Simple Harmonic Motion in Depth

Simple harmonic motion (SHM) is the most important model of oscillation in all of physics and applied mathematics — a mass on a spring, a pendulum at small angle, a floating buoy, the bob of a tuning fork, even the current in an LC circuit all obey the same defining equation. This lesson takes you well beyond the bare definition: we derive the standard kinematic results from the governing differential equation, develop the energy description, make the phase relationships between displacement, velocity and acceleration precise, introduce the phase portrait, and then open the door to damped oscillations — the bridge to the next lesson on forced motion and resonance.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M). SHM is a named topic in the Further Mechanics content, and it is one of the richest sources of Paper-3 problems because it fuses three strands the option likes to assess together: setting up an equation of motion (AO1), reasoning about energy and phase (AO2), and solving a multi-stage modelling problem in context (AO3). Paper 3 is the more problem-solving-weighted paper (AO1 40% / AO2 25% / AO3 35%), so expect questions that ask you to derive, show that, and interpret, not merely substitute. The prerequisite is the SHM you met in A-Level Mechanics / Further Mechanics 1; we build from the defining equation and assume confidence with calculus and with solving second-order linear ODEs (the further-calculus / complex-numbers toolkit).

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2 Core theory — from the defining equation to the standard results

SHM is defined by the statement that the acceleration is proportional to the displacement from a fixed equilibrium and directed back towards it:

$$\ddot{x} = -\omega^2 x, \qquad \omega > 0 .$$x¨=−ω2x,ω>0.

Here $\omega$ω is the angular frequency (units $\text{rad s}^{-1}$rad s−1). This is a second-order linear ODE with constant coefficients; its auxiliary equation is $\lambda^2 + \omega^2 = 0$λ2+ω2=0, giving $\lambda = \pm i\omega$λ=±iω, so the general solution is a combination of $\cos\omega t$cosωt and $\sin\omega t$sinωt. It is cleanest to write it in amplitude–phase form:

$$x = a\sin(\omega t + \varepsilon),$$x=asin(ωt+ε),

where $a>0$a>0 is the amplitude and $\varepsilon$ε the phase constant fixed by the initial conditions. (Some boards write $x = a\cos(\omega t + \phi)$x=acos(ωt+ϕ); the two differ only by the choice of where $t=0$t=0 sits and by $\varepsilon = \phi + \tfrac{\pi}{2}$ε=ϕ+2π​.) Differentiating,

$$v = \dot{x} = a\omega\cos(\omega t + \varepsilon), \qquad \ddot{x} = -a\omega^2\sin(\omega t + \varepsilon) = -\omega^2 x \ \checkmark$$v=x˙=aωcos(ωt+ε),x¨=−aω2sin(ωt+ε)=−ω2x ✓

so the assumed form indeed satisfies the defining equation. The motion is periodic with period and frequency

$$T = \frac{2\pi}{\omega}, \qquad f = \frac{1}{T} = \frac{\omega}{2\pi}.$$T=ω2π​,f=T1​=2πω​.

A crucial relation eliminates time. From $x = a\sin(\omega t+\varepsilon)$x=asin(ωt+ε) and $v = a\omega\cos(\omega t+\varepsilon)$v=aωcos(ωt+ε), square and add using $\sin^2+\cos^2=1$sin2+cos2=1:

$$\frac{x^2}{a^2} + \frac{v^2}{a^2\omega^2} = 1 \ \Longrightarrow\ \boxed{\,v^2 = \omega^2\left(a^2 - x^2\right)\,}.$$a2x2​+a2ω2v2​=1 ⟹ v2=ω2(a2−x2)​.

This is the single most useful SHM result: the speed is maximal $v_{\max} = a\omega$vmax​=aω at the centre $(x=0)$(x=0) and zero at the extremes $(x=\pm a)$(x=±a), and it lets you find a speed at a given position without ever solving for $t$t.

There is a second, calculus-only route to the same identity that is worth seeing because it works for any force law, not just the linear one. Write the acceleration as $\ddot{x} = v\dfrac{\mathrm{d}v}{\mathrm{d}x}$x¨=vdxdv​ (the chain rule applied to $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}v}{\mathrm{d}x}\dfrac{\mathrm{d}x}{\mathrm{d}t}$dtdv​=dxdv​dtdx​). Then the defining equation becomes a separable first-order ODE in $v$v and $x$x:

$$v\frac{\mathrm{d}v}{\mathrm{d}x} = -\omega^2 x \ \Longrightarrow\ \int v\,\mathrm{d}v = -\omega^2\!\int x\,\mathrm{d}x \ \Longrightarrow\ \tfrac12 v^2 = -\tfrac12\omega^2 x^2 + C .$$vdxdv​=−ω2x ⟹ ∫vdv=−ω2∫xdx ⟹ 21​v2=−21​ω2x2+C.

