Chapter 45 · Oscillations · Lesson 23.1
Simple Harmonic Motion
One condition defines the motion
A swing, a plucked guitar string, a mass bobbing on a spring — pull any of them away from rest and something pulls them back. When that pull-back grows in exact proportion to the displacement, the motion that follows is the most important oscillation in physics.
Read the relationship. Check its limits.
Lesson Overview
Simple harmonic motion is any oscillation in which the acceleration is proportional to the displacement from equilibrium and directed back towards it: a = −ω²x.
Displace a trolley between two springs by 4 cm and release it: it accelerates back through the centre, overshoots, and repeats — and doubling the pull-back to 8 cm changes the amplitude but not the time per cycle.
Learning Objectives
- Identify the restoring-acceleration condition
- Relate period and angular frequency
- Read the phase relationships of x, v and a
- Calculate periods for springs and small-angle pendulums
Interactive SHM Simulation
Change one input. Compare the graph and its readout before drawing a conclusion.
Angular frequency ω4.47 rad/s
Period T1.40 s
Frequency f0.71 Hz
Displacement x0.1 m
Velocity v0 m/s
Acceleration a-2 m/s²
Total oscillation energy0.1 J
Ideal massless spring: displacement is measured from the gravitational equilibrium. The restoring force shown is the net force. Apparatus size is schematic. The graph uses x/A, so physical values are in the readouts. Motion starts paused.
Ideal undamped, linear oscillations. Springs must remain within their proportional range; the pendulum uses the small-angle approximation. The apparatus views show two periods; the separate Graphs view covers four seconds. Normalised curves and physical readouts are labelled separately.
Key Formulae
- Defining condition
- a = −ω²xAcceleration proportional to displacement, always aimed back at equilibrium. The constant ω is the angular frequency: ω = 2πf = 2π/T.The minus sign is the whole definition
- Period and frequency
- T = 1/f · ω = 2πfPeriod is seconds per cycle; frequency is cycles per second — each is the reciprocal of the other. ω counts the cycle in radians: one full cycle is 2π.T in seconds per cycle; f in Hz
- Mass on a spring
- T = 2π√(m/k)A heavier mass is harder to turn around, so T grows with m; a stiffer spring pulls back harder, so T shrinks with k. The amplitude appears nowhere.Stiffer spring → shorter period
- Simple pendulum
- T = 2π√(L/g)Gravity supplies the restoring force, and for small swings it is very nearly proportional to displacement. Longer pendulum, longer period; the mass of the bob cancels out.Small angles only — and no mass in sight
Use SI units in calculations: metres, seconds, hertz and radians. Use the units shown by each control. Angular frequency is in rad/s; phase degrees are converted to radians.
Key Concepts
Practice Questions
Build the physical model.
An oscillation is a vibration that repeats. Displacement x is measured from the equilibrium position, and the amplitude A is the largest displacement reached. Simple harmonic motion is the special case defined by one condition: the acceleration is proportional to the displacement and points back at equilibrium, a = −ω²x. Any restoring force F ∝ −x produces it — Hooke’s law F = −kx gives a mass on a spring a = −(k/m)x, so the spring system is SHM exactly. The motion traces perfect sinusoids, with velocity zero at the extremes and largest at the centre, and its period is set by the system alone: T = 2π√(m/k) for a mass–spring, T = 2π√(L/g) for a small-angle pendulum.
Restoring forces make SHM
Stretch a spring right and it pulls left; compress it left and it pushes right. Hooke’s law F = −kx captures this opposition, and Newton’s second law turns it into a = −(k/m)x — acceleration proportional to minus the displacement. Any force with this shape, whatever its origin, drives simple harmonic motion.
Three graphs, one motion
If x follows a cosine, the velocity is a negative sine and the acceleration a negative cosine. At the extremes the object is momentarily at rest (v = 0) while the acceleration is largest; through the centre the speed peaks at v(max) = ωA while the acceleration passes through zero. The a-graph is always the x-graph flipped upside down — that is a = −ω²x drawn out in time.
Isochronism — why clocks swing
For ideal SHM the period is independent of amplitude: a small swing and a large one take exactly the same time, because a bigger displacement earns a proportionally bigger restoring pull. A pendulum only approximates this at small angles — push it to large angles and the restoring force falls behind, the period stretches, and the isochronism that once ran the world’s clocks slips away.
