Waves

1 Plan and risk assessment

Wave speed = frequency × wavelength. The frequency is how many waves pass a point each second; the wavelength is the distance from one crest to the next. The speed is set by what the wave travels through (the depth of the water, the tension in the string), so a higher frequency means a shorter wavelength.

v = f λ

Prediction

2 Apparatus

• ✓Ripple tank with a glass floor Holds the water the waves travel through.
• ✓Wooden bar on a motor, with its power unit Makes straight waves at a steady frequency.
• ✓Lamp on a stand above the tank Shines through the waves to make a pattern of bright lines.
• ✓Large sheet of white card The screen under the tank that the pattern falls on.
• ✓Two metre rules (resolution 1 mm) Measure distances on the card and along the string.
• ✓10 cm plastic ruler A scale in the tank to find how much the pattern is magnified.
• ✓Phone with a camera Photographs the pattern, or films it in slow motion.
• ✓Tally counter and stopwatch Counts the waves passing a mark in 10 s.
• ✓Signal generator and vibration generator (resolution 0.1 Hz) Vibrates the string at a frequency you set and read.
• ✓String, pulley and bench clamp The string the waves travel along.
• ✓100 g hanger and one 100 g slotted mass Keeps the tension in the string the same.
• ✓Safety goggles Protect your eyes near the vibrating string.

3 Method: your choices

• Which version are you doing?: AQA required practical 8: a ripple tank and a stretched stringWater waves in a ripple tank, then standing waves on a string with a signal generator.
• Level the ripple tank: DoneWater waves travel faster in deeper water, so a sloping tank changes the wavelength from one end to the other.
• Choose the water depth: About 5 mm deep (the AQA method)Shallow water: waves about 22 cm/s. Results match other classes who follow the AQA method.
• Put a scale in the tank: DoneLight spreads out from the lamp, so the pattern on the card is bigger than the real waves.
• Choose how to hold the pattern still: Photograph the card with the metre rule in shotA sharp, still picture: you can judge the middle of each bright line to about 1 mm.
• Choose how many waves to measure across: Across as many waves as possible, then divideThe error at each end is shared between all the waves, so each wavelength is several times more precise.
• Choose how to count the waves: Film 10 s in slow motion and count from the videoSlowed down, every wave can be counted: the count is right to about ±1 wave.
• Put on safety gogglesSafety: DoneA string under tension can snap and whip, or the hanger can jump off.
• Choose how to measure the loops: Across all the loops, from the vibration generator to the pulleyThe judging error at the two ends is under 1 % of nearly a metre of string.
• Fine-tune for the steadiest loops: DoneThe loops start to show a little before the exact frequency at which the string resonates.

Anomaly: No anomalous reading.

6 Analysis

1. Wavelength on the card

   wavelength = distance across the waves ÷ number of waves

   Your mean distance across 9 waves was 52.47 cm. Calculate the wavelength of the pattern on the card.

   5.83 cm Right first time

2. Wavelength of the water waves

   real length = length on the card ÷ magnification

   The 10 cm ruler’s shadow was 19.1 cm long, so the pattern is magnified 19.1 ÷ 10 = 1.91 times. Calculate the real wavelength of the water waves, in metres.

   0.0305 m Right first time

3. Frequency of the water waves

   frequency = number of waves ÷ time

   Your mean count was 74.7 waves in 10 s. Calculate the frequency.

   7.47 Hz Right first time

4. Speed of the water waves

   wave speed = frequency × wavelength (v = f λ)

   Calculate the speed of the water waves, in m/s.

   0.23 m/s Right first time

5. Wavelength on the string (3 loops)

   one loop = half a wavelength

   With 3 loops, the length across the loops was 87.4 cm. Calculate the wavelength, in metres.

   0.583 m Right first time

6. Speed of the waves on the string

   wave speed = frequency × wavelength (v = f λ)

   Calculate the speed of the waves on the string, using your frequency for 3 loops.

   34.0 m/s Right first time

7. Shape of the graph

   What does your graph show?

   As the frequency goes up, the wavelength goes down: doubling the frequency halves the wavelength. Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[1 mark]

   Write down the equation that links frequency (f), wavelength (λ) and wave speed (v).

