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Quick subject briefing
Before this lesson: Consolidation after pupils have met wave quantities, simple graphs and v = fλ. Pupils need a calculator; all data and figures are supplied. No demonstration or experiment is needed.
Wave speed is how fast the disturbance travels. Frequency counts cycles each second; wavelength is the distance per cycle, so multiplying them gives distance per second. Period is time per cycle and is the reciprocal of frequency. A distance snapshot gives wavelength; a time trace gives period. Amplitude is displacement from the undisturbed position, not wavelength or speed.
Words and symbols you will need
- Wavelength, λ (lambda), m
- Distance between adjacent matching points, such as crests. Four crests enclose three wavelength intervals.
- Frequency, f, Hz
- Complete cycles per second. Count cycles, then divide by elapsed seconds.
- Period, T, s
- Time for one complete cycle. f = 1 / T and T = 1 / f.
- Amplitude, m
- Maximum displacement from the undisturbed position. Peak-to-trough height is twice amplitude.
- Wave speed, v, m/s
- Speed of the travelling disturbance, not the oscillating buoy/particle.
Help pupils choose a method
- Identify the required quantity and read the axis quantity/unit before using a graph.
- Convert centimetres to metres or milliseconds to seconds before substituting.
- Choose v = fλ, λ = v/f or f = v/λ. If given period, first use f = 1/T. Include units and check the product fλ.
Use the supplied model — pupil page 2
Use pupil page 2: 25 cycles each second, each 0.080 m long, gives 2.0 m each second. For the separate rearrangement, divide 2.0 m/s by 5.0 cycles/s to get 0.40 m per cycle, then multiply back to check.
A wave has frequency 25 Hz and wavelength 0.080 m. Find its speed. Then find the wavelength of a different wave travelling at 2.0 m/s with frequency 5.0 Hz.
Worked answer: First, v = f × wavelength = 25 × 0.080 = 2.0 m/s. For the second test, wavelength = v / f = 2.0 / 5.0 = 0.40 m. Check: 5.0 × 0.40 = 2.0 m/s.
If pupils are still stuck
Use the matching Say/Ask/Expect coaching and named fresh retry in its replacement time. If still insecure, help the pupil label f, wavelength and units or set up the correct fraction, then record that exact difficulty for the usual teacher. Do not add an unscheduled practical.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Print the complete 8-page pupil booklet in order, one per pupil. Staple each booklet; double-sided printing is fine.
- Read the subject briefing on teacher page 1, then use the run sheet on page 2. Keep this guide separate from pupil copies.
- Pupils need pens, a ruler and a calculator. All figures, context and reading are supplied; no login, devices or apparatus are needed.
0-5 minPupil page / Q
1 / 1-3
Read aloud / check: Start from memory on page 1. At 5 minutes check the three responses.
Next: If insecure, use Try again A in the first 2 reading minutes.
Answers: 1, 2, 3.
Open check 1 and its coaching →
5-15 minPupil page / Q
2 / reading
Read aloud / check: Read the quantity table, equations and both model calculations. Read each graph-axis rule.
Next: Check A if used, then continue reading.
15-20 minPupil page / Q
3 / 4
Read aloud / check: Choose a letter and show the conversion and division. Stop at check 2.
Next: If insecure, Try again B replaces questions 5-6.
Answers: 4.
Open check 2 and its coaching →
20-30 minPupil page / Q
3-4 / 5-11
Read aloud / check: Complete the separate-test calculations, then use the harbour snapshot.
Next: If B was used, leave 5-6 blank. Require SI units and wavelength intervals.
Answers: 5, 6, 7, 8, 9, 10, 11.
30-33 minPupil page / Q
5 / 12
Read aloud / check: Pause for the independent space-and-time check.
Next: If insecure, Try again C replaces questions 13-14.
Answers: 12.
Open check 3 and its coaching →
33-45 minPupil page / Q
5-7 / 13-20
Read aloud / check: Continue with the harbour graphs, explanations and fixed-speed prediction.
Next: If C was used, leave 13-14 blank. Both figures are repeated on page 6.
Answers: 13, 14, 15, 16, 17, 18, 19, 20.
45-55 minPupil page / Q
8 / 21
Read aloud / check: Correct practice in a different colour; explain one correction.
Next: Share only practice answers. Keep exit answers hidden.