Phase difference
Two identical oscillators can run at the same T and f yet be out of step. Release one from the right as you release the other from the left and they stay half a cycle apart forever — a time offset of T/2, or phase difference of π radians. Starting one at equilibrium with a push while the other starts stretched gives a quarter-cycle difference. Phase measures where in its cycle each oscillator is.
Know what the model can tell you.
Assumptions & boundaries
Ideal undamped, linear oscillations. Springs must remain within their proportional range; the pendulum uses the small-angle approximation. The apparatus views show two periods; the separate Graphs view covers four seconds. Normalised curves and physical readouts are labelled separately.
Amplitude does not set the period. In ideal SHM a louder oscillation — twice the amplitude — covers twice the distance but with proportionally larger accelerations, and the cycle time comes out identical. The period belongs to the system (m and k, or L and g), not to how hard you started it. And a pendulum’s period does not depend on the mass of its bob at all.
At the extreme ends of a simple harmonic oscillation, the object has…
Choose an answer to test the idea.
Worked Examples
State the assumptions, convert units, calculate and check the physical meaning. Check which ideal-oscillator assumptions apply.
EasyA mass on a spring completes one full oscillation every 24 s. Find the period and the frequency.
- The period is the time for one cycle: T = 24 s.
- f = 1/T = 1/24 ≈ 0.042 Hz — about four hundredths of a cycle each second.
T = 24 s; f ≈ 0.042 Hz
MediumA spring of constant k = 125 N m⁻¹ carries a 5.0 kg mass. Show that a = −25x, and find the acceleration at x = −2.0 m.
- Hooke’s law F = −kx with Newton’s second law F = ma gives a = −(k/m)x.
- a = −(125/5.0)x = −25x — acceleration proportional to minus displacement, so this is SHM with ω² = 25.
- At x = −2.0 m: a = −25 × (−2.0) = +50 m s⁻² — displaced left, accelerating right, back towards equilibrium.
a = −25x; a = +50 m s⁻² at x = −2.0 m
HardFind the length of a simple pendulum with a period of exactly 2.0 s (a ‘seconds pendulum’), taking g = 9.81 m s⁻².
- T = 2π√(L/g) → L = gT²/(4π²).
- L = 9.81 × 4.0 ÷ 39.48 ≈ 0.99 m.
- Almost exactly one metre — a historical candidate for defining the metre itself.
L ≈ 0.99 m
ChallengingA 0.50 kg mass on a spring (k = 80 N m⁻¹) oscillates with amplitude 6.0 cm. Find ω, the period, the maximum speed, and the maximum acceleration.
- ω = √(k/m) = √(80/0.50) = √160 ≈ 12.6 rad s⁻¹, so T = 2π/ω ≈ 0.50 s.
- v(max) = ωA = 12.6 × 0.060 ≈ 0.76 m s⁻¹, reached passing through equilibrium.
- a(max) = ω²A = 160 × 0.060 = 9.6 m s⁻², reached at the extremes — nearly one g of pull-back.
ω ≈ 12.6 rad s⁻¹, T ≈ 0.50 s, v(max) ≈ 0.76 m s⁻¹, a(max) = 9.6 m s⁻²
Exam diagrams for this topic3 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
- 01
InspectRead the figure comment.
- 02
TraceFollow labels, arrows and axes.
- 03
AnswerWork one part at a time.
- 04
CheckReveal hints and marking points.
A Level
01Fig. 9.1OscillationsA Level
Figure comment
Fig. 9.1A spring hangs vertically from a rigid horizontal support drawn with hatching above it, and a block labelled m is attached to the lower end of the spring. A dashed horizontal line level with the centre of the block is labelled equilibrium position. Two further dashed horizontal lines, one above it and one below it, mark the highest and lowest positions the block reaches, and the distance from the equilibrium line to each of them is marked 4.0 cm.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. Each 4.0 cm runs from the equilibrium line to one extreme, so it is the amplitude; the block travels the 8.0 cm between the outer dashed lines twice in each cycle.
aState State the total distance travelled by the block in one complete oscillation between the dashed lines of Fig. 9.1.
Check answer 2 marks
- distance = 4 × amplitude
- = 16 cm (0.16 m)
bCalculate The period of the oscillation is 0.80 s. Calculate the maximum acceleration of the block, and state where on Fig. 9.1 it occurs.
