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C.1 Simple harmonic motion · SL + HL · Part 2 of 3 · student investigation · 80 minutes

Graphs of simple harmonic motion

Predict where an oscillating object is fastest, what the acceleration–displacement graph looks like, and how often the energy repeats. Then test each prediction with a simulation, read period, amplitude, speed and acceleration from graphs, and use the gradient of the a–x graph to find ω and T.

1 Predict · 7
2 x, v, a graphs · 16
3 The a–x graph · 18
4 Read a graph · 14
5 Energy · 15
6 Show + reflect · 10

Learning goals: by the end you can…

sketch and link the x–t, v–t and a–t graphs of SHM.
say where the speed and the acceleration are largest and zero.
explain from an a–x graph why a motion is SHM, and find ω from its gradient (−ω²).
read the period and amplitude from an x–t graph, and find v_max = ωx₀ and a_max = ω²x₀.
describe the energy changes in one oscillation, and explain why energy graphs repeat every T/2.

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Your typed answers are saved on this device only.

Everything you need is on this page. Each simulation uses the same numbers as the questions next to it, and stays locked until you have made your prediction. Type powers of ten like 3.2×10^-19 or 3.2e-19.

📘 From the data booklet (C.1)

a = −ω²xT = 1/f = 2π/ωT = 2π√(m/k)

Velocity is the gradient of the x–t graph; acceleration is the gradient of the v–t graph. v_max = ωx₀ and a_max = ω²x₀ are in the notes (HL equations).

2. The x, v and a graphs: test prediction 1.1

16 minPairs → individual

ATL Thinking: link three representations of one motion. Research: read values from a moving model.

Simulation A · Displacement, velocity and acceleration the block and its three graphs, with a moving cursor

amplitude x₀ 0.040 mperiod T 0.80 s

x now—

v now—

a now—

v_max / a_max—

2.6Play the simulation and pause when the block is at an end (x = +x₀). What are v and a there?

v = 0 and a is largest, pointing towards the centre v is largest and a = 0 v = 0 and a = 0

2.1x₀ = 0.040 m, T = 0.80 s. ω =

rad s⁻¹

2.2v_max = ωx₀ =

m s⁻¹

2.3a_max = ω²x₀ =

m s⁻²

How the three graphs are linked

v is the gradient of the x–t graph: zero at a peak, largest where x crosses zero.
a is the gradient of the v–t graph.
Also a = −ω²x: the a–t graph is the x–t graph upside down (scaled by ω²).
Each graph is a quarter of a period (π/2) after the one before: v peaks first, then x, then a.

Find the mistake

A student reads the amplitude from an x–t graph that goes from +4.0 cm to −4.0 cm:

amplitude = 4.0 − (−4.0) = 8.0 cm

What is wrong?

The amplitude is measured from the equilibrium position (the time axis) to a peak: 4.0 cm. 8.0 cm is 2x₀. The amplitude should be negative. Nothing.

Hinge question

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

An object in SHM is moving towards the equilibrium position. Its velocity and acceleration are…

A. in the same direction: it speeds up B. in opposite directions: it slows down C. both zero

2.7Was prediction 1.1 right? Explain with the graphs.

What a good answer looks like

B, at the equilibrium position. There the x–t graph is steepest, so v is largest (v_max = ωx₀ = 0.314 m s⁻¹). At the ends the graph is flat, so v = 0: the object stops to turn round. The acceleration is the other way round: zero at the centre and largest at the ends.

3. The a–x graph: test predictions 1.2 and 1.3

18 minPairs → individual

ATL Thinking: match an equation to a graph (y = mx). Communication: explain with two features of a graph.

Simulation B · The acceleration–displacement graph amplitude 0.050 m; change k and m and watch the line

spring constant k 32 N m⁻¹mass m 0.50 kg

gradient of a–x—

ω²—

period T—

a_max—

The key graph of C.1a = −ω²x has the form y = mx: the a–x graph is a straight line through the origin (a ∝ x) with a negative gradient (a opposite to x). Gradient = −ω², so ω = √(−gradient), T = 2π/ω and f = ω/2π. The line stops at ±x₀.

3.1k = 32 N m⁻¹, m = 0.50 kg. Gradient of the a–x line =

s⁻²

3.2From the gradient: T =

Back check

3.bLargest acceleration (the end of the line), a_max =

m s⁻²

3.3Another a–x line passes through (x = −0.10 m, a = 2.5 m s⁻²). f² =

s⁻²

3.4Change k and m so that the period doubles (for example m from 0.50 kg to 2.00 kg). What happens to the gradient?

it becomes ¼ as steep: gradient = −ω² and ω halves it becomes ½ as steep it does not change

Find the mistake

From an a–x graph with gradient −64 s⁻², a student writes:

ω = 64 rad s⁻¹, so T = 2π/64 = 0.098 s

What is wrong?

The gradient is −ω², not ω: ω = √64 = 8.0 rad s⁻¹ and T = 0.785 s. The minus sign was dropped: ω = −64. Nothing.

