C.1 Concept Builder Part 1 · HL Physics Notebook

1. Start with predictions

7 minPairs

ATL Thinking: predict and justify. Social: compare with a partner. You will test every prediction later with a simulation.

1.1A mass on a spring oscillates with an amplitude of 10 cm. It is pulled to 5 cm instead and released. The period…

A. halves B. stays the same C. doubles

You test it in stage 2.

1.2Which force makes an object oscillate in SHM?

A. a force of constant size that always points towards the centre B. a force that grows in proportion to the displacement and points towards the centre C. a force that grows in proportion to the displacement and points away from the centre

You test it in stage 2.

1.3The mass on a spring is made four times larger. The period…

A. becomes 4 times larger B. becomes 2 times larger C. stays the same

You test it in stage 3.

1.4A pendulum bob is replaced by one with twice the mass (same length). The period…

A. gets longer B. stays the same C. gets shorter

You test it in stage 4.

1.5Explain the reasons for one of your predictions.

What a good answer looks like

A good reason uses an idea, for example: “A bigger stretch gives a bigger force, so…” or “A heavier bob is pulled harder by gravity, but…”. You will see in the simulations which reasons hold.

2. The restoring force: test predictions 1.1 and 1.2

18 minPairs → individual

ATL Thinking: compare two models with a test. Communication: state the conditions in words, not only with symbols.

Simulation A · Restoring force a 0.50 kg block on a smooth table; choose the force law

starting displacement x 0.10 m

force F now—

acceleration a now—

period measured—

2.1Spring law, k = 20 N m⁻¹, x = 0.10 m. Size of the restoring force =

N

2.2Size of the acceleration at x = 0.10 m =

m s⁻²

From Hooke’s law to a = −ω²x

Hooke’s law: the spring pulls back with F = −kx. The minus sign: the force is opposite to x.
It is the only horizontal force, so it is the net force: ma = −kx.
Divide by m: a = −(k/m) x.
Compare with the defining equation of SHM, a = −ω²x: the bracket is ω². So ω² = k/m.
And T = 2π/ω = 2π√(m/k).

2.3For this block, ω² =

s⁻²

2.4Period T =

s

2.5Keep the spring law. Release from 0.10 m, then from 0.05 m. What happens to the period?

it stays 0.99 s: the period does not depend on the amplitude it halves, because the distance halves it gets longer, because the force is smaller

2.6Now choose the constant force (2.0 N towards the centre). Release from 0.10 m and from 0.05 m. What do you see?

the period changes with the amplitude (0.89 s and 0.63 s): this is not SHM the same period, so it is SHM the block does not oscillate

Find the mistake

A question asks: “State the conditions for simple harmonic motion.” [2] A student writes:

a = −ω²x

What is wrong?

Nothing: the equation says everything. The answer must be in words: the acceleration is proportional to the displacement, and directed towards the equilibrium position. The equation alone scores 0. The student should write F = −kx instead.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Which force–displacement graph gives SHM?

A. a horizontal line above the x-axis B. a straight line through the origin with a positive gradient C. a straight line through the origin with a negative gradient

2.7Were predictions 1.1 and 1.2 right? Explain both with the simulation.

What a good answer looks like

1.1: B, the same. With the spring law, 0.10 m and 0.05 m both gave T = 0.99 s: half the distance, half the force, the same time. 1.2: B. Only a force proportional to x and directed towards the centre gives SHM. A constant force also makes the block oscillate, but its period changed with the amplitude (0.89 s and 0.63 s), so it is not SHM.

3. The mass–spring system: test prediction 1.3

15 minPairs → individual

ATL Research: collect data and linearise a relationship. Thinking: check that an answer makes sense.

Simulation B · Timing a mass on a spring change m and k, time the oscillations, plot T² against m

mass m 0.50 kgspring constant k 20 N m⁻¹

period T—

T²—

gradient of T² against m—

k from the gradient—

Changing k clears the graph: every line on it must use the same spring.