Applying $v = 0$v=0 at $x = a$x=a gives $C = \tfrac12\omega^2 a^2$C=21​ω2a2, and rearranging recovers $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2) once more. Multiplying through by $m$m turns this very line into the energy equation of the next subsection — the $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx trick is the work–energy theorem in disguise.

Energy in SHM

Take the spring model $m\ddot{x} = -kx$mx¨=−kx, so $\omega^2 = k/m$ω2=k/m. At displacement $x$x:

$$\text{KE} = \tfrac12 m v^2 = \tfrac12 m\omega^2(a^2 - x^2), \qquad \text{PE} = \tfrac12 k x^2 = \tfrac12 m\omega^2 x^2 ,$$KE=21​mv2=21​mω2(a2−x2),PE=21​kx2=21​mω2x2,

(the PE is the elastic energy $\int_0^x kx'\,\mathrm{d}x' = \tfrac12 kx^2$∫0x​kx′dx′=21​kx2). Their sum is constant:

$$E = \text{KE} + \text{PE} = \tfrac12 m\omega^2(a^2 - x^2) + \tfrac12 m\omega^2 x^2 = \boxed{\tfrac12 m\omega^2 a^2}.$$E=KE+PE=21​mω2(a2−x2)+21​mω2x2=21​mω2a2​.

Energy sloshes back and forth between kinetic and potential while the total stays fixed — the mathematical signature of an undamped conservative oscillator.

Quantity	In terms of $x$x	Maximum	Where
Kinetic energy	$\tfrac12 m\omega^2(a^2-x^2)$21​mω2(a2−x2)	$\tfrac12 m\omega^2 a^2$21​mω2a2	$x=0$x=0
Potential energy	$\tfrac12 m\omega^2 x^2$21​mω2x2	$\tfrac12 m\omega^2 a^2$21​mω2a2	$x=\pm a$x=±a
Total energy	$\tfrac12 m\omega^2 a^2$21​mω2a2	constant	everywhere

A plot of KE against $x$x is a downward parabola peaking at the centre; PE is an upward parabola; total energy is a horizontal line, the two parabolas always summing to it. Against time, both KE and PE are sinusoidal with period $T/2$T/2 — half the period of the displacement — because they depend on the squares $\sin^2$sin2 and $\cos^2$cos2. Explicitly, with $x = a\sin\omega t$x=asinωt,

$$\text{PE} = \tfrac12 m\omega^2 a^2\sin^2\omega t = \tfrac14 m\omega^2 a^2\bigl(1 - \cos 2\omega t\bigr),$$PE=21​mω2a2sin2ωt=41​mω2a2(1−cos2ωt), $$\text{KE} = \tfrac12 m\omega^2 a^2\cos^2\omega t = \tfrac14 m\omega^2 a^2\bigl(1 + \cos 2\omega t\bigr),$$KE=21​mω2a2cos2ωt=41​mω2a2(1+cos2ωt),

using the double-angle identities $\sin^2\theta = \tfrac12(1-\cos2\theta)$sin2θ=21​(1−cos2θ) and $\cos^2\theta = \tfrac12(1+\cos2\theta)$cos2θ=21​(1+cos2θ). The two share a common mean value $\tfrac14 m\omega^2 a^2$41​mω2a2 (exactly half the total energy) and oscillate about it in antiphase at angular frequency $2\omega$2ω — which is why their period is $T/2$T/2. This is a favourite "show that" target: a question will give the displacement and ask you to show that the kinetic energy varies at twice the frequency of the motion.

The small-angle pendulum

The most famous physical realisation of SHM is the simple pendulum. A bob of mass $m$m on a light string of length $\ell$ℓ, displaced by angle $\theta$θ from the vertical, experiences a tangential restoring force $-mg\sin\theta$−mgsinθ, so Newton's second law along the arc $s = \ell\theta$s=ℓθ gives $m\ell\ddot{\theta} = -mg\sin\theta$mℓθ¨=−mgsinθ. For small $\theta$θ, $\sin\theta \approx \theta$sinθ≈θ (radians), so

$$\ddot{\theta} \approx -\frac{g}{\ell}\theta ,$$θ¨≈−ℓg​θ,

which is SHM with $\omega = \sqrt{g/\ell}$ω=g/ℓ​ and hence $T = 2\pi\sqrt{\ell/g}$T=2πℓ/g​ — independent of both the mass and (to this approximation) the amplitude. The approximation $\sin\theta\approx\theta$sinθ≈θ is the entire reason the period is amplitude-independent; the cubic correction $\sin\theta = \theta - \theta^3/6 + \cdots$sinθ=θ−θ3/6+⋯ is what reintroduces a (tiny) amplitude dependence at large swings, the same effect discussed in the stretch section.