   Your answer

   wave speed = frequency × wavelength

   1/1 self-marked

   • wave speed = frequency × wavelength, v = f λ (or any correct rearrangement, in words or symbols)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: wave speed = frequency × wavelength v = f λ

2. 01.2[1 mark]

   Another student, at the same ripple tank, counted the waves passing a mark on the card in 10 s. In one trial they lost count.

   Table 1 shows their results.

   Which trial is anomalous?

   Select one row.

   Table 1

   Trial	Number of waves counted in 10 s
   1	75
   2	53
   3	76

   Your answer

   trial 2

   1/1

   • Trial 2: much lower than the other two, which agree

   Full marks. That is exactly what the mark scheme is looking for. Trial 2 (53 waves) is far below the other two, which agree with each other.

   Model answer: Trial 2. It is much lower than the other two, which agree with each other.

3. 01.3[2 marks]

   Use Table 1 to calculate the frequency of the water waves.

   Do not use the anomalous result.

   Your answer

   7.55

   Working: Mean = (75 + 76) ÷ 2 = 75.5 waves f = 75.5 ÷ 10 f = 7.55 Hz

   2/2

   • Mean of the other two counts ÷ 10 s: 75.5 ÷ 10
   • f = 7.55 Hz

   Full marks. Clear and precise.

   Model answer: Mean = (75 + 76) ÷ 2 = 75.5 waves f = 75.5 ÷ 10 f = 7.55 Hz

4. 01.4[3 marks]

   Table 2 shows your measurements of the distance across 9 waves on the card.

   Use the frequency worked out from the counts in Table 1.

   Table 2

   Trial	Distance across 9 waves in cm
   1	52.4
   2	52.7
   3	52.3

   Calculate the wave speed that the wavelength on the card gives. Use the distances in Table 2 as measured: do not correct for the magnification.

   Give your answer in metres per second.

   Your answer

   0.44

   Working: Mean distance = 52.467 cm λ = 52.467 ÷ 9 = 5.83 cm = 0.0583 m v = 7.55 × 0.0583 v = 0.44 m/s

   3/3

   • λ = 52.467 ÷ 9 = 5.83 cm (= 0.0583 m)
   • v = 7.55 × 0.0583 (give this mark even if the wavelength was left in cm)
   • v = 0.44 m/s (allow their frequency × 0.0583)

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: Mean distance = 52.467 cm λ = 52.467 ÷ 9 = 5.83 cm = 0.0583 m v = 7.55 × 0.0583 v = 0.44 m/s

5. 01.5[2 marks]

   The method measures the distance across as many waves as possible (here, 9 waves) and divides by the number of waves, rather than measuring one wave.

   Explain why this gives a more accurate value for the wavelength.

   Your answer

   The error in judging each end is the same size but it is shared between all the waves, so the percentage uncertainty in the wavelength is smaller.

   2/2 self-marked

   • The error in judging each end (reading the metre rule) is the same size, but is shared between all the waves (one wave is very short to measure)
   • So the percentage uncertainty (error) in the wavelength is smaller (ignore "more precise")

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The error in judging where each bright line is stays the same size, but it is shared between all the waves, so the percentage uncertainty in the wavelength is much smaller than for one short wave.

6. 01.6[2 marks]

   The lamp is above the water and the card is below the tank.

   A teacher says the speed calculated from measurements on the card is larger than the true speed of the water waves.

   Explain why.

   Your answer

   Light spreads out from the lamp, so the pattern on the card is magnified. So the measured wavelength is too large, which makes the speed too large.

   2/2 self-marked

   • Light spreads out from the lamp, so the pattern on the card is magnified (bigger than the real waves)
   • So the measured wavelength is too large, and so v = f λ is too large

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: Light spreads out from the lamp, so the pattern on the card is magnified: it is bigger than the real waves. So the wavelength measured on the card is too large, and so the speed v = f λ is too large.

7. 01.7[1 mark]

   A stroboscope is a light that flashes quickly. It can make the waves look still.

   A teacher decides not to use a stroboscope with the class.

   Give one reason why.

   Your answer

   The flashing light could trigger an epileptic seizure.

   1/1 self-marked

   • The flashing light can trigger a seizure (fit) in a person with (photosensitive) epilepsy

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The flashing light can trigger a seizure in someone with photosensitive epilepsy.

8. 02.1[5 marks]

   Another student made slow water waves in a large tank.

