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Educerie · IB Diploma · Physics
Theme C Wave behaviour · C.1 Simple harmonic motion
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| State the conditions that lead to simple harmonic motion | SL, HL | "State the conditions for SHM" (2 marks) |
| Use a = −ω²x and explain its minus sign | SL, HL | "Outline the significance of the negative sign" (1 mark); ω from an a–x graph |
| Describe an oscillation by T, f, ω, amplitude, equilibrium position and displacement, with T = 1/f = 2π/ω | SL, HL | Paper 1A graph reading; "determine the frequency" (1–2 marks) |
| Use T = 2π√(m/k) for a mass–spring system | SL, HL | "Calculate the spring constant" (2 marks); predict the effect of changing m |
| Use T = 2π√(l/g) for a simple pendulum | SL, HL | Paper 1B: plot T² against l and find g from the gradient (3–4 marks) |
| Describe qualitatively how kinetic, potential and total energy change during one cycle | SL, HL | "Describe the energy changes…" (3 marks); sketch energy against x or t |
| Describe an oscillation using its phase angle, in radians | HL only | "State the phase angle"; phase difference between two graphs |
| Solve problems with x = x₀ sin(ωt + ϕ), v = ωx₀ cos(ωt + ϕ) and v = ±ω√(x₀² − x²) | HL only | Paper 2 calculation, 2–4 marks, calculator in radians |
| Calculate energies with Eₜ = ½mω²x₀² and Eₚ = ½mω²x² | HL only | Paper 2: total energy, then the share that is kinetic at a given x |
Before you start
You need Newton's second law, F = ma, and the idea of a resultant force from A.2, because SHM is nothing more than Newton's second law applied to one particular force. You need kinetic energy, gravitational potential energy and elastic potential energy from A.3. HL students also need circular motion from A.2, since ω is an angular speed and the reference circle in section 7 is circular motion seen from the side. Have your calculator ready to switch into radians.
1The idea in one paragraph
Pull something away from a stable resting place and let go, and it swings back, overshoots, and returns. When the force pulling it back is proportional to how far it has been displaced, and always points back to the resting place, the motion has a special and very tidy form called simple harmonic motion (SHM): the displacement follows a sine curve in time, and the period does not depend on how big the swing is. That one condition is written as a = −ω²x. From it come the period of a mass on a spring and of a pendulum, the shapes of the displacement, velocity and acceleration graphs, and the steady swap between kinetic and potential energy. Almost any system disturbed a small amount from a stable equilibrium behaves this way, which is why the model reaches from a swinging child to a vibrating molecule.
2Describing an oscillation
An oscillation is a repeated back-and-forth motion about a fixed point. Every oscillation you meet in this course is described with the same six words, and Figure 1 puts them on one graph.
- The equilibrium position is where the object would rest if it were not oscillating. It is where the resultant force on it is zero.
- The displacement, x, is the distance from the equilibrium position in a stated direction. It is a vector, so it is positive on one side and negative on the other.
- The amplitude, x₀, is the maximum displacement. It is measured from the equilibrium position to one extreme, not from one extreme to the other.
- The time period, T, is the time for one complete oscillation, measured in seconds.
- The frequency, f, is the number of oscillations per second, measured in hertz (Hz).
- The angular frequency, ω, is 2π times the frequency, measured in radians per second (rad s⁻¹).
The three timing quantities are tied together by one line from the data booklet:
T = 1/f = 2π/ω
Why 2π? One complete oscillation counts as one full turn, 2π radians, of a point going round a circle. You meet that circle properly in section 7 at HL. At SL it is enough to know that ω is the frequency counted in radians per second rather than in cycles per second.
Here is the buoy in Figure 1 as a calculation. A buoy on a gentle swell completes 12 oscillations in 30.0 s.
In an experiment, always time many oscillations and divide, as here: the reaction-time uncertainty at each end of the timing is then shared across all of them.
3What makes motion simple harmonic
Not every oscillation is SHM. A ball bouncing on a hard floor repeats itself, but the only force on it in the air is its weight, which is the same size wherever the ball is. That is periodic motion and not SHM. For SHM the resultant force must meet two conditions:
- its size is proportional to the displacement from the equilibrium position, and
- its direction is always towards the equilibrium position, opposite to the displacement.
A force that meets both conditions is called a restoring force. Figure 2 shows the two oscillators the course uses, and in each the restoring force is drawn in clay.
