OxfordAQA International AS and A-level Physics: Simple harmonic motion: equations, graphs and energy (9630)

This guide is aligned to the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. It covers section 3.6.2, simple harmonic motion. Section 3.6 is marked International A-level only, so this content is not part of the International AS course. Sections 3.1 to 3.5 are the shared International AS and A-level content.

For condensed recall use the revision notes. To test yourself use the practice questions. The OxfordAQA A-level Physics hub and the printable checklist list the rest of the course, and the free diagnostics show which topics to revise first.

What this unit covers

Spec bullet	What you must be able to do
Condition for SHM	State and use a = −ω²x, and recognise it from a graph or a force law
x = A cos ωt and v = ±ω√(A² − x²)	Calculate displacement and speed; find times between positions
Graphs of x, v, a and t	Sketch and read them; use gradients (v is the gradient of x–t, a is the gradient of v–t)
Maximum speed ωA, maximum acceleration ω²A	Identify where in the cycle each occurs and calculate them
Period of mass–spring system and simple pendulum	Derive T = 2π√(m/k) and T = 2π√(l/g) and apply them
Ek, Ep and total energy graphs	Sketch against displacement and against time; use total energy ½mω²A²

Angular speed ω = 2πf is met in section 3.6.1 (circular motion). Here it describes oscillations, with ω = 2π/T.

1. The condition for SHM

An oscillation is simple harmonic when the acceleration is directly proportional to the displacement from equilibrium and is always pointed back towards it. In symbols:

a = −ω²x

The minus sign says that acceleration and displacement point in opposite directions. The constant ω² links the two, so ω is the angular frequency, in rad s⁻¹. The period is T = 2π/ω and the frequency is f = 1/T = ω/2π.

Because F = ma, the force law that gives SHM is F = −(mω²)x, a restoring force proportional to displacement. Any system with a restoring force of the form F = −(constant)x oscillates in SHM, and the constant in brackets equals mω².

Worked example 1: identifying SHM

A trolley of mass 0.80 kg is held between springs so that the resultant force on it is F = −(45 N m⁻¹)x, where x is the displacement from the centre.

1. The force is proportional to displacement and opposes it, so the motion is SHM.
2. a = F/m = −(45/0.80)x = −(56.25 s⁻²)x, so ω² = 56.25 s⁻².
3. ω = √56.25 = 7.50 rad s⁻¹.
4. T = 2π/ω = 2π/7.50 = 0.838 s, and f = 1/T = 1.19 Hz.

Notice that the amplitude never appears. For SHM the period does not depend on amplitude.

2. Displacement and velocity equations

Start timing when the oscillator is at maximum positive displacement. Then

x = A cos ωt

where A is the amplitude. Velocity is the gradient of the displacement–time graph, so differentiating gives v = −Aω sin ωt. Acceleration is the gradient of the velocity–time graph, so a = −Aω² cos ωt = −ω²x. This confirms that x = A cos ωt satisfies the condition for SHM.

Deriving the speed equation

Use sin²ωt + cos²ωt = 1. From x = A cos ωt, cos ωt = x/A, so sin ωt = √(1 − x²/A²). Then

v = −Aω sin ωt = ±Aω √(1 − x²/A²) = ±ω√(A² − x²)

The ± shows that the oscillator passes each position (other than the extremes) in both directions. Speed is largest when x = 0, where v = ±ωA. At x = ±A the speed is zero.

Maximum values

Quantity	Where it occurs	Value
Maximum speed	Equilibrium, x = 0	ωA
Maximum acceleration	Either extreme, x = ±A	ω²A
Zero speed	Extremes, x = ±A	0
Zero acceleration	Equilibrium, x = 0	0

Worked example 2: a floating marker

A marker buoy bobs vertically in SHM with amplitude 0.060 m and period 2.4 s. Timing starts at its highest point.

(a) ω = 2π/T = 2π/2.4 = 2.618 rad s⁻¹.

(b) Displacement at t = 0.50 s: x = 0.060 cos(2.618 × 0.50) = 0.060 cos(1.309) = 0.060 × 0.2588 = 0.0155 m. The angle 1.309 is in radians.

(c) Speed when x = 0.040 m: v = ω√(A² − x²) = 2.618 × √(0.060² − 0.040²) = 2.618 × √0.0020 = 2.618 × 0.04472 = 0.117 m s⁻¹.

(d) Maximum speed = ωA = 2.618 × 0.060 = 0.157 m s⁻¹. Maximum acceleration = ω²A = 2.618² × 0.060 = 0.411 m s⁻².

(e) Time for the buoy to first reach x = −0.030 m (half the amplitude, below equilibrium): cos ωt = −0.5, so ωt = 2π/3 and t = (2π/3)/ω = T/3 = 0.80 s.

Part (e) shows a useful habit. Keep the answer as a fraction of a period where you can, then convert.

3. Graphs of x, v and a against t

With x = A cos ωt, the three graphs are all sinusoidal with the same period T.

Time	x	v	a
0	+A	0	−ω²A
T/4	0	−ωA	0
T/2	−A	0	+ω²A
3T/4	0	+ωA	0
T	+A	0	−ω²A

Read the table as three graphs:

• The x–t graph is a cosine curve starting at its maximum.
• The v–t graph is a negative sine curve. Its phase leads that of x by a quarter of a cycle (π/2 rad): velocity peaks a quarter cycle before displacement does.
• The a–t graph is a negative cosine curve. It is a mirror image of x–t (antiphase, phase difference π), which is the graph form of a = −ω²x.

