← All notesPhysics · C.1 Simple harmonic motion⬇ Download PDF
IB DP Physics · Theme C Wave behaviour · C.1 · ~6 min read

C.1 Simple harmonic motion

The conditions for SHM and a = −ω²x, period, frequency and angular frequency, the mass–spring system and the simple pendulum, energy changes, and (HL) phase angle, the x, v and energy equations.

Conditions for SHMa = −ω²xPeriod & angular frequencyMass–springPendulumEnergy in SHMPhase angle (HL)
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Simple harmonic motion happens when the acceleration is proportional to the displacement from equilibrium and always directed back towards it: a = −ω2x.
  • The period does not depend on the amplitude. For a spring T = 2π√(m/k); for a pendulum (small swings) T = 2π√(l/g).
  • Energy swaps between kinetic and potential twice per cycle; the total stays constant (no damping).
  • HL: x = x0 sin(ωt + φ), v = ±ω√(x02 − x2) and ET = ½mω2x02.

By the end you can

  • State the conditions for SHM and explain the minus sign in a = −ω2x.
  • Use T, f and ω, and the mass–spring and pendulum period equations.
  • Describe energy changes during an oscillation.
  • Use phase angle and the displacement and velocity equations. HL
  • Calculate kinetic, potential and total energy in SHM. HL

1What makes motion simple harmonic

Equilibrium position — where the resultant force on the oscillator is zero.

Displacement x — distance from equilibrium, with a sign. Amplitude x0 is the maximum displacement.

  • Conditions for SHM: (1) a restoring force (and so acceleration) proportional to the displacement; (2) always directed towards the equilibrium position.
a = −ω2x DATA BOOKLET
ω = angular frequency in rad s−1; the minus sign means a is opposite to x
xa−x₀+x₀straight line through 0gradient = −ω²The minus sign: acceleration always points back towards equilibrium.
The defining graph of SHM: a straight line through the origin with negative gradient.
T = 1f = 2πω DATA BOOKLET
period T in s, frequency f in Hz
Worked example

An oscillator has frequency 2.0 Hz and amplitude 5.0 cm. Find its maximum acceleration.

  1. ω = 2πf = 12.6 rad s−1 M1
  2. amax = ω2x0 = 12.62 × 0.050 M1
  3. = 7.9 m s−2, at the ends of the motion A1

2Mass–spring and pendulum

mxmass–springT = 2π√(m/k)θlsimple pendulum (small θ)T = 2π√(l/g)
Two classic oscillators. Neither period depends on the amplitude (for small pendulum swings).
T = 2π√mk DATA BOOKLET
mass on a spring
T = 2π√lg DATA BOOKLET
simple pendulum, small angles (below about 10°)
Worked example

A 0.40 kg mass hangs on a spring with k = 25 N m−1. Find the period.

  1. T = 2π√(0.40/25) M1
  2. = 0.79 s A1
Worked example

What length of pendulum has a period of 2.0 s?

  1. l = gT2/4π2 = 9.8 × 4.0/39.5 M1
  2. = 0.99 m A1
Common mistakeThinking a heavier pendulum bob swings faster or slower. Mass cancels: the pendulum period depends only on l and g. (For a spring, mass does matter.)

3Graphs of SHM

txtvtax = x₀ sin ωtv leads xby π/2a = −ω²xopposite to xx max: v = 0, a max
Where displacement is greatest, velocity is zero and acceleration is greatest (and opposite).
  • v is the gradient of the x–t graph and a is the gradient of the v–t graph. v is a quarter of a cycle (π/2) ahead of x; a is half a cycle (π) out of phase with x.
  • Energy (qualitative): at the ends, all potential energy (Ek = 0); at equilibrium, all kinetic energy (maximum speed). With damping, the total slowly falls (C.4).

4Phase and the SHM equations HL

Phase angle φ — tells you where in the cycle the oscillator starts. Two oscillators are in phase if φ is the same; in antiphase if they differ by π.

x = x0 sin(ωt + φ) DATA BOOKLET
displacement (φ = 0 starts at equilibrium moving in +x; φ = π/2 starts at +x0)
v = ωx0 cos(ωt + φ) DATA BOOKLET
velocity; maximum speed vmax = ωx0
v = ±ω√(x02 − x2) DATA BOOKLET
speed at any displacement
Worked example

For the 2.0 Hz, 5.0 cm oscillator, find the maximum speed and the speed at x = 3.0 cm.

  1. vmax = ωx0 = 12.6 × 0.050 = 0.63 m s−1 A1
  2. v = ω√(0.0502 − 0.0302) = 12.6 × 0.040 M1
  3. = 0.50 m s−1 A1

5Energy in SHM HL

xEETEpEk−x₀x₀Ep = ½mω²x², Ek = ½mω²(x₀² − x²): they add to ET = ½mω²x₀²
Two parabolas that always add to the same flat total.
ET = ½mω2x02 DATA BOOKLET
total energy, proportional to amplitude squared
Ep = ½mω2x2 DATA BOOKLET
potential energy at displacement x
Worked example

A 0.20 kg mass oscillates at 2.0 Hz with amplitude 5.0 cm. Find the total energy and the kinetic energy at x = 3.0 cm.

  1. ET = ½ × 0.20 × 12.62 × 0.0502 = 0.039 J M1A1
  2. Ep = ½ × 0.20 × 12.62 × 0.0302 = 0.014 J
  3. Ek = 0.039 − 0.014 = 0.025 J A1
Exam tipDoubling the amplitude doubles vmax but quadruples the total energy (E ∝ x02). The period is unchanged.
Linking questionSHM is the source of every wave in C.2: each particle of a medium oscillates in SHM as the wave passes. Orbits seen edge-on (D.1) also look like SHM.
Idea for your IA / explorationMeasure how the period of a pendulum depends on length, plot T2 against l and find g from the gradient (4π2/g). Extend it to large angles to see where SHM breaks down.

Test yourself — cover the answers first

Quick check

  1. State the two conditions for SHM.
  2. Where is the speed greatest in SHM?
  3. A pendulum's length is quadrupled. What happens to its period?
  4. A spring oscillator's mass is quadrupled. What happens to T?
  5. What is the gradient of an a–x graph for SHM?
  6. HL An oscillator starts at its maximum positive displacement. What is φ in x = x0 sin(ωt + φ)?

Exam-style

  1. A pendulum has a period of 2.0 s on Earth. What would its period be on the Moon (g = 1.6 m s−2)?
  2. HL A 0.50 kg mass on a spring (k = 50 N m−1) is pulled 4.0 cm from equilibrium and released. Find ω, vmax and the total energy.
Show answers
  1. acceleration ∝ displacement, and directed towards equilibrium
  2. at the equilibrium position
  3. it doubles
  4. it doubles
  5. −ω2
  6. π/2
  7. T ∝ 1/√g
    T = 2.0 × √(9.8/1.6)
    = 4.9 s
  8. ω = √(k/m) = 10 rad s−1
    vmax = ωx0 = 0.40 m s−1
    ET = ½kx02 = ½ × 50 × 0.0402 = 0.040 J
Practise this chapter
IB Physics practice papers
Next chapterC.2 Wave model →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.