Phase relationships

Write $x = a\sin\omega t$x=asinωt (taking $\varepsilon=0$ε=0 for clarity). Then

$$v = a\omega\cos\omega t = a\omega\sin\!\left(\omega t + \tfrac{\pi}{2}\right), \qquad \ddot{x} = -a\omega^2\sin\omega t = a\omega^2\sin\!\left(\omega t + \pi\right).$$v=aωcosωt=aωsin(ωt+2π​),x¨=−aω2sinωt=aω2sin(ωt+π).

So velocity leads displacement by $\tfrac{\pi}{2}$2π​ (a quarter period — velocity is greatest as the particle sweeps through the centre, where displacement is zero) and acceleration is in antiphase with displacement (phase difference $\pi$π — it always points back towards equilibrium). Two separate oscillations are in phase if their phase difference is $0$0 (or $2n\pi$2nπ), in antiphase if it is $\pi$π (or $(2n+1)\pi$(2n+1)π), and in quadrature if it is $\tfrac{\pi}{2}$2π​.

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3 Worked examples with mark scheme

Example 1 — energy of an oscillator

A particle of mass $0.4\,\text{kg}$0.4kg performs SHM with amplitude $0.1\,\text{m}$0.1m and period $0.5\,\text{s}$0.5s. Find (a) the total energy, (b) the kinetic energy when $x = 0.06\,\text{m}$x=0.06m.

$$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \ \text{rad s}^{-1}. \quad (\textbf{M1})$$ω=T2π​=0.52π​=4π rad s−1.(M1)

(a) Using $E = \tfrac12 m\omega^2 a^2$E=21​mω2a2:

$$E = \tfrac12 (0.4)(4\pi)^2(0.1)^2 = 0.2 \times 16\pi^2 \times 0.01 = 0.032\pi^2 \approx 0.316\ \text{J}. \quad (\textbf{M1 A1})$$E=21​(0.4)(4π)2(0.1)2=0.2×16π2×0.01=0.032π2≈0.316 J.(M1 A1)

(b) Using $\text{KE} = \tfrac12 m\omega^2(a^2 - x^2)$KE=21​mω2(a2−x2):

$$\text{KE} = 0.2 \times 16\pi^2 \times (0.01 - 0.0036) = 3.2\pi^2 \times 0.0064 = 0.02048\pi^2 \approx 0.202\ \text{J}. \quad (\textbf{M1 A1})$$KE=0.2×16π2×(0.01−0.0036)=3.2π2×0.0064=0.02048π2≈0.202 J.(M1 A1)

(M1 for $\omega = 2\pi/T$ω=2π/T; M1 for the correct energy formula; A1 $\approx 0.316\,\text{J}$≈0.316J; M1 for $a^2-x^2$a2−x2; A1 $\approx 0.202\,\text{J}$≈0.202J. Note $0.032\pi^2$0.032π2, not $0.32\pi^2$0.32π2 — a power-of-ten slip here is the commonest lost A-mark.)

Example 2 — building the equation and finding a speed

A particle of mass $0.5\,\text{kg}$0.5kg is attached to a spring of stiffness $k = 18\,\text{N m}^{-1}$k=18N m−1 and oscillates on a smooth horizontal table with amplitude $0.2\,\text{m}$0.2m. Find (a) the period, (b) the maximum speed, (c) the speed when the particle is $0.12\,\text{m}$0.12m from equilibrium.

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{18}{0.5}} = \sqrt{36} = 6\ \text{rad s}^{-1}. \quad (\textbf{M1 A1})$$ω=mk​​=0.518​​=36​=6 rad s−1.(M1 A1)

(a) $\displaystyle T = \frac{2\pi}{\omega} = \frac{2\pi}{6} = \frac{\pi}{3} \approx 1.05\ \text{s}.$T=ω2π​=62π​=3π​≈1.05 s. (A1)

(b) $v_{\max} = a\omega = 0.2 \times 6 = 1.2\ \text{m s}^{-1}.$vmax​=aω=0.2×6=1.2 m s−1. (M1 A1)

(c) Using $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2):

$$v^2 = 36\,(0.2^2 - 0.12^2) = 36\,(0.04 - 0.0144) = 36 \times 0.0256 = 0.9216,$$v2=36(0.22−0.122)=36(0.04−0.0144)=36×0.0256=0.9216, $$v = \sqrt{0.9216} = 0.96\ \text{m s}^{-1}. \quad (\textbf{M1 A1})$$v=0.9216​=0.96 m s−1.(M1 A1)

(M1 for $\omega=\sqrt{k/m}$ω=k/m​; A1 $\omega=6$ω=6; A1 period; M1 A1 $v_{\max}$vmax​; M1 for $v^2=\omega^2(a^2-x^2)$v2=ω2(a2−x2), A1 $0.96\,\text{m s}^{-1}$0.96m s−1.)

