8.1.1 x-t Equation

A mass hung on a spring (often called a spring-mass system) is the simplest example of a simple harmonic motion (SHM). When displaced from the equilibrium position and released, the mass will oscillate periodically about the equilibrium position.

The variation with time t of the displacement x of the mass can be described by the equation

$\displaystyle x={{x}_{0}}\sin \omega t$

where x_o is the amplitude,

T is the period, and

ω is the angular frequency

Angular Frequency

The angular frequency ω is a rather abstract concept. If you are new to SHM, just think of it as the frequency multiplied by $\displaystyle 2\pi$ for the time being. The appendix contains a fuller discussion when you are ready.

$\displaystyle \omega =2\pi f$

Since $\displaystyle f=\frac{1}{T}$ , ω can also be written as

$\displaystyle \omega =\frac{{2\pi }}{T}$

You should also realize that mathematically speaking, we need the $\displaystyle \omega =\frac{{2\pi }}{T}$ term in $\displaystyle x={{x}_{0}}\sin \omega t$ to “scale the graph horizontally” so that one sinusoidal cycle corresponds to one period T.

Must it be the “sine” function?

No, it does not have to be $\displaystyle \sin \omega t$ . It is often written as $\displaystyle x={{x}_{0}}\cos \omega t$ too. It all depends on when we choose to start time t from. In fact, the general equation for SHM is $\displaystyle x={{x}_{0}}\sin (\omega t+\phi )$ . When we say “sinusoidal”, we can mean any time-shifted version of $\displaystyle x={{x}_{0}}\sin \omega t$ .

Why is SHM called SHM?

SHM is “simple harmonic” because its motion is described by a single sinusoidal function. A square wave, for example, requires the summation of an infinite number of sinusoidal functions

$\displaystyle \text{square wave}=\sin (x)+\frac{1}{3}\sin (3x)+\frac{1}{5}\sin (5x)+...$ .

The Half-Amplitude Position

SHM is not constant speed motion. It is slower when it is near the extreme positions and faster when it is near the equilibrium position.

It is very useful to “memorize” that $\displaystyle \cos 60{}^\circ =0.5$ or $\displaystyle \sin 30{}^\circ =0.5$ . Because then you can “see” without any calculations that

The time taken to travel between $\displaystyle x={{x}_{0}}$ and $\displaystyle x=0.5{{x}_{0}}$ is exactly $\displaystyle \frac{{60{}^\circ }}{{360{}^\circ }}T=\frac{1}{6}T$

but

The time taken to travel between $\displaystyle x=0$ and $\displaystyle x=0.5{{x}_{0}}$ is exactly $\displaystyle \frac{{30{}^\circ }}{{360{}^\circ }}T=\frac{1}{{12}}T$