IB DP Physics: Simple harmonic motion: equations, period and energy (DP Physics)

This study guide covers sub-topic C.1, Simple harmonic motion, in the IB Physics guide (first assessment 2025). The core ideas are for both SL and HL. The phase angle, the sinusoidal equations for displacement and velocity, and the quantitative treatment of energy are additional higher level content, so they are marked “HL only” below. For quick recall, use the revision notes. To test yourself, use the practice questions. The course hub and the printable checklist show where C.1 sits in the course, and the free 10-minute diagnostics help you find weak spots.

Earlier mechanics helps here: see kinematics, forces and momentum and work, energy, power and efficiency.

What this unit covers

Guide reference	What you must be able to do	Level
C.1	State the conditions that lead to simple harmonic motion (SHM)	SL and HL
C.1	Use the defining equation a = –ω²x and explain the minus sign	SL and HL
C.1	Describe an oscillation using displacement, equilibrium position, amplitude, period, frequency and angular frequency	SL and HL
C.1	Link period, frequency and angular frequency: T = 1/f = 2π/ω	SL and HL
C.1	Find the period of a mass-spring system, T = 2π√(m/k), and of a simple pendulum, T = 2π√(l/g)	SL and HL
C.1	Describe energy changes over one cycle in words	SL and HL
C.1	Use a phase angle and x = x₀ sin(ωt + φ), v = ωx₀ cos(ωt + φ)	HL only
C.1	Use v = ±ω√(x₀² – x²)	HL only
C.1	Calculate energy: total E_T = ½mω²x₀² and potential E_p = ½mω²x²	HL only

Conditions for simple harmonic motion

An object oscillates when it moves back and forth about an equilibrium position, the place where the net force on it is zero. Oscillation is only simple harmonic when two things are true:

• The net force (the restoring force) always points towards the equilibrium position.
• The size of that force is proportional to the displacement from equilibrium.

In symbols, F = –kx for a constant k. Newton’s second law gives ma = –kx, so the acceleration is proportional to displacement and opposite in direction. Any motion that obeys this is SHM, whatever the system is.

Two real examples show how the conditions are met.

• Mass on a spring. The spring force is proportional to the extension or compression, as long as the spring is not stretched past its proportional limit. It always points back to the natural-length position (for a horizontal spring).
• Simple pendulum. The restoring force is the component of the weight along the arc. It is proportional to the displacement only when the angle is small. For large angles the motion is still periodic but is no longer simple harmonic.

The defining equation a = –ω²x

Since a = –(k/m)x for a mass on a spring, write ω² = k/m and the equation becomes

a = –ω²x

This is the defining equation of SHM. Any oscillator whose acceleration obeys it has a constant ω set by the system.

The minus sign. The acceleration is always directed towards equilibrium, while the displacement is measured away from it. When the object is at positive displacement the acceleration is negative, and when it is at negative displacement the acceleration is positive. Without the minus sign the object would be pushed away from equilibrium and the motion would not repeat.

Three consequences are worth knowing:

• At equilibrium (x = 0) the acceleration is zero.
• At maximum displacement (x = ±amplitude) the acceleration has its largest size, ω² times the amplitude.
• The acceleration is not constant, so the equations for uniform acceleration do not apply.

Describing an oscillation

Quantity	Meaning	Unit
Displacement x	Distance from the equilibrium position, with direction (a vector)	m
Amplitude x₀	Maximum displacement from equilibrium	m
Time period T	Time for one complete oscillation	s
Frequency f	Number of oscillations per second	Hz
Angular frequency ω	2π times the frequency, in radians per second	rad s⁻¹

The links are

T = 1/f = 2π/ω

so ω = 2πf. The angular frequency is the rate of change of the phase and has nothing to do with the speed of the object.

For SHM the period does not depend on the amplitude. A larger swing means a longer path, but the speeds are correspondingly larger, and the time per cycle stays the same. This is called isochronous motion.

Period of a mass-spring system

Comparing ω² = k/m with ω = 2π/T gives

T = 2π√(m/k)

A stiffer spring (larger k) gives a shorter period. A heavier mass gives a longer period. To halve the period you need four times the stiffness or one quarter of the mass.

Worked example 1. A trolley of mass 0.64 kg is joined to a horizontal spring of spring constant 26 N m⁻¹. Find T, f and ω.

1. T = 2π√(m/k) = 2π√(0.64/26) = 2π × 0.1569 = 0.986 s
2. f = 1/T = 1/0.986 = 1.01 Hz
3. ω = 2π/T = 6.37 rad s⁻¹ (this also equals √(k/m))

Note that gravity does not appear in the formula. A mass hanging from a vertical spring has the same T = 2π√(m/k), because gravity only shifts the equilibrium position down by mg/k.

Period of a simple pendulum

Model a simple pendulum as a compact heavy bob hung from a light, inextensible string of length l and displaced by only a few degrees. Its restoring force is proportional to displacement along the arc, which is equivalent to ω² = g/l and gives

T = 2π√(l/g)

The mass of the bob does not appear. A longer pendulum has a longer period. The period is smaller where g is larger.

