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NotesPhysics HLTopic 3.1SHM graphs, phase and timing
Back to Physics HL Topics
3.1.38 min read

SHM graphs, phase and timing (Physics HL)

IB Physics · Unit 3

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Contents

  1. 1The three SHM graphs
  2. 2Quarter-cycles and timing
  3. 3Exam-style question
The big idea: Set a mass bouncing on a spring and film it: it dips, rises, dips again, the same trip over and over. That is simple harmonic motion (SHM).
QuantityWhat it means
displacement xhow far from the centre
velocity vhow fast, and which way
acceleration ahow its velocity is changing
Three shifted waves: All three change as it swings. Plotted against time, all three are smooth waves (sinusoids) — but shifted relative to one another.
New words — equilibrium, phase, antiphase: Equilibrium = the centre, where the object would rest if it weren't moving (x = 0).

Phase = where you are in the cycle. Two curves are in phase if they peak together, and antiphase if one peaks exactly as the other troughs (half a cycle apart).

Phase angle φ (HL) = where in the cycle the motion is at t = 0, as an angle in radians (one cycle = 2π).

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The two phase rules to remember: Velocity leads displacement by a quarter-cycle (90°) — v is biggest at the centre, zero at the ends.

Acceleration is antiphase to displacement (180°) — a always points back toward the centre, opposite to x. That's the SHM rule a = -ω²x.

Sketching v from x

An x–t sine curve above and the v–t graph below, built from where x–t is flat (v = 0) and steepest (v largest)
v is zero where x–t is flat and largest where it is steepest: the v–t graph is a cosine, a quarter period ahead.

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Each full oscillation takes one period T. Because the curve is symmetric, one cycle splits into four equal quarters, and each quarter takes T/4.

The defining rule for SHM links the acceleration to the displacement, and it is given in the data booklet:

Given in the data booklet — the SHM defining condition. The minus sign means a points opposite to x (back toward the centre).
acceleration of the oscillator (m s⁻²)
angular frequency (rad s⁻¹)
displacement from equilibrium (m)
Given in the data booklet — links the period, the frequency and the angular frequency.
period — time for one full oscillation (s)
frequency — oscillations per second (Hz)
angular frequency (rad s⁻¹)
Point in the cycleDisplacement xVelocity vAcceleration a
At the centre (equilibrium)zeromaximumzero
At a turning point (the ends)maximumzeromaximum
Quarter-cycle timings: Centre → end takes T/4. End → centre takes another T/4. Centre → opposite end → back to centre is T/2 (half a cycle).

These fractions of T are the heart of most timing questions.
IB-style questionCalculate[2 marks]

A mass on a spring completes one full oscillation every 0.50 s. Find its angular frequency ω.

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The SHM equations (HL)

The graphs have equations. Put in a time and they give the displacement and the velocity at that instant; put in a position and the third gives the speed there. All three are given in the data booklet.

Displacement at time t. φ = 0 if it starts at equilibrium moving the + way.
amplitude (m)
angular frequency (rad s⁻¹)
time (s)
phase angle: where the cycle starts (rad)
displacement (m)
Velocity at time t: the cosine, a quarter period ahead of x.
amplitude (m)
angular frequency (rad s⁻¹)
time (s)
phase angle: where the cycle starts (rad)
velocity (m s⁻¹)
Speed at displacement x, with no time needed. + or − is the direction of travel.
velocity (m s⁻¹)
angular frequency (rad s⁻¹)
amplitude (m)
displacement (m)
Choosing φ, and radians: Starts at equilibrium moving the + way: φ = 0, so x = x₀ sin ωt.

Released from +x₀ (the end): φ = π/2, so x = x₀ cos ωt.

ωt + φ is an angle in radians: set the calculator to radians.
The two maximums: v max = ωx₀, at the centre (x = 0 in v = ω√(x₀² − x²)).

a max = ω²x₀, at the ends (x = x₀ in a = −ω²x).
IB-style questionCalculate[2 marks]

An object oscillates with SHM of amplitude 0.12 m and frequency 2.0 Hz. Calculate its speed when it is 0.060 m from the equilibrium position.

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How this is tested — these graphs almost always come up as a quick timing or phase question:

Paper 1A

  • Read a time to reach a point in the cycle — each quarter takes T/4 (e.g. equilibrium → maximum displacement).
  • Or identify the phase between x, v and a.

Paper 2

  • Describe how v and a change as the object moves, using a = −ω²x.
  • Use x = x₀ sin(ωt + φ), v = ωx₀ cos(ωt + φ) or v = ±ω√(x₀² − x²) for a value at a time or a position (HL).
The classic trap: Thinking velocity is biggest at the ends — it is biggest at the centre, and zero at the ends.
Split the cycle into quarters: Mark the cycle: centre → end → centre → other end → centre. Each step is T/4.

So equilibrium to maximum displacement is the first quarter = T/4.

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IB-style questionCalculate[2 marks]

A particle moves with simple harmonic motion of period 0.80 s. It starts at the equilibrium position. Find the time it takes to first reach maximum displacement.

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Try an IB Exam Question — Free AI Feedback

Test yourself on SHM graphs, phase and timing. Write your answer and get instant AI feedback — just like a real IB examiner.

A buoy bobs up and down with simple harmonic motion of amplitude 0.25 m and period 3.0 s.

the maximum speed of the buoy.
[2 marks]

Related Physics HL Topics

Continue learning with these related topics from the same unit:

3.1.1Conditions for simple harmonic motion3.1.2Period and frequency of SHM oscillators3.1.4Energy in simple harmonic motion3.1.5Energy transformations in oscillations (HL)
View all Physics HL topics
Improve your exam techniqueCommand terms, paper structure, and mark-scheme tips for Physics HL
Previous3.1.2Period and frequency of SHM oscillatorsNextEnergy in simple harmonic motion3.1.4

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