Module 5 · Newtonian World and Astrophysics

Simple Harmonic Motion

Specification: OCR A H556 | Section: 5.3.1 & 5.3.2 | Teaching time: ~6 hours

By the end of this topic you should be able to…

Define displacement, amplitude, period, frequency, angular frequency and phase difference for oscillations
Use the defining condition of SHM: acceleration is proportional to displacement from equilibrium and directed towards it
Apply the equations x = A cos(ωt), x = A sin(ωt), v = ±ω√(A² − x²) and v_max = ωA
Describe and sketch the relationship between displacement, velocity and acceleration graphs
Explain energy interchange between kinetic and potential energy during SHM, including energy–displacement graphs

Part 1

Oscillations and the Defining Condition of SHM

An oscillation is any motion that repeats itself about a central equilibrium position. Think of a pendulum swinging, a mass bouncing on a spring, or a guitar string vibrating. Before defining SHM, we need the vocabulary.

Oscillation Terminology

Displacement x — distance from the equilibrium position (positive or negative)
Amplitude A — maximum displacement from equilibrium
Period T — time for one complete oscillation (s)
Frequency f — number of oscillations per unit time (Hz = s⁻¹)
Angular frequency ω — rate of change of phase (rad s⁻¹)
Phase difference — fraction of a cycle by which one oscillation leads or lags another, measured in radians or degrees

Period–Frequency Relationship T = 1/f and f = 1/T

Angular Frequency ω = 2π/T = 2πf

The Defining Condition

Motion is simple harmonic if and only if the acceleration is:

Proportional to the displacement from equilibrium, and
Directed towards the equilibrium position.

Defining Equation of SHM a = −ω²x

The minus sign is crucial — it guarantees the restoring force always pushes the object back towards equilibrium. When displacement is positive, acceleration is negative; when displacement is negative, acceleration is positive.

💡 Key Point

To prove a system undergoes SHM, start from Newton's second law and show that the resulting acceleration has the form a = −ω²x. You must identify what ω represents for that specific system (e.g. ω = √(k/m) for a mass–spring system).

Examples of SHM

Any system where the restoring force obeys Hooke's law produces SHM (for small oscillations):

Mass on a spring — restoring force F = −kx, so a = −(k/m)x giving ω² = k/m
Simple pendulum (small angles) — restoring force F ≈ −(mg/L)x giving ω² = g/L
Atoms in a crystal lattice — interatomic forces approximate SHM for small vibrations
Buoyancy oscillator — a floating object bobbing up and down

Figure 1 — Mass–spring system. The restoring force F always acts towards the equilibrium position. Amplitude A is the maximum displacement in either direction.

Check your understanding

Knowledge Check

A mass oscillates with a frequency of 2.5 Hz. Calculate its angular frequency.

2 marks

ω = 2πf ✓
ω = 2π × 2.5 = 5π ≈ 15.7 rad s⁻¹ ✓

State two conditions that must be satisfied for an oscillation to be simple harmonic.

2 marks

Acceleration (or restoring force) is proportional to displacement from equilibrium ✓
Acceleration (or restoring force) is always directed towards the equilibrium position ✓

A pendulum has a period of 1.8 s. Calculate its frequency and angular frequency.

3 marks

f = 1/T = 1/1.8 = 0.556 Hz ✓
ω = 2π/T = 2π/1.8 ✓
ω = 3.49 rad s⁻¹ ✓

Part 2

Displacement, Velocity and Acceleration

Solutions to a = −ω²x

Solving the defining equation gives us two standard forms for displacement as a function of time:

Displacement (two forms) x = A cos(ωt) or x = A sin(ωt)

Which form you use depends on the initial conditions:

x = A cos(ωt) — object starts at maximum displacement (+A) at t = 0
x = A sin(ωt) — object passes through equilibrium (x = 0) at t = 0

The general form with a phase constant is x = A sin(ωt + φ), where φ accounts for the starting position.

Velocity

Differentiating displacement with respect to time gives velocity:

📐 Derivation of v = ω√(A² − x²)

Starting from x = A cos(ωt):

v = dx/dt = −Aω sin(ωt)

Using sin²θ + cos²θ = 1: sin(ωt) = √(1 − cos²(ωt)) = √(1 − x²/A²)

So v = ±Aω × √(1 − x²/A²) = ±ω√(A² − x²)

Velocity v = ±ω√(A² − x²) and v_max = ωA

The ± sign reflects that the object moves in both directions. Maximum speed occurs at the equilibrium position (x = 0) where v = ωA. The object is momentarily stationary at x = ±A (the turning points).