Answers: 21.
55-60 minPupil page / Q
8 / 22
Read aloud / check: Cover earlier pages. Complete question 22 independently.
Next: Collect every booklet and check afterwards. D is next lesson only.
Answers: 22.
Open check 4 and its coaching →
Estimated 60 minutes. Practice is 10 minutes, a 3-minute check, then 12 minutes. A uses reading time; B replaces 5-6; C replaces 13-14; D is next lesson only. No early-finisher extra worksheet parts are assigned. Timings are not classroom-trialled.
Check the starter, hinge and mid-practice at the marked pauses. Share only practice answers at 45 minutes. Keep question 22 and next-lesson answers hidden until submission.
Download teacher guide & answers ↓If several pupils are stuck
Read the matching coaching and page 2 rule aloud, then use the named retry within its replacement time. If still insecure, help pupils set up the correct calculation and record who needs follow-up.
Estimated 60 minutes. Practice is 10 minutes, a 3-minute check, then 12 minutes. A uses reading time; B replaces 5-6; C replaces 13-14; D is next lesson only. No early-finisher extra worksheet parts are assigned. Timings are not classroom-trialled.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Question 1: 0.45 m; question 2: 4.0 Hz; question 3: 5 m, a distance.
If secure: Continue to the reading and model on pupil page 2.
If: 45 cm becomes 45 m or 4500 m, or question 2 uses time divided by count.Open coaching
Likely issue: SI conversion or frequency per second is insecure.
- Say
- One metre contains 100 centimetres. Frequency counts complete oscillations in one second, so divide count by elapsed time.
- Ask
- For questions 1-2, what are 45 divided by 100 and 12 divided by 3?
- Expect
- 0.45 m and 4.0 Hz. Hertz means complete oscillations per second.
Next: Use the units/frequency table on page 2. Direct Try again A on page 1 in the first 2 reading minutes. Check all three fresh answers.
Pupil page 2, quantities and Convert first; Try again A on page 1.
If: Question 3 gives 10 m or a time unit.Open coaching
Likely issue: The pupil uses the full span or confuses space with time.
- Say
- Wavelength is the distance from one crest to the next. Three crests enclose two intervals.
- Ask
- In question 3, how far is it from the crest at 2 m to the next crest at 7 m? What unit is on those coordinates?
- Expect
- 5 m, a distance. The 10 m total spans two wavelengths.
Next: Use page 2, wavelength and the horizontal-axis rule. Use Try again A in reading time.
Pupil page 2, Check the horizontal axis; Try again A on page 1.
HINGE CHECK
Check 2: Question 4
Look for: B, with 60 cm = 0.60 m and f = 3.0 / 0.60 = 5.0 Hz. A letter alone is insufficient.
If secure: Start questions 5-11. Require conversion and an equation with substitution.
If: A uses 3 divided by 60.Open coaching
Likely issue: Centimetres were used with speed in m/s.
- Say
- The speed uses metres, so the wavelength must also use metres before division.
- Ask
- In question 4, what is 60 cm in metres? Now divide 3.0 by that value.
- Expect
- 0.60 m; 3.0 / 0.60 = 5.0 Hz, option B.
Next: Use page 2, Convert first, then Try again B on page 3 instead of questions 5-6.
Pupil page 2, Convert first; Try again B on page 3.
If: C multiplies speed and wavelength or D reverses the division.Open coaching
Likely issue: Rearranging the wave equation is insecure.
- Say
- Start with v = f × wavelength. To leave f alone, divide speed by wavelength. Check by multiplying your f by wavelength.
- Ask
- Which value multiplied by 0.60 m gives 3.0 m/s in question 4?
- Expect
- 5.0 Hz. 5.0 × 0.60 = 3.0, so f = 3.0 / 0.60.
Next: Use the rearrangement and model on page 2. Direct Try again B instead of questions 5-6.
Pupil page 2, rearrangements and worked model; Try again B on page 3.
If: B is chosen without a valid conversion and equation.Open coaching
Likely issue: The correct letter may be a guess.
- Say
- A correct letter needs a checkable method: convert, choose the rearrangement, substitute and check.
- Ask
- Explain how the two supplied quantities in question 4 give your chosen frequency.
- Expect
- 60 cm = 0.60 m; f = 3.0 / 0.60 = 5.0 Hz. Multiplying by 0.60 recovers 3.0 m/s.