Hinge question

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

An a–x graph is a straight line with a negative gradient, but it crosses the x-axis at x = 2 cm, not at 0. The motion is…

A. SHM about x = 0 B. not SHM about x = 0: a is not proportional to x there C. not an oscillation at all

3.5Write a 2-mark answer: “Explain, using the a–x graph, why the motion is simple harmonic.” Then say whether predictions 1.2 and 1.3 were right.

What a good answer looks like

✔ The graph is a straight line through the origin, so the acceleration is proportional to the displacement. ✔ The gradient is negative, so the acceleration is opposite to the displacement (towards equilibrium). 1.2: A was right. 1.3: C, ¼ as steep: doubling T halves ω and quarters ω².

4. Reading an x–t graph

14 minPairs → individual

ATL Research: read data from a graph and use a tangent. Thinking: link a gradient to a physical quantity.

Simulation C · Reading a displacement–time graph move the cursor; the red line is the tangent (its gradient is v)

time t 0.000 s

x—

v (gradient)—

a—

4.1Read the period from the graph. T =

4.2x₀ = 3.0 cm. v_max =

m s⁻¹

4.3a_max =

m s⁻²

4.4A time between 0.20 s and 0.40 s when the velocity is positive and largest. t =

4.5A time when the acceleration is positive and largest (first time). t =

What if? · stretch

4.wStretch: first time the acceleration is zero. t =

4.6Describe how you find a time when the velocity is zero from an x–t graph, and why.

What a good answer looks like

Find a peak or a trough. There the graph is flat (the tangent is horizontal), so the gradient, which is the velocity, is zero. The object is at an end, turning round: here at t = 0, 0.20 s, 0.40 s…

5. Energy in SHM: test prediction 1.4

15 minPairs → individual

ATL Communication: describe energy transfers in words. Thinking: connect a graph of energy with the motion.

Simulation D · Energy in SHM m = 0.20 kg, k = 20 N m⁻¹, no friction

amplitude x₀ 0.10 m

kinetic energy—

elastic potential energy—

total energy—

x—

5.1x₀ = 0.10 m. Total energy (at an end, all elastic) =

5.2At x = 0.050 m, the elastic energy =

5.3Maximum speed =

m s⁻¹

5.4How often does the kinetic energy reach its maximum? Every…

5.5Double the amplitude (0.10 m → 0.20 m). The total energy…

doubles becomes 4 times larger (0.40 J) stays the same

Find the mistake

The kinetic energy of an oscillator is zero at t = 0 and zero again at t = 0.25 s. A student writes:

T = 0.25 s

What is wrong?

Zero kinetic energy means an end. 0 s and 0.25 s are the two different ends, half a cycle apart: T = 0.50 s. T = 0.125 s. Nothing.

5.6Write a 2-mark answer: “Describe the energy changes during one oscillation.” Then say whether prediction 1.4 was right.

What a good answer looks like

✔ At the ends all the energy is elastic potential energy and the kinetic energy is zero; at the equilibrium position all the energy is kinetic (maximum speed) and the potential energy is zero. ✔ In between, energy is transferred between kinetic and potential, and the total stays constant because there is no friction. 1.4: B, every T/2: the kinetic energy peaks each time the block passes the centre, twice per period.

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…

link the x–t, v–t and a–t graphs (gradients, upside down);
say where v and a are largest and zero;
use the a–x gradient −ω² correctly;
use one number from a simulation today.

Option A · Written

One paragraph: “What the a–x graph of SHM looks like, why, and how to get the period from it.”

Option B · Sketch

The x–t, v–t, a–t and E–t graphs for one period, one above the other, with the times T/4, T/2, 3T/4 marked.

Option C · Spoken

Explain to your partner in 1 minute where an oscillating object is fastest and where its acceleration is largest, and why. Your partner ticks the success criteria.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): For SHM, a = −ω²x, so the a–x graph is a straight line through the origin with a negative gradient: a is proportional to x and opposite to it. The line ends at ±x₀. The gradient equals −ω², so ω = √(−gradient) and T = 2π/ω. In the simulation, k = 32 N m⁻¹ and m = 0.50 kg gave a gradient of −64 s⁻², so ω = 8.0 rad s⁻¹ and T = 0.785 s.

Exit ticket (no notes)

1An oscillation has amplitude 0.050 m and period 0.50 s. Its maximum speed, in m s⁻¹ =

2The gradient of an a–x graph is −36 s⁻². The period, in s =

3The kinetic energy of an oscillator is zero every 0.30 s. The period, in s =

4At the equilibrium position, an object in SHM has…

zero speed and maximum acceleration maximum speed and zero acceleration maximum speed and maximum acceleration

5Compared with the x–t graph, the a–t graph of SHM is…

the same shape the same shape, upside down shifted by a quarter of a period

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 how to find the period from an a–x graph.”

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 whether a value is the gradient or ω before I calculate the period.”

TOK · 2 minutes

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

TOKThe same motion can be shown as an x–t graph, a v–t graph, an a–x graph, an energy graph or an equation. Do different representations give us different knowledge, or the same knowledge in different forms?

Points you could make

They contain the same physics: each can be worked out from the others.
But each makes a different feature easy to see: the a–x graph shows the SHM condition at once, the energy graph shows the transfers.
Choosing a representation is part of knowing: experts move between them to answer different questions.

Part 3 (HL): Equations and energy Read the notes Back to C.1