Worked example 2. Find the period of a pendulum of length 0.85 m where g = 9.81 m s⁻².

T = 2π√(0.85/9.81) = 2π × 0.2944 = 1.85 s

Worked example 3. A student times 25 oscillations of a pendulum in 38.6 s. The length from the pivot to the centre of the bob is 0.592 m. Find g.

1. T = 38.6/25 = 1.544 s (timing many oscillations reduces the effect of reaction time)
2. Rearrange T = 2π√(l/g) to g = 4π²l/T²
3. g = 4π² × 0.592 / 1.544² = 9.80 m s⁻²

Energy changes in one cycle

With no friction or air resistance, the total mechanical energy is constant. It moves between kinetic energy and a form of potential energy.

Position	Speed	Kinetic energy	Potential energy
Maximum displacement (turning point)	Zero	Zero	Maximum
Equilibrium	Maximum	Maximum	Minimum

A spring stores elastic potential energy, while a pendulum bob stores gravitational potential energy relative to its lowest point. Over one cycle, kinetic energy rises and falls twice (the object passes the equilibrium position twice), and so does potential energy. The two always add up to the same total. In a real system the total decreases as energy is dissipated, and the amplitude falls.

Common graph facts, all described in words: against time, both energies vary with half the period of the motion and never go negative; against displacement, the potential energy is a parabola with its minimum at x = 0, and the kinetic energy is an upside-down parabola, so the total is a horizontal line.

HL only: phase angle and the equations of motion

Displacement as a function of time can be written

x = x₀ sin(ωt + φ)

where φ is the phase angle, in radians. It states where in its cycle the object is at t = 0. If φ = 0 the object starts at equilibrium moving in the positive direction. Differentiating gives the velocity:

v = ωx₀ cos(ωt + φ)

The greatest speed is ωx₀, reached at equilibrium. Combining the equations to remove time gives

v = ±ω√(x₀² – x²)

Use the plus or minus sign to match the direction of motion. The guide says radians are used for phase angle calculations, so keep every angle in radians.

Worked example 4 (HL). An oscillator has amplitude 0.040 m and period 0.80 s. At t = 0 its displacement is +0.020 m and it is moving in the positive direction. Find φ, the initial velocity and the displacement and velocity at t = 0.30 s.

1. ω = 2π/0.80 = 7.85 rad s⁻¹
2. At t = 0: 0.020 = 0.040 sin φ, so sin φ = 0.5 and φ = π/6 = 0.524 rad. Positive velocity needs cos φ > 0, which is true for π/6 (the other solution, 5π/6, would be moving the other way).
3. Initial velocity: v = 7.85 × 0.040 × cos(π/6) = 0.272 m s⁻¹
4. At t = 0.30 s the argument is 7.85 × 0.30 + 0.524 = 2.88 rad
5. x = 0.040 sin(2.88) = 0.0104 m and v = 0.314 cos(2.88) = –0.303 m s⁻¹, so the object is moving back towards equilibrium.

HL only: energy in SHM, quantitatively

The total energy is the maximum kinetic energy, which occurs at x = 0, where v = ωx₀:

E_T = ½mω²x₀²

The potential energy at displacement x is

E_p = ½mω²x²

The kinetic energy is the difference:

E_K = E_T – E_p = ½mω²(x₀² – x²)

This agrees with ½mv² when v = ±ω√(x₀² – x²). For a spring, mω² = k, so these reduce to ½kx₀² and ½kx².

Worked example 5 (HL). A mass of 0.35 kg oscillates with ω = 11 rad s⁻¹ and amplitude 0.040 m. Find the total energy, and the potential energy, kinetic energy and speed at x = 0.024 m.

1. E_T = ½ × 0.35 × 11² × 0.040² = 0.0339 J
2. E_p = ½ × 0.35 × 11² × 0.024² = 0.0122 J
3. E_K = 0.0339 – 0.0122 = 0.0217 J
4. v = ω√(x₀² – x²) = 11 × √(0.040² – 0.024²) = 11 × 0.032 = 0.352 m s⁻¹

Check: ½ × 0.35 × 0.352² = 0.0217 J, which matches.

Common errors

• Writing a = +ω²x. The minus sign is part of the definition.
• Using the angle in degrees in x = x₀ sin(ωt + φ). Work in radians.
• Calling ω the speed. It is a rate of turning of the phase.
• Believing the period of a pendulum depends on the mass, or the period of a mass-spring system depends on amplitude.
• Using T = 2π√(l/g) for a large swing. The formula needs a small angle.
• Saying the acceleration is greatest at equilibrium. It is zero there.
• Forgetting that the energy graphs against time have half the period of the motion.
• Taking only the positive root for v when the object is travelling in the negative direction.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic C.1 Simple harmonic motion.