For the mass on a spring in panel (a), the spring obeys Hooke's law: stretch it by x and it pulls back with a force kx, where k is the spring constant in N m⁻¹. Squash it by x and it pushes back with kx. In both cases the force points to the equilibrium position. Written with signs, F = −kx. Put that into Newton's second law:
k/m is a positive constant for this spring and this mass. Call it ω², and you have the defining equation of SHM:
a = −ω²x
Any motion whose acceleration obeys this equation is simple harmonic, whatever the object is. For the spring, ω² = k/m.
The minus sign carries the physics. It says the acceleration always points the opposite way to the displacement, so the object is always being pulled back towards equilibrium. Without it, a = +ω²x would describe an object that is pushed further away the further it goes, which runs away and never comes back. An examiner who asks for "the significance of the negative sign" wants exactly that: the acceleration, and so the force, is always directed towards the equilibrium position.
Writing the constant as ω² keeps it positive, so the minus sign alone fixes the direction, and it is the same ω as in T = 2π/ω.
The pendulum in panel (b) meets the conditions only approximately. The restoring force is the component of the weight along the arc, mg sin θ. For small angles, sin θ ≈ θ in radians, and θ is the arc length divided by l, so the restoring force is close to proportional to the displacement along the arc. Beyond about 10° the approximation starts to fail and the motion is no longer quite SHM. That is why the pendulum formula comes with the words "small amplitude" attached.
Figure 3 is the graph that identifies SHM at a glance. Plot acceleration against displacement and SHM gives a straight line through the origin with a negative gradient, stopping at ±x₀.
Read the angular frequency from it. The gradient is −1.6 ÷ 0.10:
The last line is the minus sign at work: a displacement on the negative side gives a positive acceleration, back towards the middle.
Why the model is everywhere. Near any stable equilibrium, a small displacement produces a restoring force that is very nearly proportional to it. So SHM is how almost every stable system responds to a small disturbance: a car on its suspension, a swaying tower, atoms in a crystal, atoms vibrating along the bonds of a molecule (the greenhouse gases of B.2 absorb infrared at their own vibration frequencies), and every particle of a medium carrying a wave in C.2.
Real oscillations lose energy to friction and air resistance, so their amplitude shrinks. That is damping, treated in C.4; here there is none unless a question says so.
4Following one cycle: displacement, velocity and acceleration
Start the mass at equilibrium moving in the positive direction, and follow it round one cycle. Figure 4 stacks the three graphs on one time axis.
Read down any dashed line and the three graphs tell one story.
- At the equilibrium position (t = 0, T/2, T): displacement, restoring force and acceleration are all zero, and the speed is greatest.
- At an extreme (t = T/4, 3T/4): displacement is ±x₀, the acceleration is greatest and points back to equilibrium, and the mass is momentarily at rest while it turns round.
Three shape rules follow.
Velocity is the gradient of the displacement graph. Where the x–t graph is steepest, at x = 0, the velocity is greatest. Where the x–t graph is flat, at the peaks, the velocity is zero. So the velocity graph is the same wave shape shifted a quarter of a cycle along.
Acceleration is the gradient of the velocity graph, and, by a = −ω²x, it is also the displacement graph turned upside down and scaled by ω². When x is at its positive peak, a is at its negative peak.
The peak values are x₀, ωx₀ and ω²x₀. The maximum acceleration comes straight from a = −ω²x with x = x₀. The maximum speed, ωx₀, is shown here for the shape; HL students derive and use it in section 7.
A useful check on any SHM sketch: the acceleration graph must always have the opposite sign to the displacement graph at the same instant.
5The two oscillators you must know
The data booklet gives two periods. Both come from ω² for that system, put into T = 2π/ω.
Mass–spring system. From section 3, ω² = k/m, so
T = 2π√(m/k)
A heavier mass is harder to accelerate, so it oscillates more slowly. A stiffer spring pulls back harder, so it oscillates faster. The period of a mass on a vertical spring is the same as on a horizontal one: gravity moves the equilibrium position down but does not change the restoring force about it.
Simple pendulum. A small, heavy bob on a light, inextensible string of length l, measured from the pivot to the centre of the bob, swinging through a small angle. For small angles ω² = g/l, so
T = 2π√(l/g)
Notice what is missing. The bob's mass: a heavier bob feels a bigger restoring force but is harder to accelerate, and the two cancel. The amplitude, from both equations: a bigger swing has further to go but is pulled back harder, and in SHM those cancel exactly, which is what made pendulums good clocks.
