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SHM Graphs (Displacement, Velocity, Acceleration)

Key idea: Use and interpret x–t, v–t and a–t relationships in simple harmonic motion, including phase and phase difference (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use oscillation quantities and describe free oscillations and their investigation.
  • Relate displacement, velocity, acceleration and phase in simple harmonic motion.
  • Identify and analyse simple harmonic motion using its defining equation and sinusoidal solutions.

1. Definitions (Must Know)

  • Displacement, x (m): distance from the equilibrium position (positive or negative).
  • Amplitude, x₀ (m): maximum magnitude of displacement.
  • Period, T (s): time for one complete cycle.
  • Frequency, f (Hz): cycles per second, f = 1/T.
  • Angular frequency, ω (rad s⁻¹): ω = 2π f = 2π/T.
  • Phase, θ (rad): the “position” within a cycle, e.g. θ = ω t + φ.
  • Phase constant, φ (rad): the phase at t = 0 (sets where the motion starts).
  • Phase difference, Δφ (rad): how far one oscillation leads or lags another.

2. Key Ideas (What Earns Marks)

  • For SHM:
    • x = x₀ sin(ω t + φ)
    • v = dx/dt = ω x₀ cos(ω t + φ)
    • a = -ω²x₀ sin(ω t + φ) so a = -ω²x
  • Phase relationships (must know):
    • velocity is π/2 ahead of displacement
    • acceleration is π out of phase with displacement (opposite sign)
  • Useful “no-time” relation:
    • v = ±ω square root of (x₀²-x²)
Exam pitfall: phase confusion between x, v, and a

For SHM, velocity leads displacement by π/2, and acceleration is π out of phase with displacement. Marking these phase links on the graph avoids sign mistakes.

3. Detailed Explanations

A. One set of graphs that shows everything

To compare shapes without worrying about units, plot normalised quantities:

  • x/x₀ = sin(2π t/T)
  • v/(ω x₀) = cos(2π t/T)
  • a/(ω² x₀) = - sin(2π t/T)

SHM graphs (normalised)

SHM graphs (normalised). x/x0, v/(ωx0), a/(ω²x0) plotted as Normalised value against t/T.

Scroll across the graph to read all labels.

SHM graphs (normalised). x/x0, v/(ωx0), a/(ω²x0) plotted as Normalised value against t/T.SHM graphs (normalised). x/x0, v/(ωx0), a/(ω²x0) plotted as Normalised value against t/T.
SHM graphs (normalised). Exact plotted values are available in the data table.
Open full-size graph
View figure data
Values for SHM graphs (normalised)
t/T (unitless)x/x0v/(ωx0)a/(ω²x0)
0010
0.1250.70710.7071-0.7071
0.2510-1
0.3750.7071-0.7071-0.7071
0.50-10
0.625-0.7071-0.70710.7071
0.75-101
0.875-0.70710.70710.7071
1010

Read these directly:

  • At maximum displacement (x = ± x₀): v = 0 and |a| is maximum.
  • At equilibrium (x = 0): |v| is maximum and a = 0.
  • v reaches its maximum a quarter cycle (T/4) before x reaches its maximum.

B. “Experimental + graphical” approach (what practical questions want)

If you measure a displacement–time graph x(t) (e.g. motion sensor / light gate + data logger):

  • Period T: time between successive peaks (or troughs).
  • Frequency f = 1/T and ω = 2π/T.
  • Velocity is the gradient of the x–t graph: v = dx/dt.
  • Acceleration comes from the gradient of the v–t graph (or curvature of x–t).

4. Common Mistakes

  • Writing x = x₀ sin(ω t) + φ (phase must be inside the sine/cosine).
  • Mixing degrees and radians.
  • Forgetting the minus sign in a = -ω²x (acceleration points towards equilibrium).

5. Exam Tips

  • If you’re asked for phase difference between x and v: answer π/2 (velocity leads).
  • If you’re asked to “sketch v–t given x–t”: same frequency, shifted left by T/4.
  • Use the “equilibrium/extreme” checkpoints to sanity-check sketches.

6. Worked Examples

Modelled example 1

Extract SHM parameters from an equation

Core

Problem

An oscillator has displacement x = 0.040 sin(10π t) in SI units. Find x₀, ω, f, T, vₘₐₓ and aₘₐₓ.
Study the worked solution

  1. Read amplitude and angular frequency

    Method

    x₀ = 0.040 m and ω = 10π rad s⁻¹.

    Reason

    These are the displacement coefficient and the coefficient of t.

    Working

    Compare with x = x₀ sin(ω t).

  2. Find cycle quantities

    Method

    f = 5.0 Hz and T = 0.20 s.

    Reason

    Use f = ω/(2π) and T = 1/f.

    Working

    f = 10π/2π = 5.0 Hz, T = 1/5.0 = 0.20 s

  3. Find maximum speed

    Method

    vₘₐₓ = 1.26 m s⁻¹.

    Reason

    Velocity amplitude is ω x₀.

