DSE Physics | Simple Harmonic Motion: A Tutor Guide | Period First, Then Speed and Energy

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students can recite that acceleration equals minus omega squared times displacement, then freeze when asked for the period or the maximum speed, because omega gets confused with the period and the amplitude with a speed. This lesson does three things only: get omega from k and m, get the period and frequency from omega, then get speed and energy from the amplitude. One set of numbers runs throughout: 50 N/m, 0.50 kg and 4.0 cm.

Objectives — Students can: (1) compute omega as the square root of k over m, giving 10 rad/s from 50 N/m and 0.50 kg; (2) compute a period of 0.63 seconds and a frequency of 1.59 Hz; (3) compute a maximum speed of 0.40 m/s, a maximum acceleration of 4.0 m/s squared and a total energy of 0.040 J from a 4.0 cm amplitude; (4) state that speed peaks at equilibrium and acceleration peaks at the ends; (5) compute 2.5 minutes per question and 0.8 marks per minute for 12 questions worth 24 marks in 30 minutes.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Angular frequency	Omega equals the square root of k over m. What is the unit of k?	Newtons per metre, which is kilograms per second squared, so omega is in radians per second
Period and frequency	If the period is 0.63 seconds, what is the frequency?	One divided by the period, about 1.59 Hz
Circular motion	What is the speed in uniform circular motion?	v equals r omega, and its projection on a diameter is simple harmonic motion
Energy	What is the elastic potential energy formula?	Half k x squared, which equals half k A squared at an end point

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Start with the projection	Project circular motion onto a diameter and mark equilibrium and the two ends	Say where speed and acceleration peak
8-16 min	Derive omega and the period	Compare a equals minus omega squared x with F equals minus k x to get omega	Derive it from Newton second law, not memory
16-24 min	Substitute	50 over 0.50 gives 10, then 2 pi over 10 gives 0.63 seconds	Units and decimals correct
24-32 min	Speed, acceleration, energy	A omega gives 0.40 m/s, A omega squared 4.0, half k A squared 0.040 J	The three numbers agree
32-40 min	Timed practice	Three questions at 2.5 minutes each, the last one on graphs	Graph matches numbers
40-45 min	Exit check	Four oral answers with notes closed	All four values correct

03 Teaching script (read it out)

Scene 1: Turn the standard equation into the spring version

The textbook gives a equals minus omega squared x, which students memorise without knowing where omega comes from. Substitute F equals minus k x into F equals m a and a equals minus k over m times x appears, so comparing with the standard form gives omega squared equals k over m. This step is the root of the whole chapter, so the student must write it once rather than watch it.

Ask the student to derive omega from F equals m a and to say which way k and m each push.

Scene 2: Omega is not the period; 2 pi over omega is

The most common error is to compute 10 and write a period of 10 seconds. Omega is in radians per second, so one second covers 10 radians, and one turn is 6.28 radians, giving 0.63 seconds and 1.59 Hz. Insist that every answer carries its unit, because a period written as 10 s exposes the confusion immediately.

Ask the student why a larger omega means a shorter period, then check against 0.63 seconds.

Scene 3: Speed at equilibrium, acceleration at the ends

At equilibrium the displacement is zero, the spring force vanishes and speed peaks at A omega; at an end the speed is zero and acceleration peaks at A omega squared. Students must mark speed and acceleration arrows at all three positions on one diagram, otherwise the numbers are right and the direction is not.

Ask the student which point has the greatest speed and which the greatest acceleration, and why.

Scene 4: Energy only moves between two stores

The total energy, half k A squared, is constant: all potential at the ends, all kinetic at equilibrium, and a mix in between. With 50 N/m and 0.040 m the total is 0.040 J, so an answer of 0.40 J means a missing square or a misplaced decimal. Three checks: order of magnitude, unit in joules, kinetic energy at equilibrium equal to the total.

Ask the student for the kinetic energy at equilibrium and why it equals the total.

04 Question types and worked steps

Type	Example	Steps	Answer
Period and frequency	k and m given	Get the square root of k over m, then 2 pi over omega, then one over the period	Omega is not the period
Maximum speed and acceleration	Amplitude and omega given	A omega, then A omega squared, state where each occurs	The square is often dropped
Energy	k and amplitude given	Half k A squared for the total, then split by position	Answer in joules
Graphs and phase	A displacement time graph given	Read amplitude and period, read speed from slope, read acceleration from curvature	Zero slope means maximum acceleration

Sanity check (teach this): Order check: k over m is 50 over 0.50, which is 100, so the square root is 10; an answer of 100 means the root was skipped. Time check: 2 pi over 10 is about 0.63 seconds, so 6.3 means a misplaced decimal. Speed check: A omega is 0.040 times 10, which is 0.40 m/s, slower than walking and therefore plausible; 4 m/s means centimetres were never converted. Energy check: half of 50 times 0.0016 is 0.040 J, and 0.40 J means the amplitude was used in centimetres.

05 Independent practice (3 questions)

Question	Answer	What to watch
k is 50 N/m and m is 0.50 kg, what is omega	10 rad/s	Square root taken
Same set, what is the period	About 0.63 seconds	Omega not treated as the period
Amplitude 4.0 cm, what is the maximum speed	0.40 m/s	Centimetres converted

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating omega as the period	“Omega is 10, so the period is 10 seconds.”	“10 is radians per second. One turn is 6.28 radians, so the period is 2 pi over 10, which is 0.63 seconds.”
Leaving the amplitude in centimetres	“A is 4, so the speed is 4 times 10, which is 40 m/s.”	“4.0 cm is 0.040 m, so the speed is 0.40 m/s. Without the conversion the answer is a hundred times too big.”
Believing speed peaks at the ends	“At the furthest point the speed must be largest.”	“At the furthest point the spring is pulling hardest but the speed is zero, and acceleration is largest. Speed peaks as it passes equilibrium.”
Dropping the square in energy	“Half of 50 times 0.040 is 1 joule.”	“Half k A squared needs the square: half of 50 times 0.0016 is 0.040 J. Without it the answer is twenty five times too big.”

07 Homework (paper, section and question numbers)

Physics chapter index — Revise circular motion, then do practice questions 1 and 2
S4 Series and parallel circuits — Question 3 uses the same habit: find the relation, then substitute
S5 Gas laws — Questions 4 and 5 compare reading a straight line with reading a curve

08 Exit check (last 5 minutes)

Question	Answer	Decision
k is 50 and m is 0.50, what is omega	10 rad/s	Error: no square root
Omega is 10, what is the period	About 0.63 seconds	Error: answered 10 seconds
Amplitude 4.0 cm, maximum speed	0.40 m/s	Error: units not converted
Where is acceleration greatest	At the two ends	Error: answered equilibrium

09 Tutor FAQ

How many formulas must be memorised?

Two: omega equals the square root of k over m, and speed equals A omega. Everything else follows from the circular projection, which is safer under exam pressure.

Why not use the period formula directly?

You can, but students cannot recall whether k or m sits on top. Getting omega first costs one step and removes one memorised error.

Does the period change if the amplitude changes?

Not for an ideal spring: the period depends only on k and m. Show it once with data, because students rarely believe it otherwise.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.