Getting g from data. Paper 1B favours this. The equation T = 2π√(l/g) is not a straight line if you plot T against l. Square it and rearrange:
So a graph of T² against l is a straight line through the origin, and its gradient is 4π²/g. Figure 5 is a class's invented data set drawn this way.
Take the gradient from the best-fit line using a triangle that spans most of it, never from two data points. The same method works for the spring: T² against m is a straight line of gradient 4π²/k.
6Energy in one cycle
With no damping, the total energy of an oscillator stays constant. What changes is how it is shared between kinetic energy and potential energy, and Figure 6 draws the sharing two ways.
Follow a pendulum released from rest at one side. At the extreme the bob is momentarily at rest and at its highest point: kinetic energy zero, gravitational potential energy maximum. Swinging down, potential energy becomes kinetic. At the equilibrium position it is fastest and lowest: kinetic energy maximum, potential energy minimum. Swinging up the other side, kinetic becomes potential again until it stops. A mass on a horizontal spring tells the same story with elastic potential energy in the spring.
Panel (a) plots energy against displacement: kinetic energy is an upside-down parabola with its peak at x = 0, potential energy is an upright parabola with its peaks at ±x₀, and the total is a horizontal line. Panel (b) plots energy against time, and it holds the trap in this section: the energies go through two complete cycles for every one oscillation, because the kinetic energy is at its maximum twice per period, once passing through the middle in each direction. Kinetic energy depends on speed, not on direction, so it is never negative.
If the oscillation is damped, the total-energy line slopes gently downwards and the amplitude shrinks with it.
7HLPhase angle and the equations of SHM
SL students can skip to section 9.
A sine curve can be started at any point in its cycle. The phase angle, ϕ, says where. It is an angle in radians, and it fixes the displacement at t = 0. The data booklet gives the displacement and velocity of any SHM as:
x = x₀ sin(ωt + ϕ) and v = ωx₀ cos(ωt + ϕ)
Figure 7 shows where these come from. Imagine a point going round a circle of radius x₀ at a constant angular speed ω. Its height above the centre is x₀ sin(angle), and the angle at time t is ωt + ϕ, where ϕ is the angle it started at. Watch only the height, as if you were seeing the circle edge-on, and you see SHM. That is why ω is called an angular frequency: it is the angular speed of the reference point.
Three starting positions cover almost every question:
| At t = 0 the object is… | ϕ | The equation becomes |
|---|---|---|
| at equilibrium, moving in the positive direction | 0 | x = x₀ sin ωt |
| at the positive extreme, at rest (released from +x₀) | π/2 | x = x₀ sin(ωt + π/2) = x₀ cos ωt |
| at equilibrium, moving in the negative direction | π | x = x₀ sin(ωt + π) = −x₀ sin ωt |
The phase difference between two oscillators of the same frequency is the difference between their phase angles. In Figure 8, B reaches its peak a quarter of a period before A, so B leads A by π/2 rad. In general, a time gap Δt between matching points gives a phase difference of 2π × Δt ÷ T.
The third equation links speed directly to displacement, with no time in it at all:
v = ±ω√(x₀² − x²)
The ± is there because at any displacement the object may be moving either way; the equation gives the speed and the question gives the direction. At x = 0 it gives the maximum speed, v = ωx₀. At x = ±x₀ it gives zero.
Now a full example. A 0.40 kg mass on a spring is pulled 0.050 m from equilibrium and released at t = 0. Its period is 0.80 s. Calculator in radians.
Questions also run the other way: how long after release does the mass first reach x = +0.025 m?
That is T/6, not T/8: the mass moves slowly near the extreme, so the first half of the distance takes longer.
8HLEnergy in numbers
At HL the energy of section 6 becomes quantitative. For a mass m oscillating with angular frequency ω and amplitude x₀, the data booklet gives:
Eₜ = ½mω²x₀² and Eₚ = ½mω²x²
The kinetic energy is whatever is left: Eₖ = Eₜ − Eₚ = ½mω²(x₀² − x²). Put v = ω√(x₀² − x²) into ½mv² and you get exactly the same thing, which is a good check that the equations belong together. For a spring, mω² = k, so these are the familiar ½kx₀² and ½kx².
Continue the example from section 7.