Example 3 — time to a given displacement

An oscillator has $\omega = 5\,\text{rad s}^{-1}$ω=5rad s−1, amplitude $0.08\,\text{m}$0.08m, and starts from the centre moving in the positive direction. Find the first time at which $x = 0.04\,\text{m}$x=0.04m.

With $x(0)=0$x(0)=0 and $\dot{x}(0)>0$x˙(0)>0 we take $x = a\sin\omega t = 0.08\sin 5t$x=asinωt=0.08sin5t. (M1)

$$0.04 = 0.08\sin 5t \ \Rightarrow\ \sin 5t = \tfrac12 \ \Rightarrow\ 5t = \tfrac{\pi}{6} \ \Rightarrow\ t = \frac{\pi}{30} \approx 0.105\ \text{s}. \quad (\textbf{M1 A1})$$0.04=0.08sin5t ⇒ sin5t=21​ ⇒ 5t=6π​ ⇒ t=30π​≈0.105 s.(M1 A1)

(M1 for choosing the sine form to match "starts from centre"; M1 for solving $\sin 5t = \tfrac12$sin5t=21​; A1 for the first root $t=\pi/30$t=π/30. Taking $5t = 5\pi/6$5t=5π/6 gives a later crossing and loses the A-mark.)

Example 4 — combining two starting data into amplitude and phase

A particle moves with SHM of angular frequency $\omega = 2\,\text{rad s}^{-1}$ω=2rad s−1. At $t = 0$t=0 it is at $x = 3\,\text{m}$x=3m moving with velocity $v = 8\,\text{m s}^{-1}$v=8m s−1. Express the motion in the form $x = a\sin(2t + \varepsilon)$x=asin(2t+ε) and hence find the amplitude.

At $t=0$t=0, $x = a\sin\varepsilon = 3$x=asinε=3 and $v = a\omega\cos\varepsilon = 2a\cos\varepsilon = 8$v=aωcosε=2acosε=8, so $a\cos\varepsilon = 4$acosε=4. (M1) Squaring and adding,

$$a^2\sin^2\varepsilon + a^2\cos^2\varepsilon = 3^2 + 4^2 = 25 \ \Rightarrow\ a^2 = 25 \ \Rightarrow\ a = 5\ \text{m}. \quad (\textbf{M1 A1})$$a2sin2ε+a2cos2ε=32+42=25 ⇒ a2=25 ⇒ a=5 m.(M1 A1)

Dividing, $\tan\varepsilon = \dfrac{a\sin\varepsilon}{a\cos\varepsilon} = \dfrac{3}{4}$tanε=acosεasinε​=43​, so $\varepsilon = \arctan\tfrac34 \approx 0.6435\,\text{rad}$ε=arctan43​≈0.6435rad (first quadrant, since both $\sin\varepsilon>0$sinε>0 and $\cos\varepsilon>0$cosε>0). Thus $x = 5\sin(2t + 0.6435)$x=5sin(2t+0.6435). (A1)

(M1 for the two initial-condition equations; M1 for the square-and-add elimination; A1 $a = 5$a=5; A1 for the correct quadrant for $\varepsilon$ε. A quick check: $v_{\max} = a\omega = 10 > 8$vmax​=aω=10>8, consistent.)

This square-and-add device — using $\sin^2\varepsilon + \cos^2\varepsilon = 1$sin2ε+cos2ε=1 to extract the amplitude and $\tan\varepsilon$tanε to extract the phase — is the workhorse for any "find the amplitude and phase from initial conditions" part, and it is exactly the $R$R-formula from A-Level trigonometry.

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4 Specimen-style exam question

(specimen-style — not from any past paper)

A light spring of natural length $0.5\,\text{m}$0.5m and stiffness $k$k hangs vertically with a particle of mass $0.25\,\text{kg}$0.25kg attached. In equilibrium the spring extends by $0.05\,\text{m}$0.05m. The particle is pulled down a further $0.03\,\text{m}$0.03m and released from rest. Taking $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2: (a) show that the motion is simple harmonic and find $\omega$ω; (b) write down the amplitude and period; (c) find the maximum speed and the speed when the particle is $0.02\,\text{m}$0.02m above the equilibrium position.