3.1m = 0.50 kg, k = 20 N m⁻¹. Period T =

s

3.2Now m = 2.0 kg (four times larger). T =

s

Why plot T² against m?

Square T = 2π√(m/k): T² = (4π²/k) m.
This has the form y = (gradient) × x, with y = T² and x = m.
So T² against m is a straight line through the origin, with gradient 4π²/k.
Then k = 4π²/gradient.

3.3k = 20 N m⁻¹. Gradient of the T²–m line =

s² kg⁻¹

Back check

3.bAnother spring gives a gradient of 2.47 s² kg⁻¹. Its k =

N m⁻¹

3.4The mass is made 4 times larger and k 2 times larger. The new period, as a multiple of T =

× T

Find the mistake

A student finds the period of a 0.50 kg mass on a spring (k = 20 N m⁻¹) like this:

T = 2π√(k/m) = 2π√(20/0.50) = 39.7 s

What is wrong?

The fraction is upside down: T = 2π√(m/k) = 0.99 s. A larger mass must give a longer period, and a stiffer spring a shorter one. The 2π should not be there. Nothing.

What if? · stretch

3.wStretch: for the 0.50 kg mass, T² =

s²

3.5Was prediction 1.3 right? Explain with the equation and your measurements.

What a good answer looks like

B, 2 times larger. T = 2π√(m/k), so T ∝ √m: 4 times the mass gives √4 = 2 times the period (0.99 s became 1.99 s). The T²–m graph is a straight line through the origin, which confirms T² ∝ m.

4. The simple pendulum: test prediction 1.4

15 minPairs → individual

ATL Thinking: spot an approximation (sin θ ≈ θ) and test its limits.

Simulation C · A simple pendulum change the length, the mass, the starting angle and the planet

length l 0.80 mmass of the bob 0.20 kgstarting angle 10°

T from 2π√(l/g)—

real period at this angle—

period measured—

The same method as the spring

The restoring force is the part of the weight along the arc: F = −mg sin θ.
For a small angle in radians, sin θ ≈ θ = x/l, so F ≈ −(mg/l) x.
ma = −(mg/l) x, so a = −(g/l) x. The mass cancels!
Compare with a = −ω²x: ω² = g/l, and T = 2π√(l/g).

4.1l = 0.80 m on Earth. T =

s

4.2Change the mass of the bob from 0.05 kg to 1.00 kg. What happens to the period?

it increases nothing: the mass cancels it decreases

4.3Take the same pendulum to the Moon (g = 1.6 m s⁻²). T =

s

4.4Set the starting angle to 10°, then 30°, then 60°. What happens to the difference between the formula and the real period?

it stays zero it grows: about 0.2 % at 10°, 1.7 % at 30° and 7 % at 60°. The formula is only for small angles the real period becomes shorter

4.5Find the length of a pendulum with a period of 2.0 s on Earth. l =

m

What if? · stretch

4.wStretch: length of a pendulum with T = 1.0 s on the Moon =

m

4.6Was prediction 1.4 right? Explain why the mass does not matter.

What a good answer looks like

B, the same. The restoring force mg sin θ is proportional to m, and so is the inertia (ma). The mass cancels: a = −(g/l)x, so T = 2π√(l/g) has no m in it. The simulation showed the same period for 0.05 kg and 1.00 kg.

5. Any system: why we use ω²

15 minPairs → individual

ATL Thinking: transfer one method to new systems. Communication: explain a general rule.

Simulation D · Any system in SHM four systems: find the bracket in a = −(…)x

defining equation—

ω²—

period T—

Important noteω² = k/m is true only for a mass on a spring. A pendulum gives g/l, two springs give 2k/m, a U-tube gives 2g/L. Every system gives its own bracket, so the general rule of SHM is written a = −ω²x. In a question about any other system, find ω from the data: ω = 2π/T, or from the gradient of an a–x graph.

