Simple harmonic motion in HSC Mathematics Extension 2: the defining equation, displacement and velocity functions, amplitude, period and the velocity-displacement relation

What this dot point is asking

NESA wants you to model simple harmonic motion (SHM), the oscillation of a particle whose acceleration is proportional to its displacement and directed towards a fixed centre. You must use the defining equation, write displacement and velocity as functions of time, and find amplitude, period and the velocity-displacement relation.

The defining equation

A particle moves in simple harmonic motion if its acceleration satisfies

where $x$ is displacement from a fixed centre and $n > 0$ is a positive constant. The minus sign is doing all the work. It says the acceleration always points opposite to the displacement: whenever the particle is on the positive side of the centre it is being pushed back negatively, and whenever it is on the negative side it is being pushed back positively. The force is a restoring force, proportional to how far the particle has strayed, so the further out it goes the harder it is pulled back. That single property, a restoring acceleration proportional to displacement, is the entire definition of SHM, and it is exactly the equation a spring (Hooke's law) or a small-angle pendulum produces. Drop the minus sign and the solution is exponential, not oscillatory; that is the textbook contrast worth carrying.

Displacement as a function of time

The general solution of $\ddot{x} = -n^2 x$ is

where $a$ is the amplitude (maximum displacement) and $\alpha$ is a phase constant fixed by initial conditions. Equivalently it can be written $x = A\cos nt + B\sin nt$. The motion repeats every time $nt$ increases by $2\pi$, so the period is

The frequency (oscillations per unit time) is $\dfrac{1}{T} = \dfrac{n}{2\pi}$. A subtle point that earns marks: the period is independent of the amplitude. A particle swinging through a wide arc takes exactly as long per cycle as one barely moving, because a larger amplitude brings a proportionally larger restoring acceleration. This is the property that makes SHM useful for timekeeping.

Building the displacement curve, stage by stage

The figures below build the displacement-time graph one feature at a time, in the order the quantities appear in a question: first the geometry of the oscillation, then the curve, then the things you measure off it.

Stage 1, the centre and the amplitude band. Every SHM is organised around a centre (here $x = 0$) and two extremes a distance $a$ either side. The particle never leaves the band between $x = a$ and $x = -a$; it swings back and forth across the centre forever. Fixing these three lines first stops the usual error of letting the curve drift.

Stage 2, draw the displacement curve. Solving $\ddot{x} = -n^2 x$ with the particle starting at the extreme $x = a$ gives $x = a\cos nt$, the cosine curve below. It begins at the top of the band, falls through the centre, reaches the bottom extreme, and returns, smoothly and forever. Choosing the cosine here matches the common starting condition "released from rest at the extreme"; a different start just shifts the curve sideways by the phase $\alpha$.

Stage 3, read off amplitude and period. The two numbers every SHM question wants are now visible on the curve. The amplitude $a$ is the height from the centre to a peak; the period $T = \dfrac{2\pi}{n}$ is the horizontal gap between successive peaks (one full cycle). These are what you extract or are given, and what the standard formulas connect.

Stage 4, where speed is zero and where it is greatest. The velocity is the slope of this curve. At the extremes (the peaks and troughs) the curve is momentarily flat, so the velocity is zero: the particle stops to turn around. At the centre crossings the curve is steepest, so the speed is greatest, equal to $an$. This is the single most counter-intuitive fact in the topic and a favourite trap: maximum speed is at the centre, not at the extremes.

Velocity as a function of time

Differentiating $x = a\cos(nt + \alpha)$ gives

The speed reaches its maximum $an$ when $\sin(nt + \alpha) = \pm 1$, which occurs as the particle passes through the centre $x = 0$. The velocity is zero at the extremes $x = \pm a$, where the particle momentarily stops before reversing. Differentiating once more returns $\ddot{x} = -an^2\cos(nt + \alpha) = -n^2 x$, which is a quick way to confirm a proposed $x(t)$ really is SHM and to recover $n$.