Acceleration

Differentiating again gives acceleration:

Acceleration a = −Aω² cos(ωt) and a_max = ω²A

Maximum acceleration occurs at the turning points (x = ±A) where a = ∓ω²A. At the equilibrium position (x = 0), acceleration is zero.

⚠️ Common Mistake

Maximum velocity and maximum acceleration happen at different points in the cycle. v is maximum at x = 0; a is maximum at x = ±A. Don't confuse them in exam questions.

Worked Example

A mass oscillates with SHM of amplitude 0.12 m and angular frequency 8.0 rad s⁻¹. Calculate the maximum speed and the speed when the displacement is 0.06 m.

v_max = ωA = 8.0 × 0.12 = 0.96 m s⁻¹

v = ω√(A² − x²) = 8.0 × √(0.12² − 0.06²)

v = 8.0 × √(0.0144 − 0.0036) = 8.0 × √0.0108

v = 8.0 × 0.104 = 0.83 m s⁻¹

Figure 2 — Displacement, velocity and acceleration graphs for SHM on a shared time axis. Velocity leads displacement by π/2; acceleration is in antiphase (π) with displacement.

Check your understanding

Knowledge Check

An object in SHM has amplitude 0.15 m and angular frequency 5.0 rad s⁻¹. Calculate its maximum acceleration.

2 marks

a_max = ω²A ✓
a_max = (5.0)² × 0.15 = 25 × 0.15 = 3.75 m s⁻² ✓

At what displacement does an SHM oscillator have half its maximum speed?

3 marks

v = ½v_max → ω√(A² − x²) = ½ωA ✓
√(A² − x²) = A/2 → A² − x² = A²/4 ✓
x² = 3A²/4 → x = ±A√3/2 = ±0.866A ✓

Describe how the displacement, velocity and acceleration graphs of SHM are related in terms of phase difference.

3 marks

Velocity leads displacement by 90° (π/2 rad) ✓
Acceleration is 180° (π rad) out of phase with displacement (antiphase) ✓
All three graphs have the same period / frequency ✓

Part 3

Energy of a Simple Harmonic Oscillator

Throughout SHM, energy is continuously exchanged between kinetic energy (E_k) and potential energy (E_p), while the total energy remains constant (assuming no damping).

Kinetic and Potential Energy

Kinetic Energy E_k = ½mv² = ½mω²(A² − x²)

Potential Energy E_p = ½mω²x²

Total Energy E_total = E_k + E_p = ½mω²A²

Energy at Key Positions

Position	E_k	E_p	Speed
x = 0 (equilibrium)	Maximum	Zero	v_max = ωA
x = ±A (turning points)	Zero	Maximum	Zero

💡 Key Point

The total energy ½mω²A² is always constant for an undamped SHM system. This is a direct consequence of the restoring force being conservative. If you see total energy changing in a calculation, something is wrong.

Figure 3 — Energy–displacement graphs for SHM. E_k + E_p = E_total at all displacements. E_k is maximum at x = 0; E_p is maximum at x = ±A.

Interactive

SHM Oscillator Simulation

x = — m

v = — m s⁻¹

a = — m s⁻²

E_k = — J

E_p = — J

Part 4

Isochronous Oscillators and Practical Methods

Isochronous Oscillators

A key feature of SHM is that the period is independent of amplitude. This property is called isochronous oscillation. A pendulum clock keeps good time whether it swings in a small arc or a slightly larger one (provided the small-angle approximation holds).

⚠️ Important Limitation

The isochronous property breaks down for large amplitude oscillations. A simple pendulum is only approximately SHM for angles below about 10°. For a mass–spring system, the period is truly independent of amplitude as long as the spring obeys Hooke's law.

Determining the Period Experimentally (PAG10)

To measure the period of SHM accurately:

Set up the oscillating system (mass on a spring or simple pendulum)
Displace the mass by a small amount and release from rest
Use a fiducial marker at the equilibrium position to improve timing accuracy
Time multiple oscillations (typically 10–20) and divide by the number of oscillations
Repeat for different amplitudes to confirm the period is independent of amplitude (isochronous)

💡 Practical Tip

Always measure the time for at least 10 complete oscillations, then divide. This reduces the percentage uncertainty in the timing measurement. The random uncertainty in T is the uncertainty in the total time divided by the number of oscillations.

Period Formulas

Mass–Spring System T = 2π√(m/k)

Simple Pendulum (small angles) T = 2π√(L/g)

Notice that for a mass–spring system, T depends on mass but not on g — so the period would be the same on the Moon. For a pendulum, T depends on g but not on mass.