   They used a stopwatch to time 10 waves passing a point. They did this three times.

   Their times were 7.3 s, 7.6 s and 7.0 s.

   Calculate the frequency of the waves.

   Give your answer to 2 significant figures.

   Your answer

   1.4

   Working: Mean = (7.3 + 7.6 + 7.0) ÷ 3 = 7.3 s T = 7.3 ÷ 10 = 0.73 s f = 1 ÷ 0.73 f = 1.4 Hz

   5/5

   • Mean = (7.3 + 7.6 + 7.0) ÷ 3 = 7.3 s
   • 7.3 ÷ 10 = 0.73 s (the period)
   • f = 1 ÷ 0.73
   • f = 1.37 Hz
   • Correctly rounded to 2 significant figures: 1.4 Hz

   Full marks. Clear and precise.

   Model answer: Mean = (7.3 + 7.6 + 7.0) ÷ 3 = 7.3 s T = 7.3 ÷ 10 = 0.73 s f = 1 ÷ 0.73 f = 1.4 Hz

9. 02.2[1 mark]

   The student repeated the timing three times and calculated a mean.

   Give one reason why.

   Your answer

   To reduce the effect of random errors.

   1/1 self-marked

   • To reduce the effect of random errors (allow: gives a more accurate mean; an answer only about anomalies is not enough)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: To reduce the effect of random errors.

10. 03.1[1 mark]

    A student sets up a standing wave on a string about 1 m long.

    Which instrument should the student use to measure across the loops?

    Tick (✓) one box.

    • 30 cm ruler
    • ✓ Metre rule (your choice)
    • Vernier callipers
    • Micrometer

    Your answer

    Metre rule

    1/1

    • Metre rule

    Full marks. Well done.

    Model answer: Metre rule: it is long enough to measure across all the loops in one go, to the nearest millimetre.

11. 03.2[3 marks]

    You turned the signal generator up until 3 loops stood still on the string.

    The frequency was 58.4 Hz. You measured the length across all 3 loops as 87.4 cm.

    Calculate the speed of the waves on the string.

    Give your answer in metres per second.

    Your answer

    34

    Working: λ = 2 × 87.4 ÷ 3 = 58.27 cm = 0.5827 m v = 58.4 × 0.5827 v = 34 m/s

    3/3

    • λ = 2 × 87.4 ÷ 3 = 58.27 cm = 0.5827 m (each loop is half a wavelength)
    • v = 58.4 × 0.5827 (give this mark even if the wavelength was converted wrongly or left in cm)
    • v = 34 m/s

    Full marks. That is exactly what the mark scheme is looking for.

    Model answer: λ = 2 × 87.4 ÷ 3 = 58.27 cm = 0.5827 m v = 58.4 × 0.5827 v = 34 m/s

12. 03.3[2 marks]

    A student uses the same apparatus to investigate how the tension in the string affects the speed of the waves.

    Give the independent variable and one control variable.

    Your answer

    The independent variable is the tension in the string. A control variable is the length of the string between the vibration generator and the pulley.

    2/2 self-marked

    • Independent: the tension in the string (the weight or number of masses on the hanger)
    • Control: the same string (type, thickness) or the same length of string between the vibration generator and the pulley

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: Independent variable: the tension in the string (the masses on the hanger). Control variable: the length of the string between the vibration generator and the pulley (or the same string).

13. 04[6 marks]

    A student has a ripple tank, a lamp, a motor with a wooden bar, a large sheet of white card, a metre rule and a stopwatch.

    Describe how the student could use this apparatus to determine the speed of water waves.

    Your answer

    1. Pour water about 5 mm deep into the ripple tank and put white card on the floor under it. 2. Switch on the lamp above the tank and the motor, at a low frequency, so a pattern of bright lines shows on the card. 3. Lay the metre rule across the tank at the water surface, so its shadow is on the card with the pattern. Photograph the pattern with the shadow in shot (or use a slow-motion video). 4. Measure the distance across as many waves as possible on the photo, using the rule’s shadow as the scale (so the magnification cancels), then divide by the number of waves to get the wavelength of the real waves. 5. Count the number of waves passing a point in 10 seconds, then divide by 10 to get the frequency. 6. Repeat the measurements and calculate means. 7. Calculate the wave speed using wave speed = frequency × wavelength.