Check answer 3 marks
- ω = 2π/0.80 = 7.85 rad s⁻¹
- a₀ = ω²x₀ = 7.85² × 0.040 = 2.5 m s⁻²
- at the two outer dashed lines, directed towards the equilibrium line
cDetermine The block has a mass of 0.25 kg. Determine the spring constant of the spring and the maximum resultant force on the block.
Check answer 3 marks
- for a mass on a spring ω² = k/m, so k = mω² = 0.25 × 61.7
- k = 15 N m⁻¹
- F = ma₀ = 0.25 × 2.47 = 0.62 N
dDeduce Deduce the extension of the spring when the block is at the equilibrium line marked in Fig. 9.1, and explain why that line is not level with the lower end of the unloaded spring.
Check answer 4 marks
- at the equilibrium position the spring force balances the weight: kx = mg
- x = (0.25 × 9.81) / 15.4 = 0.16 m
- the spring is already stretched by 16 cm in supporting the block, so the equilibrium line lies 16 cm below the unloaded end
- the oscillation takes place about this stretched position, not about the natural length
Transfer challenge
A simple pendulum, for which T = 2π√(L/g), is to swing with the same period, 0.80 s, as the block in Fig. 9.1. Calculate its length, and state what that length has in common with the extension of the loaded spring.
Check answer 3 marks
- L = gT²/4π²
- L = 9.81 × 0.80² / 4π² = 0.16 m
- it is equal to the static extension of the spring, since both are equal to g/ω²
IB
02Figure 4Simple harmonic motionIB
Figure comment
Figure 4Side view of the apparatus on a bench. A vertical rod rises from the heavy base of a clamp stand, and a horizontal clamp arm projects from the top of the rod. A helical spring hangs from the end of the arm and is labelled 'spring, spring constant k'. A rectangular block labelled m hangs from the lower end of the spring. Beside the mass a vertical double-headed arrow labelled 'oscillation' shows that it moves up and down about its hanging position.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The oscillation arrow is centred on the hanging position, not the spring's natural length: x is measured from there, where the spring already carries an extension mg/k.
aState State the point of the oscillation drawn at which the block's acceleration is zero, and the points at which its magnitude is greatest.
Check answer 2 marks
- acceleration is zero at the hanging position, at the middle of the double-headed arrow
- acceleration is greatest at the two ends of the arrow, at maximum displacement
bShow (that) Show that, before it is set oscillating, the block stretches the spring by mg/k, and hence calculate this extension for m = 0.20 kg and k = 25 N m⁻¹.
Check answer 3 marks
- at rest the upward spring tension balances the weight: ke = mg
- rearranging gives e = mg / k
- e = 0.20 × 9.8 / 25 = 7.8 × 10⁻² m (7.8 cm)
cDetermine For the same block and spring, determine the period of the oscillation, and determine the magnitude of the block's acceleration at the top of the arrow when the amplitude is 3.0 cm.
Check answer 4 marks
- ω = sqrt(k/m) = sqrt(25 / 0.20) = 11.2 rad s⁻¹
- T = 2π / ω = 0.56 s
- using a = -ω² x, the magnitude at maximum displacement is ω² x0 = 125 × 0.030
- a = 3.8 m s⁻², directed downward, back towards the hanging position
dExplain Explain why, for this apparatus, the motion stops being simple harmonic once the amplitude exceeds the extension found earlier.
Show a hint
What force can act on the block once the spring has returned to its natural length?
Check answer 4 marks
- at an amplitude equal to the static extension e, the maximum acceleration is ω² e = (k/m)(mg/k) = g
- at the top of that swing the spring has returned to its natural length and exerts no force on the block
- a larger amplitude would demand a downward acceleration greater than g, which gravity alone cannot supply
- the spring goes slack instead, the restoring force is no longer proportional to displacement, and the block briefly falls freely
Transfer challenge
A trolley of mass 0.50 kg rests on a horizontal frictionless track between two identical springs of spring constant 25 N m⁻¹ each. Both springs are attached to the trolley, their far ends are fixed to walls, and both are initially at their natural lengths. Determine the period of the trolley's oscillation, and explain why g appears nowhere in the answer although it fixed the hanging position in the figure.