Model solution. In equilibrium the tension balances the weight: $k(0.05) = (0.25)(9.8)$k(0.05)=(0.25)(9.8), so $k = \dfrac{2.45}{0.05} = 49\ \text{N m}^{-1}$k=0.052.45​=49 N m−1. Displace the particle a distance $x$x below equilibrium; the extension is $0.05 + x$0.05+x, so the net downward force is

$$F = mg - k(0.05 + x) = \underbrace{mg - 0.05k}_{=\,0} - kx = -kx .$$F=mg−k(0.05+x)==0mg−0.05k​​−kx=−kx.

Hence $m\ddot{x} = -kx$mx¨=−kx, i.e. $\ddot{x} = -\dfrac{k}{m}x = -\dfrac{49}{0.25}x = -196x$x¨=−mk​x=−0.2549​x=−196x. This is SHM with $\omega^2 = 196$ω2=196, so $\omega = 14\ \text{rad s}^{-1}$ω=14 rad s−1. (a)

The particle is released from rest $0.03\,\text{m}$0.03m from equilibrium, so amplitude $a = 0.03\,\text{m}$a=0.03m and $T = \dfrac{2\pi}{14} = \dfrac{\pi}{7} \approx 0.449\,\text{s}$T=142π​=7π​≈0.449s. (b)

Maximum speed $v_{\max} = a\omega = 0.03 \times 14 = 0.42\ \text{m s}^{-1}$vmax​=aω=0.03×14=0.42 m s−1. At $x = 0.02\,\text{m}$x=0.02m (the sign is irrelevant since only $x^2$x2 enters),

$$v^2 = \omega^2(a^2 - x^2) = 196\,(0.03^2 - 0.02^2) = 196\,(0.0009 - 0.0004) = 196 \times 0.0005 = 0.098,$$v2=ω2(a2−x2)=196(0.032−0.022)=196(0.0009−0.0004)=196×0.0005=0.098, $$v = \sqrt{0.098} \approx 0.313\ \text{m s}^{-1}. \quad \textbf{(c)}$$v=0.098​≈0.313 m s−1.(c)

The "show that" in (a) is won by deriving $F = -kx$F=−kx and stating that this is the SHM condition — the cancellation $mg = 0.05k$mg=0.05k is the key line examiners look for.

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5 Synoptic links

• Second-order ODEs (further calculus / complex numbers). $\ddot{x} = -\omega^2 x$x¨=−ω2x is solved by the auxiliary equation $\lambda^2+\omega^2 = 0\Rightarrow\lambda=\pm i\omega$λ2+ω2=0⇒λ=±iω; the complex roots produce the sine/cosine solution. Damped SHM (below) is the same machinery with real or complex roots.
• Circular motion. The projection of uniform circular motion onto a diameter is SHM: a point moving round a circle of radius $a$a at angular speed $\omega$ω has shadow $x = a\sin\omega t$x=asinωt. This is why $\omega$ω is called the angular frequency.
• Energy methods (mechanics). The constancy of $\tfrac12 m\omega^2 a^2$21​mω2a2 is conservation of energy; the same $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx integration you use for variable forces re-derives $v^2=\omega^2(a^2-x^2)$v2=ω2(a2−x2).
• Trigonometric identities. Combining $x = P\cos\omega t + Q\sin\omega t$x=Pcosωt+Qsinωt into $a\sin(\omega t+\varepsilon)$asin(ωt+ε) uses the $R$R-formula $R=\sqrt{P^2+Q^2}$R=P2+Q2​, a direct A-Level Maths link, and is exactly what Example 4 carried out.
• Damped and forced oscillations (next lesson). Adding a resistance term gives $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$x¨+2γx˙+ω02​x=0; the auxiliary equation $\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0$λ2+2γλ+ω02​=0 has discriminant $4(\gamma^2 - \omega_0^2)$4(γ2−ω02​), and its sign decides between under-, critical and over-damping. Pure SHM is the special case $\gamma = 0$γ=0, where the roots are $\pm i\omega_0$±iω0​ and the oscillation never decays. Keep that connection in view — the algebra you have practised here is the $\gamma=0$γ=0 corner of the richer picture you meet next.

