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

HL: the equations of SHM and energy

Predict the velocity equation, where an object starts for a given phase angle, how fast it moves halfway out, and where the kinetic and potential energies are equal. Then test each prediction with a simulation: differentiate x = x₀ sin ωt, read the phase angle on a turning circle, use v = ±ω√(x₀² − x²), and find the energies of a vertical spring.

1 Predict · 7
2 Differentiate · 16
3 Phase angle · 17
4 Speed and x · 13
5 Energy · 17
6 Show + reflect · 10

Learning goals: by the end you can…

get v and a by differentiating x = x₀ sin ωt (and x₀ cos ωt), with the calculator in radians.
read the data booklet equation x = x₀ sin(ωt + φ): find φ from the start, using the direction of motion.
find a phase difference from a time shift, Δφ = (Δt/T) × 2π.
use v = ±ω√(x₀² − x²), v_max = ωx₀ and a_max = ω²x₀.
use E_T = ½mω²x₀² and E_p = ½mω²x², and avoid the vertical-spring trap.

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)

x = x₀ sin(ωt + φ)v = ωx₀ cos(ωt + φ)v = ±ω√(x₀² − x²)E_T = ½mω²x₀²E_p = ½mω²x²a = −ω²x

φ is in radians. The derivative of sin ωt is ω cos ωt, and the derivative of cos ωt is −ω sin ωt.

2. The equations by differentiating: test prediction 1.1

16 minPairs → individual

ATL Thinking: link a gradient to a derivative. Research: check a result two ways (a formula and a measured tangent).

Simulation A · Differentiating x = 0.050 sin(8.0t) the red line is the tangent; its gradient is v

time t 0.000 s

x—

gradient of x–t (measured)—

ωx₀ cos ωt—

a = −ω²x—

From x to v to a

x = x₀ sin ωt (the clock starts at the centre, moving +).
v = dx/dt: v = ωx₀ cos ωt (the ω comes out in front).
a = dv/dt: a = −ω²x₀ sin ωt.
x₀ sin ωt is x, so a = −ω²x: the defining equation comes back.
The largest value of sin or cos is 1: v_max = ωx₀, a_max = ω²x₀.

2.1x = 0.050 sin(8.0t). Period T =

2.2v_max =

m s⁻¹

2.3Set t = 0.10 s. x =

2.4v at t = 0.10 s =

m s⁻¹

2.5a at t = 0.10 s =

m s⁻²

Find the mistake

For x = 0.050 sin(8.0t) at t = 0.10 s, a student’s calculator gives:

x = 0.050 sin(0.8) = 0.000698 m

What is wrong?

The calculator is in degrees. ωt is an angle in radians: sin(0.8 rad) = 0.717, so x = 0.0359 m. The student should use cos. Nothing.

2.6An object starts at rest at +x₀, so x = x₀ cos ωt. Its velocity is…

v = ωx₀ sin ωt v = −ωx₀ sin ωt v = ωx₀ cos ωt

2.7Was prediction 1.1 right? Explain with the simulation.

What a good answer looks like

A. Differentiating x₀ sin ωt gives ωx₀ cos ωt: the ω comes out in front. In the simulation the gradient of the tangent at t = 0.10 s was 0.279 m s⁻¹, the same as 0.40 cos(0.80). Option B forgets the ω, so its units are wrong (m, not m s⁻¹).

3. The phase angle: test prediction 1.2

17 minPairs → individual

ATL Thinking: use a model (the turning circle) to read an equation. Communication: say where the object is and which way it moves.

Simulation B · The phase angle on a turning circle x₀ = 0.20 m, T = 2.0 s (ω = π rad s⁻¹); choose φ

phase angle φ 0 rad

x at t = 0—

motion at t = 0—

equation—

Reading x = x₀ sin(ωt + φ)x₀ = amplitude, ω = angular frequency (the number in front of t), and φ = the phase angle: the angle of the point on the circle when the clock starts. φ does not change the amplitude, the period or the energy; it only says where the object is, and which way it moves, at t = 0. To find φ: x(0) = x₀ sin φ, and the sign of v(0) = ωx₀ cos φ chooses between the two angles with the same sine.

3.1x(0) = +0.10 m and moving in the + direction. φ =

rad

3.2x(0) = +0.10 m but moving in the − direction (towards the centre). φ =

rad

3.3The object starts at rest at −x₀. φ (between 0 and 2π) =

rad

3.4A loudspeaker cone moves with x = 0.30 sin(4πt + π/2) m. Frequency =

3.5Two oscillations of period 0.40 s; one reaches its peaks 0.050 s after the other. Phase difference =

rad

Hinge question

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

x = 0.20 sin(πt + π). At t = 0 the object is…

A. at +x₀, at rest B. at the centre, moving in the − direction C. at the centre, moving in the + direction

3.6Was prediction 1.2 right? Explain with the circle.

What a good answer looks like

B, at +x₀, at rest. φ = π/2 puts the point at the top of the circle at t = 0: its height is the full radius, x = +x₀, and it is moving sideways, so the height is not changing: v = 0. In fact sin(ωt + π/2) = cos ωt, which is the “start at an end” equation.

4. Speed at any displacement: test prediction 1.3

13 minPairs → individual

ATL Thinking: read a graph that has no time axis. Research: compare a prediction with a calculated value.

