Simple harmonic motion
Study simple harmonic motion for Year 12 Specialist Mathematics in Queensland (QCAA). This is the back-and-forth motion of a particle whose acceleration always points to a centre and is proportional to its displacement, captured by the equation x double dot equals minus n squared x.
You will learn to read the amplitude and period from a sine or cosine model, use the velocity relation to find speed at any position, and work out the maximum speed and acceleration — a central application of calculus to motion in a straight line.
Theory
Simple harmonic motion (SHM) in Year 12 Specialist Mathematics (QCAA, Queensland) is the straight-line motion whose acceleration satisfies \(\ddot{x}=-n^2x\) — acceleration always directed back to the centre and proportional to displacement. Its solutions are \(x=a\sin(nt+\varepsilon)\) or \(x=a\cos(nt+\varepsilon)\), with amplitude \(a\) and period \(\dfrac{2\pi}{n}\). This page shows how to find the amplitude, period, speed and acceleration of an oscillation.
Simple harmonic motion is motion in a straight line in which the acceleration is always directed towards a fixed centre and is proportional to the displacement \(x\) from that centre. In symbols the defining equation is \(\ddot{x}=-n^2x\), where \(n>0\) is a constant. The negative sign makes the acceleration a restoring one — it always points back to the centre.
Every SHM can be written as \(x=a\sin(nt+\varepsilon)\) or \(x=a\cos(nt+\varepsilon)\). The amplitude \(a\) is the greatest distance reached from the centre, \(\varepsilon\) is the phase (which sets the starting position), and \(n\) fixes how fast the motion cycles.
The particle oscillates between the two endpoints \(x=-a\) and \(x=a\), turning around at each. The period is the time for one complete oscillation, \(T=\dfrac{2\pi}{n}\), and the frequency is \(f=\dfrac{1}{T}=\dfrac{n}{2\pi}\).
The speed at any position is given by the velocity relation \(v^2=n^2(a^2-x^2)\). The particle moves fastest at the centre (\(x=0\)), where the speed is \(na\), and is momentarily at rest at the endpoints (\(x=\pm a\)); its acceleration is greatest in magnitude, \(n^2a\), at those endpoints.
Simple harmonic motion is defined by the acceleration equation, whose general solutions are a sine or cosine of the same \(n\):
The amplitude, period and frequency come straight from \(a\) and \(n\):
The velocity relation connects speed to position, giving the maximum speed (at the centre) and maximum acceleration (at an endpoint):
Analysing a simple harmonic motion
- Identify \(n\): compare the acceleration with \(\ddot{x}=-n^2x\) (or read \(n\) from \(x=a\sin(nt+\varepsilon)\)); \(n\) is the positive square root of the coefficient.
- Read the amplitude \(a\): it is the coefficient of the sine or cosine, or \(\sqrt{a^2}\) from the bracket in \(v^2=n^2(a^2-x^2)\).
- Apply the results: period \(T=\dfrac{2\pi}{n}\), maximum speed \(na\) at the centre, maximum acceleration \(n^2a\) at an endpoint.
- Substitute a position into \(v^2=n^2(a^2-x^2)\) for a speed, or the conditions at \(t=0\) to fix the sine/cosine form and the phase.
Read \(a\) and \(n\) from \(x=a\sin(nt)\):
| \(x\) | \(=\) | \(6\sin(2t)\) |
| \(a\) | \(=\) | \(6\) |
| \(n\) | \(=\) | \(2\) |
Period from \(T=\dfrac{2\pi}{n}\):
| \(T\) | \(=\) | \(\dfrac{2\pi}{n}\) |
| \(=\) | \(\dfrac{2\pi}{2}\) | |
| \(=\) | \(\pi\) |
Maximum speed is \(na\) (at the centre):
| \(v_{\max}\) | \(=\) | \(na\) |
| \(=\) | \(2\times 6\) | |
| \(=\) | \(12\) |
Amplitude \(6\) m, period \(\pi\) s, maximum speed \(12\) m/s.
