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.
\( \begin{align} \displaystyle 4 \cos \left(2t-\frac{\pi}{6} \right) &= \pm 2 \\ \cos \left(2t-\frac{\pi}{6} \right) &= \pm \frac{1}{2} \\ 2t-\frac{\pi}{6} &= \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \\ 2t &= \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{9 \pi}{6}, \frac{11 \pi}{6} \\ \therefore t &= \frac{\pi}{4}, \frac{5 \pi}{12}, \frac{3 \pi}{4}, \frac{11 \pi}{12} \end{align} \)
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume
