Every formula, graph, trick, and trap for IB Physics HL Topic C.1 Simple Harmonic Motion. Hand-built by an IB-specialist Singapore tutor with 15+ years of IB Physics experience.
Topic: C.1 Simple Harmonic MotionSyllabus: SL + HL (HL extension on energy + phase)Read time: ~12 minutesLast updated: Apr 2026
Simple Harmonic Motion is the gateway to every oscillation question in IB Physics HL. It underpins waves (Topic C.2–C.5), AC circuits, atomic vibrations, and even the quantum harmonic oscillator. The HL syllabus extends the SL treatment with a quantitative energy analysis ($E_T$, $E_K$, $E_P$ in terms of $\omega$, $x_0$ and $x$) and the full sinusoidal equations of motion, including a phase angle $\varphi$ set by initial conditions.
This cheatsheet condenses the full Topic C.1 syllabus — defining equation, period and frequency formulas, mass-spring and pendulum systems, energy in SHM, phase relations, and graph interpretation — into one page you can revise from. The most common HL traps (SUVAT misuse, Moon-pendulum, $E_T \propto x_0^2$) are flagged in red. Scroll to the bottom for the printable PDF download and the full Photon Academy notes library.
Acceleration is proportional to displacement and always directed towards equilibrium. Here $\omega$ is the angular frequency in $\mathrm{rad\,s^{-1}}$.
Key positions
Position
$|a|$
$|v|$
Equilibrium ($x = 0$)
0
maximum
Amplitude ($x = \pm x_0$)
maximum
0
TrickIf an exam question gives you an $a$–$x$ graph that is a straight line through the origin with negative gradient, the motion is SHM. The gradient equals $-\omega^2$.
TrapThe acceleration is not constant in SHM — it varies linearly with displacement. Never apply SUVAT (constant-acceleration) equations to an SHM problem. SUVAT misuse is the single biggest mark loss on Paper 2 SHM questions.
The signature of SHM. Acceleration is a straight line through the origin against displacement, slope $-\omega^2$ — always restoring toward equilibrium.
From the Photon question bank
What examiners actually test on this concept:
a = −ω²xRestoring toward equilibriuma–x straight lineNo SUVAT in SHM
SHM ⇔ $a = -\omega^2 x$: acceleration is proportional to displacement and always toward equilibrium. An $a$–$x$ graph that's a straight line through the origin with slope $-\omega^2$ is SHM. very frequent
Acceleration is NOT constant — never use SUVAT on SHM. This is the single biggest Paper 2 mark loss on the topic.
At equilibrium $a=0$, $v$ is max; at amplitude $a$ is max, $v=0$.
§2 — Period, Frequency & Angular Frequency C.1 SL + HL
Period–frequency relations
Period:$T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}$
Frequency:$f = \dfrac{\omega}{2\pi}$
Angular freq:$\omega = 2\pi f$
Mass–spring system
$T = 2\pi\sqrt{\dfrac{m}{k}}$ — independent of $g$ and amplitude.
Springs in parallel: $k_\text{eff} = nk$ (stiffer, shorter $T$). Springs in series: $k_\text{eff} = k/n$ (softer, longer $T$). A mass between two identical springs of constant $k$ each behaves as if $k_\text{eff} = 2k$.
Simple pendulum
$T = 2\pi\sqrt{\dfrac{l}{g}}$ — independent of mass and amplitude.
Valid only for small angles, $\theta_{\max} \lesssim 10^\circ$.
TrickTo compare periods, write $T \propto \sqrt{m/k}$ or $T \propto \sqrt{l/g}$ and use ratios. You don't need to compute $T$ numerically when the question asks "by what factor does $T$ change?".
TrapA mass-spring system on the Moon has the same period (no $g$ in the formula). A pendulum on the Moon has a longer period (smaller $g \Rightarrow$ larger $T$). This is a classic Paper 1 multiple-choice trap.
NoteIf the angular displacement of a pendulum exceeds about $10^\circ$, the motion is no longer simple harmonic and $T = 2\pi\sqrt{l/g}$ becomes inaccurate.
Two SHM systems. A mass on a spring ($T=2\pi\sqrt{m/k}$, no $g$) and a simple pendulum ($T=2\pi\sqrt{l/g}$, no mass) — each independent of amplitude.
Mass–spring $T=2\pi\sqrt{m/k}$ is independent of $g$ and amplitude; pendulum $T=2\pi\sqrt{l/g}$ is independent of mass.frequent
On the Moon: a mass–spring period is unchanged, but a pendulum's is LONGER (smaller $g$) — a classic MCQ.
Compare by ratios: $T\propto\sqrt{m/k}$ or $\sqrt{l/g}$ — no need to compute $T$ when asked "by what factor".
§3 — Energy in SHM C.1 SL (qualitative), HL (quantitative)
Qualitative (SL + HL)
Kinetic energy $E_K$ is maximum at equilibrium; potential energy $E_P$ is maximum at amplitude. Total energy $E_T = E_K + E_P$ is constant in the absence of damping. KE and PE exchange four times per cycle.