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6 Mark-scheme literacy

• "Show that the motion is simple harmonic" demands you reach the form $\ddot{x} = -(\text{positive constant})\,x$x¨=−(positive constant)x and explicitly identify that constant as $\omega^2$ω2. Reaching $\ddot{x}=-196x$x¨=−196x and stopping loses the final mark — say "this is SHM with $\omega = 14$ω=14".
• Prefer $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2) over solving for $t$t whenever a question links a speed to a position. It is faster and avoids inverse-trig branch errors.
• Exact vs decimal. Leave $T = \pi/7$T=π/7 and convert to $0.449\,\text{s}$0.449s only at the end; premature rounding of $\omega$ω propagates into every later part (follow-through marks may rescue you, but only if your method is clearly shown).
• Units and direction. Quote $\text{rad s}^{-1}$rad s−1 for $\omega$ω; a missing or wrong unit can cost the A-mark even with the right number.

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7 Grade-band model answers

Question: "A particle moving with SHM has speed $0.3\,\text{m s}^{-1}$0.3m s−1 at a point $0.04\,\text{m}$0.04m from equilibrium and speed $0.16\,\text{m s}^{-1}$0.16m s−1 at a point $0.06\,\text{m}$0.06m from equilibrium. Find the amplitude and the period."

Mid-band. "Using $v^2 = \omega^2(a^2-x^2)$v2=ω2(a2−x2). So $0.09 = \omega^2(a^2 - 0.0016)$0.09=ω2(a2−0.0016) and $0.0256 = \omega^2(a^2 - 0.0036)$0.0256=ω2(a2−0.0036). Dividing, $\frac{0.09}{0.0256} = \frac{a^2-0.0016}{a^2-0.0036}$0.02560.09​=a2−0.0036a2−0.0016​. Solving gives $a \approx 0.07$a≈0.07 and then $\omega \approx 5.2$ω≈5.2, so $T \approx 1.2\,\text{s}$T≈1.2s."

Examiner-style commentary: The right equation and a correct elimination strategy, but the working is compressed — the intermediate algebra between "dividing" and the final $a$a is missing, so any slip cannot be followed through and the precision is unconvincing. Earns the method marks; risks an accuracy mark for un-shown work.

Stronger. Sets up both equations, then writes the elimination cleanly: $0.09(a^2 - 0.0036) = 0.0256(a^2-0.0016)\Rightarrow 0.0644\,a^2 = 0.0003176\Rightarrow a^2 = 0.004932\Rightarrow a = 0.0702\,\text{m}$0.09(a2−0.0036)=0.0256(a2−0.0016)⇒0.0644a2=0.0003176⇒a2=0.004932⇒a=0.0702m; substitutes back to get $\omega^2 = 0.09/(0.004932-0.0016)=27.0$ω2=0.09/(0.004932−0.0016)=27.0, $\omega = 5.20$ω=5.20, $T = 2\pi/\omega = 1.21\,\text{s}$T=2π/ω=1.21s.

Examiner-style commentary: Full marks. Every line of the elimination is shown, so any arithmetic slip would be followed through, and both final answers carry units and sensible precision.

Top-band. As Stronger, but first states the plan: "Two unknowns $a,\omega$a,ω; two data points give two equations in $a^2$a2 and $\omega^2$ω2; dividing eliminates $\omega^2$ω2." Carries exact values as far as possible, gives $a = 0.0702\,\text{m}$a=0.0702m and $T = 1.21\,\text{s}$T=1.21s to 3 s.f., and adds a one-line check: "$v_{\max} = a\omega = 0.365\,\text{m s}^{-1}$vmax​=aω=0.365m s−1, comfortably above both given speeds, as it must be."

Examiner-style commentary: This is how a top candidate communicates — strategy stated, exact working, units, and a sanity check that the maximum speed exceeds the quoted speeds. The reasoning is auditable end to end; it would also score the AO2 communication credit Paper 3 rewards.