5.1Two springs, k = 150 N m⁻¹ each, m = 1.5 kg. ω² =

s⁻²

5.2T =

s

5.3Liquid of total length L = 0.50 m in a U-tube: a = −(2g/L)x. T =

s

Find the mistake

A student finds the period of a 0.80 m pendulum with a 0.20 kg bob:

ω² = k/m = 9.8/0.20 = 49, T = 2π/7 = 0.90 s

What is wrong?

k/m is only for a spring. For a pendulum ω² = g/l = 9.8/0.80 = 12.25, so T = 1.80 s, and the mass does not appear. g should be 10. Nothing.

5.4A second identical spring is added side by side (in parallel) to a mass–spring system. New period ÷ old period =

5.5In your own words: why does the IB write the defining equation as a = −ω²x and not as a = −(k/m)x?

What a good answer looks like

Because k/m belongs to one system only. Every system in SHM gives a = −(something)x, with its own something: k/m for a spring, g/l for a pendulum, 2k/m for two springs, 2g/L for a U-tube. Writing ω² covers all of them, and ω can always be found from the period (ω = 2π/T) or from a graph.

6. Show it, then reflect

10 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

state the two conditions for SHM in words;
show where a = −ω²x comes from for a spring;
say what the period does and does not depend on;
use one number from a simulation today.

Option A · Written

One paragraph: “What makes an oscillation simple harmonic, and why the period of SHM does not depend on the amplitude.”

Option B · Sketch

A labelled force–displacement graph for SHM and for the constant force, with today’s numbers.

Option C · Spoken

Explain to your partner in 1 minute how ma = −kx leads to T = 2π√(m/k). Your partner ticks the success criteria.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): An oscillation is simple harmonic when the acceleration (or restoring force) is proportional to the displacement from the equilibrium position and directed towards the equilibrium position. For a spring, ma = −kx, so a = −(k/m)x, which is a = −ω²x with ω² = k/m. The period does not depend on the amplitude: twice the displacement gives twice the force and twice the acceleration, so the trip takes the same time. In the simulation, 0.10 m and 0.05 m both gave T = 0.99 s, but the constant force gave 0.89 s and 0.63 s, so that motion was not SHM.

Exit ticket (no notes)

1A 0.80 kg mass oscillates on a spring of k = 50 N m⁻¹. Its period, in s =

2The period of a 1.2 m pendulum on Earth (g = 9.8 m s⁻²), in s =

3An object of mass 0.20 kg in SHM feels a restoring force of 0.30 N at a displacement of 0.050 m. Its period, in s =

4The conditions for SHM are that the acceleration is…

constant and towards the centre proportional to the displacement and directed towards the equilibrium position proportional to the velocity

5The amplitude of a mass–spring oscillation is doubled. The period…

doubles stays the same increases by √2

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: “I am not sure when to use k/m and when to use ω² from the data.”

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: “Next time I will check which system I have before I write ω² = k/m.”

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKA pendulum clock keeps good time only because the period hardly depends on the amplitude, yet we know T = 2π√(l/g) is only an approximation (sin θ ≈ θ). How can knowledge built on an approximation be reliable enough to measure time for centuries?

Points you could make

An approximation is reliable inside a known range: at 10° the error is only 0.2 %, smaller than other errors in a clock.
Scientists state the limits of a model and test them (in the simulation the error grew to 7 % at 60°).
Engineers designed clocks to stay inside that range (small swings), so the approximation and the technology were built together.

Part 2: Graphs of SHM Read the notes Back to C.1

What makes simple harmonic motion?

Learning goals: by the end you can…

1. Start with predictions

2. The restoring force: test predictions 1.1 and 1.2

3. The mass–spring system: test prediction 1.3

4. The simple pendulum: test prediction 1.4

5. Any system: why we use ω²

6. Show it, then reflect

Exit ticket (no notes)