Quantitative (HL only)
Total:$E_T = \tfrac{1}{2} m \omega^2 x_0^2$
Potential:$E_P = \tfrac{1}{2} m \omega^2 x^2$
Kinetic:$E_K = \tfrac{1}{2} m \omega^2 (x_0^2 - x^2)$
Speed:$v = \pm\,\omega \sqrt{x_0^2 - x^2}$
Max speed:$v_{\max} = \omega x_0$
TrickWhen $E_K = E_P$, each equals $\tfrac{1}{2} E_T$, so $x = x_0/\sqrt{2} \approx 0.71\, x_0$. This occurs four times per cycle.
Trap$E_T \propto x_0^2$. Doubling the amplitude quadruples the total energy (and doubles the maximum speed). Do not confuse "$E_T$ is constant during oscillation at fixed amplitude" with what happens when amplitude itself changes.
Energy exchange in SHM. PE (up-parabola) and KE (down-parabola) swap continuously while their sum $E_T$ stays constant; they are equal at $x=\pm x_0/\sqrt2$.
From the Photon question bank
What examiners actually test on this concept:
E_T = ½mω²x₀²KE ↔ PE exchangeE ∝ x₀²E_K = E_P at x₀/√2
Total energy $E_T=\tfrac12 m\omega^2 x_0^2$ is constant; KE is max at equilibrium, PE at amplitude, exchanging four times per cycle. frequent (HL)
$E_T \propto x_0^2$: doubling the amplitude quadruples the energy (and doubles $v_{\max}=\omega x_0$).
$E_K=E_P$ when $x=x_0/\sqrt2 \approx 0.71x_0$ (each equals $\tfrac12 E_T$).
§4 — Phase Angle & Equations of Motion C.1 HL
General sinusoidal equations (HL)
Displacement:$x = x_0 \sin(\omega t + \varphi)$
Velocity:$v = \omega x_0 \cos(\omega t + \varphi)$
$\varphi$ is the initial phase angle (rad), set by the initial conditions at $t = 0$.
Initial condition
$\varphi$
Starts at $+x_0$ (cosine form)
$\varphi = \pi/2$
Starts at equilibrium, moving positive
$\varphi = 0$
Starts at $-x_0$
$\varphi = -\pi/2$
TrickThe phase difference between $x$ and $v$ is $\pi/2$ rad. The phase difference between $x$ and $a$ is $\pi$ rad (anti-phase). These hold for any SHM oscillator regardless of $\omega$ or amplitude.
TrapRadians only! Phase angles and $\omega t$ must be in radians. Mixing radians and degrees in a single calculation is the most common HL Paper 2 error here.
Note$v = \pm\omega \sqrt{x_0^2 - x^2}$ is derived from energy conservation and is often faster than the full sinusoidal expression when only the speed (not direction) is required.
The reference circle. A point going round a circle of radius $x_0$ at $\omega$ projects onto a diameter as $x = x_0\sin(\omega t+\varphi)$ — SHM is circular motion seen edge-on.
From the Photon question bank
What examiners actually test on this concept:
x = x₀sin(ωt+φ)φ from initial conditionsx–v phase π/2Radians only
$v$ leads $x$ by $\pi/2$; $a$ is antiphase to $x$ ($\pi$). These hold for any SHM oscillator. frequent (HL)
Set $\varphi$ from the $t=0$ conditions: start at $+x_0$ → cosine ($\varphi=\pi/2$); start at equilibrium moving positive → $\varphi=0$.
Radians only — mixing degrees into $\omega t$ is the most common HL slip.
§5 — Graphs of $x$, $v$ and $a$ C.1 SL + HL
$x$, $v$ and $a$ in SHM. $v$ is the gradient of $x$ (a quarter-period ahead); $a$ is the gradient of $v$ and exactly antiphase to $x$.
Take $x = x_0 \cos(\omega t)$ as the canonical "starts at amplitude" case. Differentiating once gives velocity, twice gives acceleration:
Quantity
Equation
Peak value
At $t = 0$
Displacement $x$
$x_0 \cos(\omega t)$
$x_0$
$+x_0$ (peak)
Velocity $v$
$-\omega x_0 \sin(\omega t)$
$\omega x_0$
$0$
Acceleration $a$
$-\omega^2 x_0 \cos(\omega t)$
$\omega^2 x_0$
$-\omega^2 x_0$ (trough)
Phase summary
$x$ and $a$ are in antiphase: $\Delta\varphi = \pi$.
$v$ leads $x$ by $\pi/2$.
$v$ and $a$ have a phase difference of $\pi/2$.
NoteOn any $a$–$x$ graph, SHM appears as a straight line through the origin with negative slope. The slope equals $-\omega^2$, so $\omega = \sqrt{-\text{gradient}}$.