    Working

    vₘₐₓ = (10π)(0.040) = 1.26 m s⁻¹

  4. Find maximum acceleration

    Method

    aₘₐₓ = 39.5 m s⁻².

    Reason

    Acceleration amplitude is ω²x₀.

    Working

    aₘₐₓ = (10π)²(0.040) = 39.5 m s⁻²

Guided practice 2

Find the phase constant from initial conditions

About 5 min

Problem

For x = x₀ sin(ω t + φ), the oscillator is at equilibrium at t = 0 and moving in the positive direction. Find φ in the interval 0 ≤ φ < 2π.

Try this before viewing the solution

Phase constant

Hints

Hint 1: use both initial conditions
First solve sin φ = 0, then require v(0) = ω x₀ cos φ > 0.
View solution step by step

  1. Use initial displacement

    Method

    φ = 0 or π.

    Reason

    At equilibrium, x(0) = x₀ sin φ = 0.

    Working

    sin φ = 0 ⇒ φ = 0 or π

  2. Use motion direction

    Method

    Select φ = 0.

    Reason

    v(0) = ω x₀ cos φ must be positive; cos 0 > 0 but cos π < 0.

    Working

    φ = 0

Common misconception 3

Phase difference between x and v

Find and correct the mistake

Learner claim

For x = x₀ sin(ω t), a learner says velocity lags displacement by π/2. Diagnose the direction of the phase difference.

Try this before viewing the solution

Velocity relative to displacement

View solution step by step

  1. Differentiate displacement

    Method

    v = ω x₀ cos(ω t).

    Reason

    Velocity is the gradient of the displacement–time graph.

    Working

    v = dx/dt = ω x₀ cos(ω t)

  2. Express the phase shift

    Method

    Velocity leads displacement by π/2.

    Reason

    cos θ = sin(θ + π/2); the positive shift reaches corresponding features earlier.

    Working

    v = ω x₀ sin(ω t + π/2)

Examiner practice 4

Using checkpoints to fill missing values

3 marks

Examination question

In SHM, the particle is at maximum positive displacement x = +x₀. State its velocity, write its acceleration, and describe the acceleration direction. [3 marks]

Try this before viewing the solution

View solution step by step

  1. State velocity

    1 mark

    Method

    v = 0.

    Reason

    The particle instantaneously stops before reversing at an extreme.

    Working

    v = 0.

  2. State acceleration

    1 mark

    Method

    a = -ω²x₀.

    Reason

    Apply the defining relation at x = +x₀.

    Working

    a = -ω²x₀

  3. Describe direction

    1 mark

    Method

    Acceleration points towards equilibrium and has maximum magnitude.

    Reason

    At the positive extreme, the restoring direction is negative.

    Working

    |a| = ω²x₀ towards x = 0.

Challenge 5

Find v at a displacement using the no-time relation

Minimal support

Independent transfer

An oscillator has amplitude 0.080 m and angular frequency 12 rad s⁻¹. Find its speed at x = 0.050 m and explain why the relation alone cannot determine the velocity sign.

Try this before viewing the solution

Hints

Hint 1: separate magnitude from direction
Use |v| = ω square root of (x₀²-x²), then consider the two passages through the same displacement.
View solution step by step

  1. Calculate the magnitude

    Method

    Obtain |v| = 0.749 m s⁻¹ ≈ 0.75 m s⁻¹.

    Reason

    The no-time relation combines the amplitude, displacement and angular frequency.

    Working

    |v| = 12 square root of (0.080²-0.050²) = 12 square root of 0.0039 = 0.749 m s⁻¹

  2. Interpret the sign ambiguity

    Method

    The velocity may be + 0.749 or -0.749 m s⁻¹.

    Reason

    The particle passes the same displacement once in each direction per cycle; a direction or phase condition is needed.

    Working

    Speed = 0.75 m s⁻¹; velocity sign is not fixed.

7. Mind Stretchers

Mind stretcher 1: Why is acceleration “opposite” to displacement?Extension

Explain why a = -ω²x automatically means the motion is always pulled back toward equilibrium.

Show Answer

If x > 0 (to the right of equilibrium), then a = -ω²x < 0 (acceleration to the left), so the particle is accelerated back toward equilibrium.

If x < 0, then a > 0, again pointing back toward equilibrium.

Mind stretcher 2: Convert phase difference to a time differenceExtension

Two oscillations have the same period T = 0.50 s. One oscillation leads the other by a phase difference of Δφ = π/2.

Find the time lead.

Show Answer

Δ t = (Δφ/2π)T

Δ t = ((π/2)/2π)(0.50) = 0.125 s

Mind stretcher 3: Simulation Bridge: SHM ExplorerExtension

Concept Explorer: SHM Explorer

Adjust amplitude, period, and phase to connect x-v-a checkpoints and test SHM equations with guided prompts.

BetaA LevelOscillationsBest for: A Level oscillations
  • x-v-a Checkpoints
  • Phase Relation
  • vₘₐₓ = ωA
  • a = −ω²x

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

Compare the graphs in the SHM Explorer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027