Two results to carry into the exam. The total energy is proportional to the square of the amplitude, so doubling the amplitude of the same oscillator quadruples its energy. And kinetic and potential energy are equal where x² = x₀²/2, that is at x = ±x₀/√2 = ±0.71x₀, not at half the amplitude. That is where the two parabolas cross in Figure 6(a).
9Where marks are lost
Missing half of the definition. "The force is proportional to the displacement" is only one condition. The force must also be directed towards the equilibrium position. Both are needed for both marks.
Explaining the minus sign as "the acceleration is negative". It is not always negative: at negative displacement it is positive. The minus sign means the acceleration is always in the opposite direction to the displacement, towards equilibrium.
Mixing up f and ω. ω = 2πf. Putting f into a = −ω²x is the most common numerical slip. Likewise the time for 20 oscillations is not T, and the amplitude is not the extreme-to-extreme distance.
Putting the extremes in the wrong place. At the equilibrium position the speed is greatest and the acceleration is zero. At the extremes the speed is zero and the acceleration is greatest. Swapping these is the classic Paper 1A trap.
Thinking the period depends on the amplitude or on the bob's mass. In SHM it depends on neither. The pendulum period depends on l and g only; the spring period on m and k only.
Drawing the energy–time graph with period T. Kinetic and potential energy each go through two full cycles per oscillation, and neither ever goes below zero.
HL: calculator in degrees, or ϕ = 0 by habit. sin(ωt + ϕ) needs radians. And an object released from rest at maximum displacement has ϕ = π/2, not 0.
10Draw it right
- x against t: a sine or cosine curve of constant amplitude, ±x₀ labelled, T marked peak to peak.
- v against t: shifted a quarter-period, so v is zero where x peaks and peaks where x is zero.
- a against t: the x–t graph upside down; a always has the opposite sign to x.
- a against x: a straight line through the origin with a negative gradient, stopping at x = ±x₀. Not a curve, and not continuing beyond the amplitude.
- Energy against x: two parabolas, kinetic peaking at x = 0 and potential peaking at ±x₀, crossing at ±x₀/√2, with a horizontal total-energy line touching both peaks.
- Energy against t: two curves that never go negative, completing two cycles per period, adding to a constant.
- Linearised data: T² on the vertical axis, l or m on the horizontal, both with units (s², m, kg), a ruled best-fit line, and a large gradient triangle.
11Try it
Marks in brackets. Answers and marker's notes are at the end. Show every substitution: it carries marks of its own.
Q1. A particle performs simple harmonic motion with period 0.50 s and amplitude 3.0 cm. What is the magnitude of its maximum acceleration? 1 mark
A. 0.12 m s⁻² B. 0.38 m s⁻² C. 4.7 m s⁻² D. 470 m s⁻²
Q2. (a) State the two conditions for a body to perform simple harmonic motion. 2 marks
(b) Outline the significance of the negative sign in the equation a = −ω²x. 1 mark
Q3. A 0.60 kg mass hangs from a spring and oscillates vertically with a period of 0.90 s.
(a) Calculate the spring constant of the spring. 2 marks
(b) The mass is replaced by a 2.4 kg mass. Determine the new period. 2 marks
Q4. A student investigates how the period of a mass on a spring depends on the mass. She times 20 oscillations for each mass. Her results (invented) are:
| m / kg | 0.100 | 0.200 | 0.300 | 0.400 | 0.500 |
|---|---|---|---|---|---|
| time for 20 oscillations / s | 10.9 | 14.7 | 17.8 | 20.4 | 22.6 |
(a) Explain why she timed 20 oscillations rather than one. 1 mark
(b) Calculate T² for m = 0.300 kg. 1 mark
(c) Her graph of T² against m is a straight line of gradient 2.46 s² kg⁻¹. Determine the spring constant. 2 marks
(d) The line meets the T² axis at 0.051 s², not at the origin. Suggest a reason. 2 marks
Q5. A simple pendulum is released from rest at a small angle. Describe the energy changes of the bob during one complete oscillation, assuming no energy is lost. 3 marks
Q6 (HL). An object of mass 0.25 kg performs SHM with amplitude 0.12 m and period 1.5 s. At t = 0 it passes through the equilibrium position moving in the positive direction.