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8 Common misconceptions

1. "Amplitude is the total distance travelled in a cycle." No — amplitude is the maximum displacement from equilibrium. The distance covered in one full period is $4a$4a.
2. "Maximum speed occurs at the endpoints." The opposite: speed is greatest at the centre $(x=0)$(x=0) and zero at the extremes; acceleration is greatest at the extremes and zero at the centre.
3. "Period depends on amplitude." For SHM, $T = 2\pi/\omega$T=2π/ω is independent of amplitude (isochronism). Amplitude affects speeds and energy, not the period.
4. "$\omega$ω is the frequency." $\omega$ω is the angular frequency in $\text{rad s}^{-1}$rad s−1; the ordinary frequency is $f = \omega/2\pi$f=ω/2π in hertz. Mixing them gives answers out by $2\pi$2π.
5. "Energy is $\tfrac12 m\omega^2 x^2$21​mω2x2." That is the potential energy at displacement $x$x; the total energy is $\tfrac12 m\omega^2 a^2$21​mω2a2, a constant.
6. "KE and PE have the same period as the displacement." They vary as $\sin^2$sin2 / $\cos^2$cos2, so their period is $T/2$T/2.
7. "Acceleration is constant." SHM acceleration $\ddot{x}=-\omega^2 x$x¨=−ω2x varies with position — it is the defining feature; SUVAT (constant-acceleration) formulae do not apply. Students who reach for $v^2 = u^2 + 2as$v2=u2+2as on an SHM problem are using the wrong toolkit; the correct position–speed relation is $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2).
8. "The equilibrium position is where the spring is at natural length." For a vertical spring the equilibrium is where the net force is zero — i.e. where tension balances weight, which is at extension $mg/k$mg/k, not at natural length. Measuring $x$x from the wrong origin breaks the cancellation that produces $\ddot{x} = -\omega^2 x$x¨=−ω2x and is the single most common modelling error in vertical-SHM questions, as the specimen question above is designed to test.

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9 Common errors

• Forgetting to square $\omega$ω: writing $v^2 = \omega(a^2-x^2)$v2=ω(a2−x2) instead of $\omega^2(a^2-x^2)$ω2(a2−x2).
• A power-of-ten slip in energy (as in Example 1: $0.032\pi^2$0.032π2, not $0.32\pi^2$0.32π2).
• Working in degrees: $\omega t$ωt is in radians; set your calculator to radians for $\sin$sin, $\cos$cos of phase.
• Taking the wrong inverse-trig root and reporting a later crossing time instead of the first.
• Dropping the factor of a half in PE $=\int kx\,\mathrm{d}x = \tfrac12 kx^2$=∫kxdx=21​kx2 (writing $kx^2$kx2).
• Confusing amplitude $a$a with the instantaneous displacement $x$x in $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2).

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10 Going further (stretch)

The phase portrait — a plot of $v$v against $x$x — is the entry point to dynamical-systems thinking used throughout university physics. For SHM the trajectory is the ellipse

$$\frac{x^2}{a^2} + \frac{v^2}{a^2\omega^2} = 1,$$a2x2​+a2ω2v2​=1,

traced clockwise once per period. The area enclosed by this ellipse, by the standard $\pi \times (\text{semi-axes})$π×(semi-axes) formula, is $\pi \cdot a \cdot a\omega = \pi a^2\omega$π⋅a⋅aω=πa2ω; since the energy is $E = \tfrac12 m\omega^2 a^2$E=21​mω2a2, the enclosed area is $\dfrac{2\pi E}{m\omega}$mω2πE​ — directly proportional to the energy at fixed $\omega$ω. This area-equals-action idea is a baby version of the action variable of Hamiltonian mechanics, the quantity that becomes $(n+\tfrac12)\hbar$(n+21​)ℏ when the oscillator is quantised, giving the equally spaced energy levels of the quantum harmonic oscillator. A STEP-flavoured question might ask: if the spring is slightly non-linear, $\ddot{x} = -\omega^2 x - \beta x^3$x¨=−ω2x−βx3 with small $\beta>0$β>0, is the period still independent of amplitude? An energy argument shows the restoring force stiffens at large $x$x, so the period now decreases with amplitude — the loss of isochronism that makes the pendulum clock only approximately accurate, and the seed of perturbation theory. Even the identity "SHM is the shadow of circular motion" generalises: projecting motion on an ellipse gives two SHMs in quadrature, and combining perpendicular SHMs of commensurate frequencies traces the Lissajous figures seen on an oscilloscope.

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11 Additional A* practice

1. A particle performs SHM of period $\dfrac{\pi}{2}\,\text{s}$2π​s and amplitude $0.05\,\text{m}$0.05m. Find its maximum speed and maximum acceleration. Answer. $\omega = 2\pi/T = 4$ω=2π/T=4. $v_{\max}=a\omega = 0.2\,\text{m s}^{-1}$vmax​=aω=0.2m s−1; $|\ddot{x}|_{\max}=a\omega^2 = 0.05\times16 = 0.8\,\text{m s}^{-2}$∣x¨∣max​=aω2=0.05×16=0.8m s−2.