Simulation C · Speed at any displacement x₀ = 6.0 cm, T = 1.5 s; the oval is v against x

displacement x 3.0 cm

v = ±ω√(x₀² − x²)—

as a fraction of v_max—

v_max = ωx₀—

4.1x = 3.0 cm. Speed =

cm s⁻¹

4.2Speed at x = x₀/2, as a fraction of v_max =

4.3Why are there two dots (+ and −) at every x?

at the same displacement the object can be moving either way the speed is different on the way out and on the way back the graph is drawn twice

4.4Another SHM has v_max = 3.0 m s⁻¹ and a_max = 24 m s⁻². ω =

rad s⁻¹

Back check

4.bIts amplitude, x₀ =

What if? · stretch

4.wStretch: at what displacement is the speed half of v_max? x =

4.5Was prediction 1.3 right? Explain with the shape of the graph.

What a good answer looks like

B, about 0.87 v_max. v = ω√(x₀² − x²): at x = x₀/2 this is ωx₀√(3/4) = 0.866 v_max (21.8 of 25.1 cm s⁻¹). The oval is flat near the centre and steep near the ends: the object keeps most of its speed until it is close to the end, then slows down quickly.

5. Energy equations: test prediction 1.4

17 minPairs → individual

ATL Thinking: separate two energies that look alike. Communication: name every energy store in a transfer.

Simulation D · Energy of a vertical spring m = 0.20 kg, k = 8.0 N m⁻¹, g = 9.8 m s⁻²; pulled down by x₀ and released

amplitude x₀ 0.10 m

extension of the spring—

elastic energy ½k(e + x)²—

½mω²x²—

Eₖ + elastic + gravitational—

The energy equationsE_T = ½mω²x₀² (the kinetic energy at the centre, where v_max = ωx₀). E_p = ½mω²x², and E_k = E_T − E_p = ½mω²(x₀² − x²). For a vertical spring these are the oscillation energies, measured from the equilibrium position.

5.1Extension of the spring at the equilibrium position, e =

5.2x₀ = 0.10 m. Oscillation energy E_T = ½mω²x₀² =

5.3Elastic energy stored in the spring at the lowest point =

5.4At what displacement are E_k and E_p (= ½mω²x²) equal? x =

5.5Play the simulation. As the mass rises from the lowest point to the equilibrium position, which energies change, and how?

elastic energy falls; kinetic and gravitational energy rise; the total stays constant only elastic energy turns into kinetic energy the total energy falls

Find the mistake

For the mass at its lowest point (x₀ = 0.10 m), a student writes:

elastic energy in the spring = ½mω²x₀² = 0.040 J

What is wrong?

½mω²x² is the oscillation energy from equilibrium. The spring stores ½k(extension)², and the extension is e + x₀ = 0.345 m: 0.476 J. The student should use x₀/2. Nothing.

5.6A mass–spring system has maximum elastic potential energy E. The mass is tripled and the amplitude doubled. New maximum, as a multiple of E =

× E

5.7Was prediction 1.4 right? Explain with E_p ∝ x².

What a good answer looks like

B, x₀/√2. E_p = ½mω²x² grows with x², so at x = x₀/2 it is only ¼ of the total (and E_k is ¾). Equal shares need E_p = ½E_T: x² = x₀²/2, x = 0.71x₀ (0.0707 m for x₀ = 0.10 m).

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…

get v and a from x by differentiating (radians);
find φ from the start, using the direction of motion;
use v = ±ω√(x₀² − x²) or the energy equations correctly;
use one number from a simulation today.

Option A · Written

One paragraph: “What the phase angle in x = x₀ sin(ωt + φ) means, and how to find it.”

Option B · Sketch

Four small x–t graphs for φ = 0, π/2, π and 3π/2, each labelled with where the object starts and which way it moves.

Option C · Spoken

Explain to your partner in 1 minute why ½mω²x² is not the elastic energy of a vertical spring. Your partner ticks the success criteria.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): In x = x₀ sin(ωt + φ), the phase angle φ is the starting angle of the point on the turning circle: it tells you where the object is and which way it moves at t = 0. It does not change the amplitude, the period or the energy. To find it, use x(0) = x₀ sin φ; two angles have the same sine, so use the direction of motion (the sign of cos φ) to choose. For x(0) = +0.10 m with x₀ = 0.20 m: moving +, φ = π/6; moving −, φ = 5π/6. Then differentiate as usual: v = ωx₀ cos(ωt + φ).

Exit ticket (no notes)

1x = 0.020 sin(15t) (m, s). The maximum speed, in m s⁻¹ =

2For the same motion, the acceleration at t = 0.050 s, in m s⁻² =

3A 0.50 kg body moves in SHM with ω = 10 rad s⁻¹ and amplitude 0.10 m. Its total energy, in J =

4An object starts at rest at −x₀. In x = x₀ sin(ωt + φ), φ is…

0 π/2 3π/2

5At x = x₀/2, the potential energy is…

half of the total a quarter of the total three quarters of the total

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 choose between the two angles when I find φ.”

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 the direction of motion at t = 0 before I write down φ.”

TOK · 2 minutes

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

TOKx = x₀ sin(ωt + φ) describes the motion perfectly, but the phase angle φ only depends on when we chose to start the clock. Which parts of a scientific description belong to nature, and which belong to our choices?

Points you could make

The amplitude and the period are properties of the oscillator; φ depends on the observer’s choice of t = 0.
Many quantities depend on choices: the zero of potential energy, the direction called positive, the frame of reference.
Good physics separates what is chosen (conventions) from what is measured, and checks that real predictions do not depend on the choices.

Levelled practice Read the notes Back to C.1