The velocity-displacement relation

It is often more useful to relate speed directly to position, without time. Using $\ddot{x} = \dfrac{d}{dx}\left(\tfrac{1}{2}v^2\right)$ and integrating $\dfrac{d}{dx}\left(\tfrac{1}{2}v^2\right) = -n^2 x$ gives $\tfrac{1}{2}v^2 = -\tfrac{1}{2}n^2 x^2 + c$. Applying $v = 0$ at $x = a$ fixes the constant, yielding

This confirms maximum speed $v = an$ at the centre $x = 0$ and zero speed at $x = \pm a$. Plotted as $v$ against $x$ (the phase plane), the relation is an ellipse: the particle circulates around it once per period, fastest where the ellipse is tallest (the centre) and stationary where it crosses the $x$-axis (the extremes).

Motion not centred at the origin

Many exam problems oscillate about a centre $c$ other than the origin, with defining equation $\ddot{x} = -n^2(x - c)$. The treatment is identical once you measure displacement from the centre: the period is still $T = \frac{2\pi}{n}$, the particle oscillates between $c - a$ and $c + a$, and the velocity-displacement relation becomes $v^2 = n^2(a^2 - (x - c)^2)$. Reading the defining equation to extract $n^2$ from the coefficient and $c$ from the bracket is the first marked step in almost every SHM question, so do it carefully and state both values explicitly.

Choosing sine or cosine

The general solution can be written as $x = c + a\cos(nt + \alpha)$ or equivalently $x = c + A\cos nt + B\sin nt$. Use the cosine-with-phase form when you are given a starting position and direction, and find the phase $\alpha$ from the initial conditions, taking care to pick the value of $\alpha$ consistent with the sign of the initial velocity. The two-term form is convenient when the initial conditions are given as $x(0)$ and $\dot{x}(0)$ separately, since $A = x(0) - c$ and $B$ follows from the velocity. Both describe the same motion; choose whichever matches the data in the question.

Connecting to the velocity-displacement relation

The relation $v^2 = n^2(a^2 - x^2)$ is derived by writing acceleration as $\frac{d}{dx}\left(\frac{1}{2}v^2\right)$ and integrating, which is exactly the technique from the velocity-and-acceleration-functions dot point applied to the SHM equation. This is worth remembering because many questions ask for speed at a given position without reference to time, and the velocity-displacement relation answers them in one step, with no need to find the phase constant.

A common exam use runs the relation backwards: you are told the speed at one position and asked for the amplitude. Because $v^2 = n^2(a^2 - x^2)$ contains $a$, substituting a known $(x, v)$ pair lets you solve for $a^2$, then $a$, then the two extremes $c \pm a$. This is faster and safer than going via the displacement equation and a phase angle, so reach for it whenever time is not mentioned.

How exam questions ask about simple harmonic motion

The phrasing tells you which tool to pick up:

• "A particle moves so that $\ddot{x} = -k(x - c)$. Find the period / centre." Read $n^2 = k$ (so $n = \sqrt{k}$) and centre $c$ straight off, then $T = \frac{2\pi}{n}$. This is the one-mark opener in almost every SHM question.
• "Show that the motion is simple harmonic." Differentiate the given $x(t)$ twice and show $\ddot{x} = -n^2(x - c)$, naming $n$ and $c$. Producing the defining equation is the whole task.
• "Find the speed / velocity as it passes through ... / at $x = \ldots$" No time is mentioned, so use $v^2 = n^2\big(a^2 - (x - c)^2\big)$ in one line. Take the correct square root for the direction asked.
• "Find the amplitude" given a speed at a point. Substitute the $(x, v)$ pair into the velocity-displacement relation and solve for $a$; then the extremes are $c \pm a$.
• "Find the first time the particle is at $x = \ldots$ / reaches the centre." Now you do need time: write $x = c + a\cos(nt + \alpha)$, fix $\alpha$ from the initial position and the sign of the initial velocity, then solve for $t$, choosing the smallest positive root.
• "What distance does it travel in one period / in $n$ seconds?" In one full period a particle covers $4a$ (two extremes and back). Do not confuse distance travelled with displacement, which returns to its start each period.
• "Between which values of $x$ does it oscillate?" The extremes $c - a$ and $c + a$; find $a$ first, usually from a given speed at a given point.