2. An oscillator of mass $0.3\,\text{kg}$0.3kg has total energy $0.6\,\text{J}$0.6J and amplitude $0.1\,\text{m}$0.1m. Find $\omega$ω and the speed at $x = 0.05\,\text{m}$x=0.05m. Answer. $E=\tfrac12 m\omega^2 a^2\Rightarrow 0.6 = 0.5(0.3)\omega^2(0.01)\Rightarrow \omega^2 = 400,\ \omega = 20$E=21​mω2a2⇒0.6=0.5(0.3)ω2(0.01)⇒ω2=400, ω=20. Then $v^2 = 400(0.01-0.0025)=3$v2=400(0.01−0.0025)=3, $v=\sqrt3\approx1.73\,\text{m s}^{-1}$v=3​≈1.73m s−1.

3. A point performs SHM about $O$O. It is $0.1\,\text{m}$0.1m from $O$O moving at $0.6\,\text{m s}^{-1}$0.6m s−1, and $0.15\,\text{m}$0.15m from $O$O moving at $0.4\,\text{m s}^{-1}$0.4m s−1. Find the amplitude. Answer. $0.36 = \omega^2(a^2-0.01)$0.36=ω2(a2−0.01), $0.16 = \omega^2(a^2-0.0225)$0.16=ω2(a2−0.0225). Dividing: $\frac{0.36}{0.16}=\frac{a^2-0.01}{a^2-0.0225}\Rightarrow 2.25(a^2-0.0225)=a^2-0.01\Rightarrow 1.25a^2 = 0.040625\Rightarrow a^2=0.0325,\ a\approx0.180\,\text{m}$0.160.36​=a2−0.0225a2−0.01​⇒2.25(a2−0.0225)=a2−0.01⇒1.25a2=0.040625⇒a2=0.0325, a≈0.180m.

4. A particle starts at $x = a$x=a at $t=0$t=0 with SHM of angular frequency $\omega$ω. Show that it first reaches $x = a/2$x=a/2 at $t = \dfrac{\pi}{3\omega}$t=3ωπ​. Answer. $x = a\cos\omega t$x=acosωt (starts at amplitude). $\tfrac12 = \cos\omega t \Rightarrow \omega t = \tfrac{\pi}{3}\Rightarrow t = \pi/(3\omega)$21​=cosωt⇒ωt=3π​⇒t=π/(3ω). $\checkmark$✓

5. Show that for SHM the time-average of the kinetic energy over a complete period equals the time-average of the potential energy, each being half the total. Answer. $\langle\sin^2\rangle = \langle\cos^2\rangle = \tfrac12$⟨sin2⟩=⟨cos2⟩=21​ over a period, so $\langle\text{KE}\rangle = \tfrac12 m\omega^2 a^2\langle\cos^2\rangle = \tfrac14 m\omega^2 a^2 = \tfrac12 E$⟨KE⟩=21​mω2a2⟨cos2⟩=41​mω2a2=21​E, and likewise $\langle\text{PE}\rangle = \tfrac12 E$⟨PE⟩=21​E. $\checkmark$✓

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12 Board alignment

Written for AQA A-Level Further Mathematics 7367, Paper 3 Mechanics option (7367/3M). The same SHM content appears in Edexcel Further Mechanics 2 (9FM0/4B) and OCR(A)/OCR(B) MEI Further Mechanics; the formulae $v^2=\omega^2(a^2-x^2)$v2=ω2(a2−x2), $T=2\pi/\omega$T=2π/ω and $E=\tfrac12 m\omega^2 a^2$E=21​mω2a2 are board-independent, so this lesson transfers directly.

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13 Visual summary

$$\boxed{\;\ddot{x} = -\omega^2 x \;\Rightarrow\; x = a\sin(\omega t+\varepsilon),\quad v^2=\omega^2(a^2-x^2),\quad T=\tfrac{2\pi}{\omega},\quad E=\tfrac12 m\omega^2 a^2\;}$$x¨=−ω2x⇒x=asin(ωt+ε),v2=ω2(a2−x2),T=ω2π​,E=21​mω2a2​

Position $x$x	Speed	Acceleration	KE	PE
$0$0 (centre)	$a\omega$aω (max)	$0$0	max	$0$0
$\pm a$±a (ends)	$0$0	$\omega^2 a$ω2a (max)	$0$0	max

graph LR
  A["x = +a<br/>v = 0, accel max inward"] -->|"quarter period"| B["x = 0<br/>v = max, accel = 0"]
  B -->|"quarter period"| C["x = -a<br/>v = 0, accel max inward"]
  C -->|"quarter period"| D["x = 0<br/>v = max (other way)"]
  D -->|"quarter period"| A

Exam tip. Reach for $v^2=\omega^2(a^2-x^2)$v2=ω2(a2−x2) for any speed-at-a-position part; use $x=a\sin\omega t$x=asinωt or $a\cos\omega t$acosωt (matched to the start condition) only when you genuinely need time.