Differentiate once for the velocity:
| \(x\) | \(=\) | \(5\cos(3t)\) |
| \(\dot{x}\) | \(=\) | \(-5\sin(3t)\times 3\) |
| \(=\) | \(-15\sin(3t)\) |
Differentiate again for the acceleration:
| \(\ddot{x}\) | \(=\) | \(-15\cos(3t)\times 3\) |
| \(=\) | \(-45\cos(3t)\) |
Write \(\ddot{x}\) as a multiple of \(x\) to match \(\ddot{x}=-n^2x\):
| \(\ddot{x}\) | \(=\) | \(-9\,(5\cos(3t))\) |
| \(=\) | \(-9x\) | |
| \(n^2\) | \(=\) | \(9\) |
| \(n\) | \(=\) | \(3\) |
\(\ddot{x}=-9x\), so the motion is simple harmonic with \(n=3\).
Compare with \(v^2=n^2(a^2-x^2)\) to read \(n\) and \(a\):
| \(n^2\) | \(=\) | \(9 \Rightarrow n=3\) |
| \(a^2\) | \(=\) | \(25 \Rightarrow a=5\) |
Maximum speed is \(na\), at the centre:
| \(v_{\max}\) | \(=\) | \(na\) |
| \(=\) | \(3\times 5\) | |
| \(=\) | \(15\) |
Substitute \(x=4\) and take the positive root:
| \(v^2\) | \(=\) | \(9(25-4^2)\) |
| \(=\) | \(9(25-16)\) | |
| \(=\) | \(9(9)=81\) | |
| \(v\) | \(=\) | \(9\) |
Amplitude \(5\) m, maximum speed \(15\) m/s, and \(9\) m/s when \(x=4\).
Amplitude is half the distance between the extremes:
| \(a\) | \(=\) | \(\dfrac{10-2}{2}\) |
| \(=\) | \(\dfrac{8}{2}\) | |
| \(=\) | \(4\) |
Centre is midway between the extremes:
| \(c\) | \(=\) | \(\dfrac{10+2}{2}\) |
| \(=\) | \(\dfrac{12}{2}\) | |
| \(=\) | \(6\) |
The period is \(12\) s; solve \(T=\dfrac{2\pi}{n}\) for \(n\):
| \(12\) | \(=\) | \(\dfrac{2\pi}{n}\) |
| \(n\) | \(=\) | \(\dfrac{2\pi}{12}\) |
| \(=\) | \(\dfrac{\pi}{6}\) |
Amplitude \(4\) m, centre \(6\) m, \(n=\dfrac{\pi}{6}\).
Common pitfalls
Frequently asked questions
What is simple harmonic motion?
It is straight-line motion in which the acceleration is always directed towards a fixed centre and is proportional to the displacement, \(\ddot{x}=-n^2x\). The solutions are \(x=a\sin(nt+\varepsilon)\) or \(x=a\cos(nt+\varepsilon)\).
How do you find the period of simple harmonic motion?
Read \(n\) from the motion, then the period is \(T=\dfrac{2\pi}{n}\). For example, \(x=4\cos(2t)\) has \(n=2\), so \(T=\dfrac{2\pi}{2}=\pi\) seconds.
Where is the particle fastest in simple harmonic motion?
At the centre of the motion (\(x=0\)), where the speed reaches its maximum value \(na\). At the endpoints \(x=\pm a\) the particle is momentarily at rest.
What are the maximum speed and maximum acceleration?
The maximum speed is \(na\), reached at the centre; the maximum acceleration has magnitude \(n^2a\), reached at the endpoints. Only \(n\) is squared in the acceleration.
How do you find the speed at a given position?
Use the velocity relation \(v^2=n^2(a^2-x^2)\). Substitute the position \(x\), then take the positive square root for the speed.
How do you get \(v^2=n^2(a^2-x^2)\) from \(\ddot{x}=-n^2x\)?
Write the acceleration as \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)=-n^2x\) and integrate to get \(\tfrac12 v^2=-\tfrac12 n^2x^2+c\). The condition \(v=0\) at \(x=a\) fixes \(c=\tfrac12 n^2a^2\), giving \(v^2=n^2(a^2-x^2)\).