    6/6 self-marked

    • Water about 5 mm deep in the tank; card under the tank; lamp above; motor (bar) switched on at a low frequency
    • Freeze the pattern: photograph it (with the rule, or its shadow from the water surface, in shot as a scale) or film it in slow motion
    • Measure across as many waves as possible, against a scale at the water surface (the card pattern is magnified)
    • Divide the distance by the number of waves to get the wavelength
    • Count the number of waves passing a point in a measured time (e.g. 10 s)
    • Frequency = number of waves ÷ time
    • Wave speed = frequency × wavelength
    • Repeat the measurements and calculate means
    • Alternative: time one wave crossing a measured distance (the tank); speed = distance ÷ time

    Self-marked: Level 3, 6 of 6 marks. You ticked 9 of 9 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Pour water about 5 mm deep into the ripple tank and put white card on the floor under it. 2. Switch on the lamp above the tank and the motor, at a low frequency, so a pattern of bright lines shows on the card. 3. Lay the metre rule across the tank at the water surface, so its shadow is on the card with the pattern. Photograph the pattern with the shadow in shot (or use a slow-motion video). 4. Measure the distance across as many waves as possible on the photo, using the rule’s shadow as the scale (so the magnification cancels), then divide by the number of waves to get the wavelength of the real waves. 5. Count the number of waves passing a point in 10 seconds, then divide by 10 to get the frequency. 6. Repeat the measurements and calculate means. 7. Calculate the wave speed using wave speed = frequency × wavelength.

Unseen data: the same skills, new apparatus

1. 01[3 marks]

   A machine makes waves in a swimming pool. A student times 10 waves passing a floating buoy, three times.

   Time for 10 waves to pass the buoy

   Trial	Time in s
   1	25.1
   2	25.4
   3	26.4

   Calculate the frequency of the waves. Give your answer to 2 significant figures.

   Your answer

   0.39

   Working: (25.1 + 25.4 + 26.4) / 3 = 25.6 25.6 / 10 = 2.56 1 / 2.56

   3/3

   • Mean time = 25.6 s.
   • Time for one wave (period) = 2.56 s.
   • Frequency = 1 ÷ period, in Hz.

   Full marks. Clear and precise.

   Model answer: Mean time for 10 waves = (25.1 + 25.4 + 26.4) ÷ 3 = 25.6 s. Period = 25.6 ÷ 10 = 2.56 s. Frequency = 1 ÷ 2.56 = 0.39 Hz.

2. 02[2 marks]

   A student shakes one end of a slinky spring from side to side at 2.9 Hz. The waves on the slinky have a wavelength of 56 cm.

   Calculate the speed of the waves on the slinky in m/s.

   Your answer

   1.62

   Working: 56 cm = 0.56 m 2.9 * 0.56

   2/2

   • 56 cm = 0.56 m.
   • Wave speed = frequency × wavelength, in m/s.

   Full marks. Well done.

   Model answer: Wavelength = 56 cm = 0.56 m. Wave speed = 2.9 × 0.56 = 1.62 m/s.

3. 03[2 marks]

   A student fixes a camera above a long wave tank and photographs straight water waves. They measure the wavelength on the printed photo with a ruler.

   Which TWO changes would give a more accurate value for the wavelength of the real waves? Tick (✓) two boxes.

   • ✓ Measure across as many waves as possible on the photo and divide by the number of waves. (your choice)
   • ✓ Float a ruler of known length on the water in the photo, and use it to change lengths on the photo into real lengths. (your choice)
   • Take three photos and use the mean wavelength: this removes the error from the photo being a different size from the waves.
   • Measure from one crest to the next only: measuring across several waves adds up their errors.
   • Run the wave machine faster so that more waves fit in the photo.

   Your answer

   Measure across as many waves as possible on the photo and divide by the number of waves.; Float a ruler of known length on the water in the photo, and use it to change lengths on the photo into real lengths.

   2/2

   • Measure across as many waves as possible on the photo and divide by the number of waves.
   • Float a ruler of known length on the water in the photo, and use it to change lengths on the photo into real lengths.

   Full marks. Clear and precise.

   Model answer: Measuring across many waves makes the error in judging each end a much smaller fraction of the length. The photo is a different size from the real waves (a systematic error): a scale of known length in the photo corrects it; a mean of repeats cannot. Running the machine faster changes the wavelength being measured.