.

This relationship is more than mathematical equivalence — it’s a deep physical insight. The SHM equations ( $x = A\sin(\omega t + \phi)$ ) are directly the projections of circular motion. That’s why SHM is described using sine and cosine functions — they are projections of rotation.

Alternative Method — X-axis Projection

The projection onto the x-axis gives $x_Q = A\cos(\omega t)$ .

Its acceleration: $a = -A\omega^2\cos(\omega t) = -\omega^2 x_Q$ .

Same SHM condition. Both projections (x and y) are SHM — just 90° out of phase with each other.

Common Mistake

Some students write the acceleration as $a = -A\omega^2\sin(\omega t)$ and stop there without connecting it back to the displacement. The crucial step is recognising that $A\sin(\omega t) = y_Q$ (the displacement), so $a = -\omega^2 \cdot y_Q$ . This final substitution is what explicitly shows the $a \propto -y$ condition for SHM.

This result is used to derive SHM equations. When a particle starts at $y = 0$ at $t = 0$ (moving in positive direction), its displacement is $y = A\sin(\omega t)$ . When it starts at the maximum displacement (positive extreme), its displacement is $y = A\cos(\omega t)$ . The only difference is the initial phase — always check the initial conditions to pick the correct starting function.

Question

Solution — Step by Step

Why This Works

Alternative Method — X-axis Projection

Common Mistake

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