The Shadow of a Spinning Wheel Moves Exactly Like a Spring

A swing pauses for an instant at each end and rushes fastest through the lowest point. Its speed and acceleration change continuously, and they peak at different places.

Linking SHM to steady motion round a circle makes these changes easy to work out, and it leads to the force law that defines SHM.

This part covers SHM as projected circular motion, velocity and acceleration, the force law, and graphs with their phase relationships.

**A point moving round a circle of radius $A$ at constant angular speed $\omega$ has an angle $\omega t+\varphi$ at time $t$, so its projection on a diameter is $x=A\cos (\omega t+\varphi )$ — exactly the equation of SHM.

The radius of the circle becomes the amplitude, the angular speed becomes the angular frequency, and the angle becomes the phase.** Projecting on the perpendicular diameter gives $y=A\sin (\omega t+\varphi )$, another SHM a quarter-cycle out of step.

Worked example. A point on the rim of a $0.2$ m radius wheel turns at $2$ revolutions per second. Its shadow on a wall moves in SHM with

$$A=0.2\text{ m}\omega =2\pi \times 2=4\pi \text{ rad/s}T=0.5\text{ s}$$

The shadow's maximum speed equals the point's circular speed, $v=\omega A=4\pi \times 0.2\approx 2.51$ m/s.

An everyday example. The shadow of a bicycle pedal on the road, cast by the evening sun, slides back and forth while the pedal goes round and round.

The substance. The circle is only a reference — the body in SHM moves along a straight line, not in a circle.

**Differentiating $x=A\cos (\omega t+\varphi )$ gives $v=-A\omega \sin (\omega t+\varphi )=\pm \omega \sqrt{A^2-x^2}$ and $a=-{\omega }^{2}x$, so speed is greatest ($A\omega$) at the mean position and zero at the extremes, while acceleration is zero at the mean position and greatest (${\omega }^{2}A$) at the extremes.

Derivation.**

$$v=\frac{dx}{dt}=-A\omega \sin (\omega t+\varphi )a=\frac{dv}{dt}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x$$

Using ${\sin }^2+{\cos }^2=1$ gives $v=\pm \omega \sqrt{A^2-x^2}$.

Worked example. An SHM has $A=0.1$ m and $T=0.2$ s, so $\omega =10\pi \approx 31.4$ rad/s.

$$v_{max}=A\omega \approx 3.14\text{ m/s}{a}_{max}={\omega }^{2}A\approx 98.7{\text{ m/s}}^2$$

At $x=0.06$ m:

$$v=31.4\sqrt{0.01-0.0036}=31.4\times 0.08\approx 2.51\text{ m/s}a=-(31.4{)}^2\times 0.06\approx -59.2{\text{ m/s}}^2$$

An everyday example. The needle of a sewing machine flashes through the middle of its stroke but stops for an instant at the top and bottom, where it reverses.

The substance. Where speed is greatest, acceleration is zero, and the other way round.

**By Newton's second law, $F=ma=-m{\omega }^{2}x$, so SHM needs a restoring force proportional to displacement and directed towards the mean position, $F=-kx$ with $k=m{\omega }^2$ and $\omega =\sqrt{\frac{k}{m}}$.

The minus sign means the force always points back towards the mean position; the proportionality means twice the displacement gives twice the force.

Worked example 1 — a spring.** A $0.5$ kg block on a spring with $k=200$ N/m:

$$\omega =\sqrt{\frac{200}{0.5}}=20\text{ rad/s}T=\frac{2\pi }{20}\approx 0.314\text{ s}$$

At $x=0.03$ m the restoring force is $200\times 0.03=6$ N, towards the mean position.

Worked example 2 — is it SHM?

- $F=-5x$ — yes, with $k=5$ N/m
- $F=-5x^3$ — no, force is not proportional to $x$
- $F=-4x+2$ — yes, about the new mean position $x=0.5$, since $F=-4(x-0.5)$

An everyday example. A car's suspension spring pushes the wheel back each time it jolts up over a bump.

The substance. Almost any small oscillation about stable equilibrium is close to SHM, because the restoring force is nearly proportional to small displacements.

**For $x=A\cos \omega t$, velocity is $v=A\omega \cos \left(\omega t+\frac{\pi }{2}\right)$ and acceleration is $a=A{\omega }^2\cos (\omega t+\pi )$, so velocity leads displacement by $\frac{\pi }{2}$, acceleration leads velocity by $\frac{\pi }{2}$, and acceleration is exactly opposite in phase to displacement.

Values at quarter periods:

- $t=0$**: $x=A$, $v=0$, $a=-{\omega }^{2}A$
- **$t=\frac{T}{4}$**: $x=0$, $v=-A\omega$, $a=0$
- **$t=\frac{T}{2}$**: $x=-A$, $v=0$, $a=+{\omega }^{2}A$
- **$t=\frac{3T}{4}$**: $x=0$, $v=+A\omega$, $a=0$

Worked example. With $A=0.1$ m and $\omega =10\pi$ rad/s ($T=0.2$ s), at $t=0.05$ s $=\frac{T}{4}$:

$$x=0.1\cos \frac{\pi }{2}=0v=-0.1\times 10\pi \times \sin \frac{\pi }{2}\approx -3.14\text{ m/s}a=0$$

Other graphs. $a$ against $x$ is a straight line of slope $-{\omega }^2$ through the origin, and $v$ against $x$ is an ellipse.

An everyday example. A swinging pendulum is fastest at the bottom, where its displacement is zero, and has its largest pull back at the ends.

The substance. All three graphs have the same period — only their phases differ.