(a) State its phase angle ϕ. 1 mark
(b) Calculate its displacement at t = 0.40 s. 2 marks
(c) Calculate its speed when its displacement is 0.060 m. 2 marks
(d) Determine the fraction of its total energy that is kinetic when its displacement is 0.060 m. 2 marks
12In one breath
SHM happens when the restoring force is proportional to the displacement and always directed towards equilibrium, which Newton's second law turns into a = −ω²x; the minus sign means the acceleration always points back to the middle. T = 1/f = 2π/ω. Speed is greatest and acceleration zero at equilibrium; speed is zero and acceleration greatest at the extremes; a against x is a straight line through the origin of gradient −ω². A mass on a spring has T = 2π√(m/k), a small-angle pendulum T = 2π√(l/g), neither depending on amplitude, and T² against l or m is a straight line whose gradient gives g or k. Energy swaps between kinetic and potential twice each cycle; the total stays constant. HL: x = x₀ sin(ωt + ϕ) in radians, v = ωx₀ cos(ωt + ϕ), v = ±ω√(x₀² − x²), total energy ½mω²x₀², so energy goes as amplitude squared.
Answers
Q1. C. ω = 2π ÷ 0.50 = 12.6 rad s⁻¹, and maximum acceleration = ω²x₀ = 12.6² × 0.030 = 4.7 m s⁻². A uses f²x₀ instead of ω²x₀. B is ωx₀, the maximum speed. D leaves the amplitude in centimetres.
Q2. (a) The acceleration, or resultant force, is proportional to the displacement from the equilibrium position, and it is always directed towards the equilibrium position. (b) The acceleration is always in the opposite direction to the displacement, so the object is always pulled back towards equilibrium. 1 for proportional to displacement, 1 for directed towards equilibrium or opposite to displacement, 1 for the meaning of the sign. "The acceleration is negative" scores 0 for part (b).
Q3. (a) T = 2π√(m/k), so k = 4π²m ÷ T² = 4π² × 0.60 ÷ 0.90² = 29 N m⁻¹. (b) T is proportional to √m. The mass is multiplied by 4, so the period is multiplied by √4 = 2: T = 2 × 0.90 = 1.8 s. M1 for rearranging and substituting, A1 for 29 N m⁻¹ with unit; M1 for T ∝ √m or a recalculation with k, A1 for 1.8 s. A correct method in (b) using a wrong k from (a) keeps both marks.
Q4. (a) Timing many oscillations reduces the percentage uncertainty in T, because the reaction-time uncertainty of starting and stopping the watch is spread over 20 periods. (b) T = 17.8 ÷ 20 = 0.890 s, so T² = 0.792 s². (c) The gradient is 4π²/k, so k = 4π² ÷ 2.46 = 16 N m⁻¹. (d) The spring has mass and part of it oscillates with the load, so the oscillating mass is a little more than m at every point, which lifts every T² and the whole line. A zero error in the masses is also acceptable. (a) 1 for smaller uncertainty in T with a reason; (b) 1 for 0.792 s², not 0.890; (c) M1 for gradient = 4π²/k, A1 for 16 N m⁻¹; (d) 1 for the spring's mass or a systematic error, 1 for it raising every T² and lifting the line. "Human error" scores 0.
Q5. At release the bob is at rest at its highest point: all its energy is gravitational potential, kinetic energy zero. Swinging down, potential energy is converted into kinetic, until at equilibrium kinetic energy is maximum and potential energy minimum. Rising to the other extreme it converts kinetic back into potential, stops momentarily, and the sequence repeats on the way back. The total energy is constant throughout. 1 for all potential and zero kinetic at the extremes, 1 for maximum kinetic at equilibrium, 1 for the conversion in both directions with total energy constant. An answer that stops at the bottom of the first swing is capped at 2.
Q6 (HL). (a) ϕ = 0. (b) ω = 2π ÷ 1.5 = 4.19 rad s⁻¹. x = 0.12 sin(4.19 × 0.40) = 0.12 sin(1.68) = 0.119 m. (c) v = ω√(x₀² − x²) = 4.19 × √(0.12² − 0.060²) = 4.19 × 0.104 = 0.435 m s⁻¹. (d) Eₖ ÷ Eₜ = (x₀² − x²) ÷ x₀² = (0.0144 − 0.0036) ÷ 0.0144 = 0.75, so three-quarters of the energy is kinetic. (a) A1; (b) M1 for ω and substitution in radians, A1 for 0.119 m; 0.0035 m comes from degrees and scores M1 only; (c) M1, A1; (d) M1 for Eₖ = ½mω²(x₀² − x²) or the ratio, A1 for 0.75. Calculating 0.0237 J and 0.0316 J and dividing is equally good.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section C.1 Simple harmonic motion. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.