Next: Ask for conversion and substitution. If insecure, use Try again B instead of questions 5-6.
Pupil page 2, Convert first and f = v / λ; Try again B on page 3.
MID PRACTICE CHECK
Check 3: Question 12
Look for: Question 12: wavelength 0.80 m, period 0.20 s, frequency 5.0 Hz and speed 4.0 m/s.
If secure: Continue questions 13-20 using the axis rules, including both harbour figures.
If: Calls 0.20 s the wavelength or 0.80 m the period.Open coaching
Likely issue: The graph shape has been used without its horizontal-axis quantity.
- Say
- A distance snapshot gives metres between crests: wavelength. A time trace gives seconds between cycles: period.
- Ask
- In question 12, which supplied interval is in metres, and which is in seconds?
- Expect
- 0.80 m is wavelength; 0.20 s is period.
Next: Use page 2, Check the horizontal axis. Complete Try again C on page 5 instead of questions 13-14, then continue 15-20.
Pupil page 2, axis rule; Try again C on page 5.
If: Multiplies 0.20 by 0.80, or treats the 0.20 s period as 0.20 Hz.Open coaching
Likely issue: Period and frequency have been equated.
- Say
- Period is seconds per cycle. Frequency is cycles per second, so take its reciprocal before multiplying by wavelength.
- Ask
- In question 12, how many cycles fit in one second if each takes 0.20 s? What is 5.0 × 0.80?
- Expect
- 5.0 cycles per second = 5.0 Hz; speed = 4.0 m/s.
Next: Use f = 1 / T on page 2 before the wave equation. Complete Try again C instead of questions 13-14.
Pupil page 2, frequency/period and wave equation; Try again C on page 5.
EXIT CHECK
Check 4: Question 22
Look for: Question 22: 5.0 Hz, 0.40 m and 2.0 m/s; the time-trace spacing is period; doubled amplitude leaves speed unchanged.
If secure: Record which objectives are demonstrated independently and any exact follow-up needed.
If: Numerical work has an incorrect frequency, missing cm conversion or wrong wave-equation operation.Open coaching
Likely issue: Count/time, conversion or equation use still needs support.
- Say
- Calculate cycles per second, convert the wavelength to metres, then multiply frequency by wavelength.
- Ask
- For question 22, what are 15 / 3 and 40 / 100? Which multiplication gives wave speed?
- Expect
- 5.0 Hz and 0.40 m; 5.0 × 0.40 = 2.0 m/s.
Next: Next lesson, revisit the quantity table and model on page 2, then direct Try again D independently on page 8.
Pupil page 2, quantities, conversion and wave equation; Try again D on page 8, next lesson.
If: Names the time spacing wavelength or says doubled amplitude doubles speed.Open coaching
Likely issue: The axis meaning or dependence of wave speed remains insecure.
- Say
- A time axis measures seconds per cycle, or period. The wave equation uses frequency and wavelength; those are unchanged in this claim.
- Ask
- In question 22, which horizontal-axis unit identifies period? Which two factors set speed here?
- Expect
- Seconds identify period. Frequency and wavelength stay fixed, so speed remains 2.0 m/s.
Next: Next lesson, use the axis and fixed-speed rules on page 2 before Try again D. Keep this outside today’s 60 minutes.
Pupil page 2, Check the horizontal axis and fixed-speed rule; Try again D on page 8, next lesson.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - 45 / 100 = 0.45 m.
- Question 2
Pupil page 1 - f = 12 / 3.0 = 4.0 Hz.
- Question 3
Pupil page 1 - Adjacent crest spacing is 5 m; it is a distance. The 10 m from first to last spans two wavelengths.
- Question 4
Pupil page 3 - B. 60 cm = 0.60 m; f = v / wavelength = 3.0 / 0.60 = 5.0 Hz.
- Question 5
Pupil page 3 - f = number / time = 30 / 75 = 0.40 Hz.
- Question 6
Pupil page 3 - v = f × wavelength = 0.40 × 15 = 6.0 m/s.
- Question 7
Pupil page 3 - Wavelength = v / f = 5.0 / 0.25 = 20 m.
- Question 8
Pupil page 4 - f = v / wavelength = 18 / 0.60 = 30 Hz.
