# How to Find the Centre, Period, Velocity, Acceleration and Times of Simple Harmonic Motion Explained

Simple Harmonic Motion is a specific type of periodic motion where is any motion whose displacement-time equation, apart from the constants, is a single sine and/or cosine function. A certain kind of rectilinear motion is particularly important, known as Simple Harmonic Motion.

A particle is said to be undergoing simple harmonic motion with the centre as the origin if:

$$x=A \sin (nt + \theta) \text{ or } x = A \cos (nt + \theta)$$

where $A, n$ and $\theta$ are constants, with $A$ and $n$ positive.

The constant $A$ is called the amplitude of the motion, and the particle is confined in the interval $-A \le x \le A$.

The origin is the centre of the motion, as it is the midpoint between the two extremes of the motion, $x = – A$ and $x = A$.

The period $T$ of the motion is given by $\displaystyle T = \frac{2 \pi}{n}$.

## Example 1

A particle moves in a straight line. Its displacement, $x$ metres, after $t$ seconds is given by $\displaystyle x = 3 + 4 \cos \left(2t + \frac{\pi}{3} \right)$.

(a)   Find the centre of the motion.

\begin{align} \require{AMSsymbols} \displaystyle x &= \color{red}{3} + 4 \cos \left(2t + \frac{\pi}{3} \right) \\ \therefore x &= 3 \end{align}

(b)   Find the period of the motion.

\begin{align} \require{AMSsymbols} \displaystyle x &= 3 + 4 \cos \left(\color{red}{2}t + \frac{\pi}{3} \right) \\ T&= \frac{2 \pi}{\color{red}{2}} \\ \therefore T &= \pi \end{align}

(c)   Find the amplitude of the motion.

\begin{align} \require{AMSsymbols} \displaystyle x &= 3 + \color{red}{4} \cos \left(2t + \frac{\pi}{3} \right) \\ \therefore A &= 4 \end{align}

(d)   Find the interval of the motion.

\begin{align} \require{AMSsymbols} \displaystyle -1 &\le \cos \left(2t + \frac{\pi}{3} \right) \le 1 \\ -4 &\le 4 \cos \left(2t + \frac{\pi}{3} \right) \le 4 \\ -1 &\le 3 + 4 \cos \left(2t + \frac{\pi}{3} \right) \le 7 \\ \therefore -1 &\le x \le 7 \end{align}

(e)   Find the first time when the particle is at the centre.

\begin{align} \require{AMSsymbols} \displaystyle 3 + 4 \cos \left(2t + \frac{\pi}{3} \right) &= 3 \\ 4 \cos \left(2t + \frac{\pi}{3} \right) &= 0 \\ \cos \left(2t + \frac{\pi}{3} \right) &= 0 \\ 2t + \frac{\pi}{3} &= \frac{\pi}{2} \\ 2t &= \frac{\pi}{2}-\frac{\pi}{3} \\ &= \frac{\pi}{6} \\ \therefore t &= \frac{\pi}{12} \end{align}

(f)   Find the first time the particle is at the maximum displacement.

\begin{align} \require{AMSsymbols} \displaystyle 3 + 4 \cos \left(2t + \frac{\pi}{3} \right) &= 7 \\ 4 \cos \left(2t + \frac{\pi}{3} \right) &= 4 \\ \cos \left(2t + \frac{\pi}{3} \right) &= 1 \\ 2t + \frac{\pi}{3} &= 2 \pi \\ 2t &= 2 \pi-\frac{\pi}{3} \\ &= \frac{5 \pi}{3} \\ \therefore t &= \frac{5 \pi}{6} \end{align}

## Example 2

A particle moves in a straight line. Its displacement, $x$ metres, after $t$ seconds is given by $x = \sqrt{3} \sin 2t-\cos 2t + 3$.

(a)   Express $\ddot{x}$, the acceleration, in terms of $x$.

\begin{align} \displaystyle x &= \sqrt{3} \sin 2t-\cos 2t + 3 \\ \dot{x} &= \frac{dx}{dt} \\ &= 2 \sqrt{3} \cos 2t + 2 \sin 2t \\ \ddot{x} &= \frac{d^2x}{dt^2} \\ &= -4 \sqrt{3} \sin 2t + 4 \cos 2t \\ &= -4 \left(\sqrt{3} \sin 2t-\cos 2t \right) \\ \require{ASMsymbols} \therefore \ddot{x} &= -4 \left(x-3 \right) \end{align}

(b)   Find the centre of the motion.

$x = 3$

(c)   Find the period of the motion.

\begin{align} \displaystyle \ddot{x} &= -4(x-3) \\ &= -2^2 (x-3) \\ n &= 2 \\ T &= \frac{2 \pi}{n} \\ &= \frac{2 \pi}{2} \\ &= \pi \text{ seconds} \end{align}

(d)   Express the velocity of the particle in the form $\dot{x} = A \cos (2t – \theta)$.

\begin{align} \displaystyle A \cos (2t-\theta) &= A (\cos 2t \cos \theta + \sin 2t \sin \theta) \\ \dot{x} &= 2 \sqrt{3} \cos 2t + 2 \sin 2t \\ A \cos \theta &= 2 \sqrt{3} \\ A \sin \theta &= 2 \\ \tan \theta &= \frac{A \sin \theta}{A \cos \theta} \\ &= \frac{2}{2 \sqrt{3}} \\ &= \frac{1}{\sqrt{3}} \\ \therefore \theta &= \frac{\pi}{6} \\ A^2 \cos^2 \theta + A^2 \sin^2 \theta &= \left(2\sqrt{3} \right)^2 + 2^2 \\ A^2(\cos^2 \theta + \sin^2 \theta) &= 12 + 4 \\ A^2 &= 16 \\ A &= 4 &A\gt 0 \\ \therefore \dot{x} &= 4 \cos \left(2t-\frac{\pi}{6} \right) \end{align}

(e)   Find all times within the first $\pi$ seconds when the particle moves at 2 metres per second in either direction.

\begin{align} \displaystyle 4 \cos \left(2t-\frac{\pi}{6} \right) &= \pm 2 \\ \cos \left(2t-\frac{\pi}{6} \right) &= \pm \frac{1}{2} \\ 2t-\frac{\pi}{6} &= \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \\ 2t &= \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{9 \pi}{6}, \frac{11 \pi}{6} \\ \therefore t &= \frac{\pi}{4}, \frac{5 \pi}{12}, \frac{3 \pi}{4}, \frac{11 \pi}{12} \end{align} 