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.
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.
Simple definition
Simple harmonic motion is any oscillation in which the acceleration is proportional to the displacement from equilibrium and directed back towards it: a = −ω²x.
Example
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.
Defining conditiona = −ω²x
Acceleration 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 frequencyT = 1/f · ω = 2πf
Period 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 springT = 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 pendulumT = 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
01
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.
02
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.
03
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.
04
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 phase difference of T/2, or 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.
02
Change one variable at a time
Make the relationship visible.
x is a cosine, v a negative sine, a a negative cosine (v and a drawn to their own scales). Change A and only the x curve grows — the period refuses to move. Change ω and all three squeeze together.
Period T = 2π/ω2.09 s
Max speed v(max) = ωA3.6 m/s
Max acceleration a(max) = ω²A10.8 m/s²
03
Catch the common trap
Explain before calculating.
At the extreme ends of a simple harmonic oscillation, the object has…
Choose an answer to test the model.
04
Worked examples
State the rule, substitute, then check units.
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.
AnswerT = 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.
Answera = −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.
AnswerL ≈ 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.
Answerω ≈ 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.Open subsectionHide 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.
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.
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.
recall2 marks
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.
routine3 marks
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.
demanding3 marks
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.
top of the paper4 marks
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
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.
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.
recall2 marks
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.
routine3 marks
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.
demanding4 marks
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.
top of the paper4 marks
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
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.
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.
recall2 marks
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.
routine3 marks
Check answer · 3 marks
at that gridline x / x0 = 4.0 / 8.0 = 0.50, so x² / x0² = 0.25
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.
demanding4 marks
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.
top of the paper4 marks
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