SHM definition, equations, graphs and phase
Key idea: H2 Physics lessons on free oscillations, simple harmonic motion, energy interchange, damping, forced response and resonance.
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The core idea
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Big question: What makes an oscillation simple harmonic?
Simple harmonic motion requires acceleration proportional to displacement and opposite in direction: a = −ω²x. A sinusoidal displacement graph gives velocity a quarter-cycle out of phase and acceleration in antiphase with displacement. The relations v = ±ω√(A²−x²) and vmax = ωA connect the graph to motion at any position.
Use the defining acceleration test
Simple harmonic motion requires a = −ω²x. Acceleration is proportional to displacement and always directed towards equilibrium; periodic motion alone is not enough. An acceleration–displacement graph is therefore a straight line through the origin with gradient −ω².
A sinusoidal form such as x = A cosωt satisfies the definition because differentiating twice gives a = −Aω²cosωt = −ω²x. The equation also fixes the phase origin: at t = 0 this chosen form starts at +A.
Check your understanding: A periodic acceleration graph is curved against displacement. Is the motion SHM?
Not exactly. SHM needs a straight a–x relationship through the origin with negative gradient.
Read the three phase-related graphs
Velocity is the gradient of the displacement–time graph. If the time origin is an equilibrium crossing in the positive direction, x = A sinωt and v = Aω cosωt = v₀ cosωt. If the motion instead starts at +A, the equivalent choice x = A cosωt gives v = −Aω sinωt. In either case, velocity is a quarter-cycle out of phase with displacement and acceleration is exactly in antiphase with displacement.
At equilibrium, |v| is maximum and a = 0. At either extreme, v = 0 and |a| is maximum. Away from a time graph, v = ±ω√(A² − x²); choose the sign from the stated direction of motion.
Check your understanding: At x = +A, what are velocity and acceleration?
Velocity is zero and acceleration is −ω²A, its greatest magnitude directed towards equilibrium.
Key ideas to keep
- The minus sign in a = −ω²x shows direction toward equilibrium.
- Speed is greatest at equilibrium and zero at the turning points.
- Phase must be compared at the same time, not at different positions on unrelated graphs.
See the reasoning
Worked example
Choose the velocity sign from the motion
Question: An oscillator has A = 0.060 m and ω = 8.0 rad s⁻¹. At x = −0.036 m it is moving away from equilibrium. Find velocity and acceleration.
Step 1: Find speed from position
Why: The position form avoids needing the unknown time phase.
Working: |v| = 8.0√(0.060² − 0.036²) = 0.384 m s⁻¹.
Step 2: Assign the sign
Why: At negative x, moving away from equilibrium means moving further negative.
Working: v = −0.384 m s⁻¹.
Step 3: Apply the defining relation
Why: Acceleration must point towards equilibrium.
Working: a = −ω²x = −64(−0.036) = +2.30 m s⁻².
Answer: v = −0.384 m s⁻¹ and a = +2.30 m s⁻².
Check: Velocity points away from equilibrium while acceleration points back towards it, so the object is slowing on its way to the negative extreme.
Another worked model
Question
A particle performs SHM with amplitude 0.050 m and ω = 6.0 rad s⁻¹. At x = +0.030 m while moving towards equilibrium, find acceleration and velocity.
Check the worked solution
a = −ω²x = −36(0.030) = −1.08 m s⁻². The speed is ω√(A² − x²) = 6.0√(0.050² − 0.030²) = 0.240 m s⁻¹. Motion towards equilibrium from positive x makes v = −0.240 m s⁻¹.
Use a hint if needed
Practise with support
Try this
For x = 0.040 sin(10t), find maximum speed, acceleration at x = −0.020 m and the first time after t = 0 at which x is maximum.
Hint: Use vmax = ωA, a = −ω²x and set sinωt = 1.
Check your answer
vmax = 10(0.040) = 0.400 m s⁻¹. At x = −0.020 m, a = −100(−0.020) = +2.00 m s⁻². The first maximum occurs when 10t = π/2, so t = 0.157 s.
Now work without the hint
Practise independently
Your turn
At x = 0.025 m, an oscillator has a = −0.900 m s⁻². If its amplitude is 0.060 m, find ω, maximum speed and speed at that position.
Check your answer
ω = √(|a|/|x|) = √(0.900/0.025) = 6.00 rad s⁻¹. vmax = ωA = 0.360 m s⁻¹ and |v| = 6.00√(0.060² − 0.025²) = 0.327 m s⁻¹.
Avoid these traps
Common mistakes
Common mistake
Every periodic motion is simple harmonic motion.
What is wrong with this reasoning?
Show better thinking
SHM specifically requires a = −ω²x: acceleration must be proportional to displacement and directed towards equilibrium.
Common mistake
Velocity and acceleration are both greatest at equilibrium.
What is wrong with this reasoning?
Show better thinking
Speed is greatest at equilibrium where acceleration is zero. Acceleration magnitude is greatest at the extreme displacements where velocity is zero.
Write for the examiner
Exam guidance
Use gradient for velocity and curvature for acceleration when interpreting an x–t graph.
Exam-style practice [8 marks]
A particle follows x = 0.040 cos(5.0t). Determine amplitude, period, maximum speed and maximum acceleration. Sketch aligned x–t, v–t and a–t graphs for one cycle and state their phase relationships.
Plan before you answer
- Read A and ω from the equation.
- Use vmax = Aω and amax = Aω².
- Differentiate to fix signs and starting values.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
A = 0.040 m and ω = 5.0 rad s⁻¹, so T = 2π/ω = 1.26 s. Maximum speed Aω = 0.200 m s⁻¹ and maximum acceleration Aω² = 1.00 m s⁻². Since v = −0.200 sin(5.0t), v begins at zero and negative; a = −1.00 cos(5.0t), so a begins at its negative extreme. Velocity differs from displacement by π/2 and acceleration differs by π.
Come back in three days
Check what stayed with you
Recall question 1
State the defining equation for SHM.
Check the answer
a = −ω²x.
Recall question 2
Where is speed greatest in SHM?
Check the answer
At equilibrium.
Recall question 3
What is the phase difference between acceleration and displacement?
Check the answer
π rad, or 180°.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions and defines natural frequency as the frequency of a system in free oscillation. Ideal SHM uses a linear restoring relation and no environmental energy exchange; damping and steady forced response are introduced only when stated.
- GCE A-Level H2 PhysicsTopic 9(d) / Topic 9(e) / Topic 9(f) / Topic 9(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027