Check answer 4 marks
- displacing the trolley by x stretches one spring and compresses the other, so both forces act back towards the centre
- effective spring constant = 2 × 25 = 50 N m⁻¹
- T = 2π sqrt(0.50 / 50) = 0.63 s
- in the vertical case the weight only shifts the equilibrium position by mg/k; measured from that position the restoring force is still -kx, so g never enters the period
03Figure 6Simple harmonic motion · Doppler effectIB
Figure comment
Figure 6A pair of empty axes provided for the sketch. The horizontal axis is labelled 'displacement x / cm' and is scaled from −8.0 through 0 to +8.0, with faint gridlines at −8.0, −4.0, 0, +4.0 and +8.0 and a dashed vertical line drawn at x = 0. The vertical axis is labelled 'energy / J' and carries only a zero at its foot, so no numerical scale is imposed. No curve of any kind is drawn on the axes.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. The horizontal axis is displacement, not time: kinetic energy is an inverted parabola peaking at x = 0 here, not a cosine curve, and the vertical axis carries no scale.
aState The axes are drawn to span the whole of the motion, from one extreme to the other. State the values of x, read from the horizontal scale, at which a potential-energy curve drawn on these axes would reach the total energy of the particle, and state the kinetic energy there.
Check answer 2 marks
- at x = -8.0 cm and x = +8.0 cm, the ends of the scale, which are the amplitude
- the kinetic energy is zero at those two displacements
bDetermine The total energy of the particle is 1.23 × 10⁻² J. Determine its potential energy and its kinetic energy at the gridline x = +4.0 cm.
Check answer 3 marks
- at that gridline x / x0 = 4.0 / 8.0 = 0.50, so x² / x0² = 0.25
- potential energy = 0.25 × 1.23 × 10⁻² = 3.1 × 10⁻³ J
- kinetic energy = 1.23 × 10⁻² - 3.1 × 10⁻³ = 9.2 × 10⁻³ J
cDetermine Determine the displacement, on the scale given, at which the kinetic and potential energies of the particle are equal, and state whether this falls on the +4.0 cm gridline drawn.
Check answer 4 marks
- equal energies means each is half the total, so x² / x0² = 0.50
- x = x0 / sqrt(2) = 8.0 / sqrt(2)
- x = +/- 5.7 cm
- this lies outside the +/- 4.0 cm gridlines, so the curves do not cross at half the amplitude
dExplain The particle is restarted with amplitude 4.0 cm, the gridline on these axes, and the same period. Explain what happens to each of the two energy curves, giving the new total energy.
Check answer 4 marks
- total energy is proportional to x0², so it falls to a quarter: 3.1 × 10⁻³ J
- the potential-energy curve keeps exactly the same shape, since potential energy depends on x and not on the amplitude; it is simply followed only out to +/- 4.0 cm
- the kinetic-energy curve is a new, lower inverted parabola, still peaking at x = 0 but now at 3.1 × 10⁻³ J
- the curves still cross where each is half the total, now at x = 4.0 / sqrt(2) = +/- 2.8 cm
Transfer challenge
A trolley of mass 0.60 kg oscillates on a horizontal spring of spring constant 15 N m⁻¹ with an amplitude of 12 cm. Determine the total energy of the oscillation and the speed of the trolley at a displacement of 6.0 cm, and state the fraction of the total energy that is kinetic at that point.
Check answer 4 marks
- total energy = (1/2) k x0² = 0.5 × 15 × 0.12² = 0.108 J
- potential energy at x = 0.060 m is 0.5 × 15 × 0.060² = 0.027 J, so kinetic energy = 0.081 J
- v = sqrt(2 × 0.081 / 0.60) = 0.52 m s⁻¹
- kinetic fraction = 0.081 / 0.108 = 0.75, matching 1 - (x/x0)² at x/x0 = 0.5
Put the ideas to work.
Timekeeping
A pendulum's nearly amplitude-independent period makes it useful for timing small swings.
Vibration measurements
Displacement, velocity and acceleration describe different aspects of the same motion. Their maximum values depend differently on frequency.
Before you move on.
- Identify the restoring-acceleration condition.
- Relate period and angular frequency.
- Read the phase relationships of x, v and a.
Quick Quiz
Five questions. Answers stay in this open page only.
Downloads & Resources
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