Gradient links

• The gradient of the x–t graph at any time equals the velocity. Where the curve is steepest (crossing x = 0) the speed is ωA, and where the curve is flat (the peaks and troughs) the speed is zero.
• The gradient of the v–t graph equals the acceleration. The v–t graph is steepest when it crosses zero, which is when the oscillator is at an extreme and a = ±ω²A.
• A tangent drawn on a graph gives the instantaneous value. A straight line joining two points on a curve gives only an average.

Worked example 3: from a graph

An x–t graph shows amplitude 4.0 cm and period 1.6 s.

1. ω = 2π/1.6 = 3.93 rad s⁻¹.
2. The steepest gradient is ωA = 3.93 × 0.040 = 0.157 m s⁻¹. This is at the instants the curve crosses x = 0, so the tangent there has gradient ±0.157 m s⁻¹.
3. The greatest gradient of the matching v–t graph is ω²A = 3.93² × 0.040 = 0.617 m s⁻².

4. The period of a mass–spring system

A mass m on a spring of constant k is displaced by x from equilibrium. For a spring obeying Hooke’s law the resultant force on the mass is F = −kx, whether the spring is horizontal or vertical. In the vertical case the weight is cancelled by the extra extension at equilibrium, so only the change in extension from the equilibrium position produces the net force.

Derivation:

1. F = −kx and F = ma, so ma = −kx.
2. a = −(k/m)x, which has the form a = −ω²x, so the motion is SHM with ω² = k/m.
3. T = 2π/ω = 2π√(m/k).

The period increases with mass and decreases with stiffness. Quadrupling m doubles T.

Worked example 4: a hanging mass

A 0.30 kg mass stretches a spring by 8.0 cm when it hangs at rest.

1. At rest the weight equals the spring force: k = mg/e = (0.30 × 9.81)/0.080 = 36.8 N m⁻¹.
2. T = 2π√(m/k) = 2π√(0.30/36.8) = 0.567 s.

Check: because k/m = g/e, you can also write T = 2π√(e/g) = 2π√(0.080/9.81) = 0.567 s, matching the result.

5. The period of a simple pendulum

A bob of mass m hangs on a light string of length l and swings through a small angle θ (in radians). Take displacement x along the arc from the lowest point.

Derivation:

1. The restoring force is the component of weight along the arc: F = −mg sin θ.
2. For small θ, sin θ ≈ θ, and the arc length is x = lθ, so θ = x/l.
3. F ≈ −mg(x/l), so ma = −(mg/l)x and a = −(g/l)x.
4. This has the form a = −ω²x with ω² = g/l, so T = 2π√(l/g).

The mass cancels, so the period does not depend on the mass of the bob. The result holds only for small angles; the approximation sin θ ≈ θ must be good.

Worked example 5: pendulum length

A pendulum has period 3.0 s. Using g = 9.81 m s⁻²:

l = gT²/(4π²) = (9.81 × 3.0²)/(4π²) = 88.29/39.48 = 2.24 m.

6. Energy in SHM

The total energy of the oscillator is constant (no energy is lost to the surroundings) and swaps between kinetic and potential forms. Measure potential energy from the equilibrium position.

Quantity	Against displacement	Against time (from x = A cos ωt)
Ek	½mω²(A² − x²), an inverted parabola, zero at ±A and maximum at x = 0	½mω²A² sin²ωt
Ep	½mω²x², an upright parabola, zero at x = 0	½mω²A² cos²ωt
Total	½mω²A², a horizontal line	½mω²A², a horizontal line

Ek follows from Ek = ½mv² with v² = ω²(A² − x²). Because Ek + Ep is constant, Ep = ½mω²A² − Ek = ½mω²x². For a spring, k = mω², so Ep = ½kx² takes the familiar form.

Key features of the graphs:

• The Ek and Ep curves meet at x = ±A/√2, and at those points each equals half of the total energy.
• Against time, Ek and Ep oscillate at twice the frequency of the motion, with period T/2. Both are always zero or positive.
• At each extreme, all the energy is potential. At equilibrium, all of it is kinetic.
• The total energy is proportional to A². Doubling the amplitude gives four times the energy.

Worked example 6: energy numbers

An oscillator of mass 0.20 kg has ω = 9.0 rad s⁻¹ and amplitude 0.050 m.

1. Total energy = ½mω²A² = ½ × 0.20 × 9.0² × 0.050² = 0.0203 J.
2. At x = 0.030 m: Ek = ½mω²(A² − x²) = ½ × 0.20 × 81 × (0.0025 − 0.0009) = 0.0130 J.
3. Ep = ½mω²x² = ½ × 0.20 × 81 × 0.0009 = 0.00729 J.
4. Check: 0.0130 + 0.00729 = 0.0203 J, which equals the total.
5. Ek and Ep are equal at x = A/√2 = 0.0354 m.
6. The period of the energy graphs against time is T/2 = (2π/9.0)/2 = 0.349 s.

Common errors

• Evaluating cos ωt in degrees. The quantity ωt is in radians.
• Using x = A cos ωt with timing that starts at equilibrium. Check where the oscillator is at t = 0 before choosing the form.
• Writing a = −ω²x with the minus sign dropped when asked for direction.
• Saying the period depends on amplitude. For SHM it does not.
• Applying T = 2π√(l/g) to a large swing, where sin θ ≈ θ fails.
• Drawing Ek and Ep against time at the same frequency as the motion. Their period is T/2.
• Marking the maximum speed at an extreme. It occurs at equilibrium.
• Using the full extension of a vertical spring as x. Use the displacement from the equilibrium position.

Work through the practice questions next.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. This guide follows section 3.6.2 (International A-level only).