From the Photon question bank
What examiners actually test on this concept:
x/v/a sinusoidsDifferentiate x → v → aQuarter-period shiftsGradient of a–x
$v$ is the gradient of $x$ (leads by $\pi/2$); $a$ is the gradient of $v$ and antiphase to $x$. Peaks: $v_{\max}=\omega x_0$, $a_{\max}=\omega^2 x_0$. very frequent
Read $\omega$ off an $a$–$x$ graph: gradient $= -\omega^2$, so $\omega=\sqrt{-\text{gradient}}$ and $T=2\pi/\omega$.
Sketch by shifting each curve a quarter-period from the one above it.
§6 — Exam Attack Plan All sections
When you see this in the question — reach for that:
Question trigger
Reach for
"Show that the motion is SHM"
Derive $a = -\omega^2 x$ from forces; identify $\omega^2$.
Compare $\varphi$; convert time lag $\Delta t$ via $\Delta\varphi = \omega \Delta t$.
Worked Example — IB-Style HL SHM Problem
Question (HL Paper 2 style — 7 marks)
A 0.250 kg mass is attached to a horizontal spring of force constant $k = 16.0\ \mathrm{N\,m^{-1}}$ on a frictionless surface. The mass is pulled 8.0 cm from equilibrium and released from rest. Calculate (a) the angular frequency $\omega$, (b) the period $T$, (c) the maximum speed of the mass, and (d) the speed of the mass when it is 4.0 cm from equilibrium.
Solution
Angular frequency from $\omega = \sqrt{k/m}$: $\omega = \sqrt{16.0 / 0.250} = \sqrt{64} = 8.00\ \mathrm{rad\,s^{-1}}$ (M1)(A1)
Period from $T = 2\pi/\omega$: $T = 2\pi / 8.00 = 0.785\ \mathrm{s}$ (A1)
Maximum speed from $v_{\max} = \omega x_0$ with $x_0 = 0.080$ m: $v_{\max} = 8.00 \times 0.080 = 0.640\ \mathrm{m\,s^{-1}}$ (M1)(A1)
Speed at $x = 0.040$ m using $v = \omega\sqrt{x_0^2 - x^2}$: $v = 8.00 \times \sqrt{0.080^2 - 0.040^2} = 8.00 \times \sqrt{0.00480} = 0.554\ \mathrm{m\,s^{-1}}$ (M1)(A1)
State result with units to 3 s.f. as required by IB conventions. (R1)
Examiner's note: The most common error is using SUVAT — "$v^2 = u^2 + 2as$" — to find the speed at $x = 4$ cm. Acceleration in SHM is not constant, so SUVAT is inadmissible and zero credit is awarded. The correct route is the energy formula $v = \pm\omega\sqrt{x_0^2 - x^2}$.
Common Student Questions
Why can I never use SUVAT for SHM?
SUVAT (the constant-acceleration kinematic equations) only works when acceleration is constant. In SHM the acceleration varies linearly with displacement, $a = -\omega^2 x$, so it is never constant. Apply the SHM equations instead — for speed at a given displacement use $v = \pm\omega\sqrt{x_0^2 - x^2}$. Substituting SHM data into SUVAT is one of the most common ways HL students lose all the marks on a 6-mark question.
Does a pendulum or a mass-spring system change period on the Moon?
A mass-spring period $T = 2\pi\sqrt{m/k}$ does not depend on $g$, so it is unchanged on the Moon. A simple pendulum $T = 2\pi\sqrt{l/g}$ depends on $g$, and gravity is weaker on the Moon, so the period becomes longer. This contrast is a classic IB Paper 1 multiple-choice trap — the examiner is testing whether you actually remember the formulas.
What happens to the total energy if I double the amplitude?
Total energy $E_T = \tfrac{1}{2} m \omega^2 x_0^2$, so $E_T \propto x_0^2$. Doubling the amplitude multiplies the total energy by four, not by two. The maximum speed $v_{\max} = \omega x_0$ also doubles. This factor-of-four scaling is one of the most heavily examined HL Topic C.1 questions.
What is the phase difference between displacement, velocity, and acceleration in SHM?
Velocity leads displacement by $\pi/2$ (90°). Acceleration is in antiphase with displacement (phase difference $\pi$, or 180°) — when $x$ is at $+x_0$, $a$ is at $-a_{\max}$. Acceleration leads velocity by $\pi/2$. These phase relationships hold for any SHM oscillator regardless of mass, spring constant, amplitude, or angular frequency.
When are kinetic and potential energy equal in SHM?
$E_K = E_P$ when each equals half of $E_T$. Setting $\tfrac{1}{2} m \omega^2 (x_0^2 - x^2) = \tfrac{1}{2} m \omega^2 x^2$ gives $x = x_0/\sqrt{2} \approx 0.71\, x_0$. This happens four times per cycle: at $+x_0/\sqrt{2}$ and $-x_0/\sqrt{2}$, each visited once on the way out and once on the way back through equilibrium.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Simple Harmonic Motion library (only available to enrolled students or via the resource library subscription) adds:
Topic C.1 Notes PDF — every concept worked through in full, with derivations of $a = -\omega^2 x$ for both spring and pendulum.
Tutorial booklet — 25+ IB-style questions sequenced from foundation to AHL difficulty.
Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/R1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