Worked Example

A mass of 0.40 kg hangs from a spring of spring constant 25 N m⁻¹. Calculate the period of oscillation.

T = 2π√(m/k)

T = 2π√(0.40/25)

T = 2π × √0.016 = 2π × 0.126

T = 0.794 s

Exam Practice

Exam-Style Questions

A mass of 0.25 kg is attached to a spring and oscillates with SHM. The amplitude of oscillation is 0.080 m and the maximum kinetic energy is 0.50 J.

(a) Calculate the angular frequency of the oscillation.
(b) Calculate the maximum speed of the mass.
(c) Calculate the speed of the mass when its displacement is 0.040 m.

5 marks

(a) E_total = ½mω²A² → 0.50 = ½ × 0.25 × ω² × 0.080² ✓
0.50 = 0.125 × ω² × 0.0064 = 8.0 × 10⁻⁴ × ω² ✓
ω² = 0.50 / (8.0 × 10⁻⁴) = 625 → ω = 25 rad s⁻¹ ✓
(b) v_max = ωA = 25 × 0.080 = 2.0 m s⁻¹ ✓
(c) v = ω√(A² − x²) = 25 × √(0.080² − 0.040²) = 25 × √(0.0048) = 25 × 0.0693 = 1.73 m s⁻¹ ✓

A simple pendulum has a length of 1.20 m. A student measures the time for 20 complete oscillations and records a value of 44.2 s.

(a) Calculate the period and hence a value for the acceleration of free fall g.
(b) The student repeats the experiment with a larger amplitude. Explain whether the value of g obtained would be different.

5 marks

(a) T = 44.2/20 = 2.21 s ✓
T = 2π√(L/g) → T² = 4π²L/g ✓
g = 4π²L/T² = 4π² × 1.20 / 2.21² ✓
g = 47.37 / 4.884 = 9.70 m s⁻² ✓
(b) For larger amplitudes, the small-angle approximation breaks down — the restoring force is no longer proportional to displacement ✓
The period increases slightly for larger amplitudes, so the calculated g would be smaller than the true value ✓

An object oscillates with SHM. At displacement x = 0.030 m, its speed is 0.45 m s⁻¹ and its acceleration has a magnitude of 4.5 m s⁻².

(a) Calculate the angular frequency of the oscillation.
(b) Calculate the amplitude of the oscillation.

4 marks

(a) a = ω²x → ω² = a/x = 4.5/0.030 = 150 ✓
ω = √150 = 12.2 rad s⁻¹ ✓
(b) v² = ω²(A² − x²) → 0.45² = 150(A² − 0.030²) ✓
0.2025 = 150A² − 0.135 → 150A² = 0.3375 → A² = 0.00225 → A = 0.0474 m ✓

QWC

Extended Response (6 marks)

A student sets up a mass–spring system and displaces the mass from its equilibrium position. When released, the mass oscillates with simple harmonic motion.

Describe the energy changes that take place during one complete oscillation of the mass. In your answer, you should:
• describe how the kinetic energy and potential energy change
• explain what happens to the total energy
• refer to the motion of the mass at specific points in the cycle

[6 marks, QWC]

6 marks

At maximum displacement (the turning points), the mass is momentarily at rest — its kinetic energy is zero and its potential energy is at a maximum ✓
As the mass moves towards the equilibrium position, the potential energy decreases and the kinetic energy increases ✓
At the equilibrium position, the mass is moving at its maximum speed — kinetic energy is at a maximum and potential energy is zero ✓
The mass continues through equilibrium, losing kinetic energy and gaining potential energy as it moves towards the opposite turning point ✓
Throughout the oscillation, the total energy (kinetic + potential) remains constant (assuming no damping) ✓
Energy is continuously transferred between kinetic and potential forms — this is a characteristic feature of SHM ✓

Topic Summary

Defining Condition

SHM occurs when acceleration is proportional to displacement and directed towards equilibrium: a = −ω²x.

Key Quantities

ω = 2πf = 2π/T. v_max = ωA at x = 0. a_max = ω²A at x = ±A.

Energy

Total energy ½mω²A² is constant. E_k and E_p interchange continuously throughout each cycle.

Isochronous

Period is independent of amplitude (for true SHM). Mass–spring: T = 2π√(m/k). Pendulum: T = 2π√(L/g).

Equations to Know

a = −ω²x

x = A cos(ωt)

v = ±ω√(A² − x²)

v_max = ωA

E_total = ½mω²A²

T = 2π√(m/k)

T = 2π√(L/g)