- Question 9
Pupil page 4 - 75 cm / 100 = 0.75 m. v = 24 × 0.75 = 18 m/s.
- Question 10
Pupil page 4 - Amplitude = 0.40 m, from the undisturbed position to a crest. 0.80 m is peak-to-trough height.
- Question 11
Pupil page 4 - The four crests at 0, 24, 48 and 72 m enclose three complete wavelength intervals. Wavelength = 72 / 3 = 24 m.
- Question 12
Pupil page 5 - Wavelength = 0.80 m; T = 0.20 s. f = 1 / 0.20 = 5.0 Hz; v = 5.0 × 0.80 = 4.0 m/s.
- Question 13
Pupil page 5 - Height = 0.40 - (-0.40) = 0.80 m, or twice amplitude. Yes: 0.80 m exceeds the supplied 0.70 m threshold.
- Question 14
Pupil page 5 - Four crests enclose three complete wavelength intervals, not four. Divide the 72 m span by three to obtain 24 m.
- Question 15
Pupil page 6 - T = 4.0 s, for example the time from the crest at 0 s to the crest at 4 s. Accept 4-8 s or 8-12 s. Crest-to-trough gives half a period.
- Question 16
Pupil page 6 - f = 1 / T = 1 / 4.0 = 0.25 Hz. Also accept three complete oscillations / 12 s.
- Question 17
Pupil page 6 - Wavelength = 24 m; period = 4.0 s, so f = 0.25 Hz. v = 0.25 × 24 = 6.0 m/s. Accept wavelength / period = 24 / 4 = 6.0 m/s.
- Question 18
Pupil page 7 - 6.0 m/s is the speed of the travelling wave/disturbance horizontally. The moored buoy oscillates vertically about its undisturbed position; its changing instantaneous speed is not this horizontal wave speed.
- Question 19
Pupil page 7 - Frequency is still 3 / 12 = 0.25 Hz. Wavelength is still 24 m, so v = 0.25 × 24 = 6.0 m/s. Halving amplitude alone has not halved wave speed in these data.
- Question 20
Pupil page 7 - New wavelength = v / f = 6.0 / 0.50 = 12 m. Frequency doubles while speed stays fixed, so wavelength halves. Check 0.50 × 12 = 6.0 m/s.
- Question 21
Pupil page 8 - Accept a corrected answer with its physical or mathematical reason, not a copied final value. All-correct route: intervals are the gaps between crests, so four consecutive crests enclose three complete wavelengths.
- Question 22
Pupil page 8 - (a) f = 15 / 3.0 = 5.0 Hz. 40 cm = 0.40 m; v = 5.0 × 0.40 = 2.0 m/s. (b) Period, in seconds. (c) Speed remains 2.0 m/s because f and wavelength are unchanged; amplitude is not a factor in v = f × wavelength.
- Try again A
- (a) 0.35 m. (b) 18 / 6.0 = 3.0 Hz. (c) 4.0 m; the 8 m full span contains two intervals.
- Try again B
- 50 cm = 0.50 m; f = 4.0 / 0.50 = 8.0 Hz. Check 8.0 × 0.50 = 4.0 m/s.
- Try again C
- Wavelength = 0.60 m; T = 0.30 s; f = 1 / 0.30 = 3.33... Hz; v = (1 / 0.30) × 0.60 = 2.0 m/s. Accept 3.3 Hz and approximately 2.0 m/s; keep calculator precision when possible.
- Try again D
- f = 24 / 4.0 = 6.0 Hz; 30 cm = 0.30 m; v = 6.0 × 0.30 = 1.8 m/s. The 0.50 s interval is period. Doubling amplitude with f and wavelength fixed leaves speed unchanged. The time trace is a separate example, not the first wave.
Worksheet selection and marking
The 14 selected worksheet parts total 20 marks. Keep method and reasoning, accept valid equivalent units after consistent conversion, and use sensible follow-through for a prior incorrect reading. Wave speed is not buoy speed. No practical completion or sound-boundary requirements are assessed.
Selection: Q1(b-f), Q2(a-d), Q3(a-e); source pages 1-4.
Open the maintained worked solutions →Collect and follow up: Collect all named booklets. Use the independent exit to record frequency calculation, SI conversion, wave-speed use, axis interpretation and amplitude reasoning. Tell the usual teacher the exact remaining error and who needs Try again D.
