Wave Properties and Behaviour - Worksheets, Questions and Revision

20 original exam-style questions - 7 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 8)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

GCSE · Physics

P6a Wave Properties and Behaviour

AQA 8464 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key words and equations: Wave properties and behaviour (P6a)

Revision Library

A transverse wave is a wave in which the oscillations (vibrations) of the particles or field are at right angles (perpendicular) to the direction the wave transfers energy, for example water ripples, light and waves on a string. A longitudinal wave is a wave in which the oscillations are parallel to (along the same line as) the direction the wave transfers energy, for example sound and ultrasound waves, which travel as a series of compressions (particles pushed closer together) and rarefactions (particles spread further apart). Amplitude is the maximum displacement of a point on a wave from its undisturbed (rest) position. Wavelength (symbol lambda) is the distance from one point on a wave to the equivalent point on the next wave (e.g. peak to peak), measured in metres, m. Frequency (symbol f) is the number of complete waves passing a point per second, measured in hertz, Hz. Period (symbol T) is the time taken for one complete wave to pass a point, measured in seconds, s. Equations used in this worksheet: wave speed = frequency * wavelength (v = f * lambda); period = 1 / frequency (T = 1/f); distance = speed * time. Every equation needed is given within the question itself - you do not need to memorise it for this worksheet.

State whether each of the following waves is transverse or longitudinal.
(i) A sound wave travelling through air.
(ii) A light wave travelling through a vacuum.
(iii) A ripple travelling across the surface of a garden pond.

(Total for Question 1 is 3 marks)

Explain the difference between a transverse wave and a longitudinal wave, in terms of the direction of vibration of the particles (or field) compared with the direction the wave transfers energy.

(Total for Question 2 is 2 marks)

The diagram shows a transverse wave drawn on a displacement-distance grid. The wave trace crosses the centre line (its undisturbed position) at distances of 0 cm, 4 cm, 8 cm and 12 cm along the horizontal axis. Between 0 cm and 4 cm the trace rises to a maximum height of 3 cm above the centre line (at 2 cm), and between 4 cm and 8 cm it falls to a maximum of 3 cm below the centre line (at 6 cm), before repeating this pattern.

Figure (to be drawn): A transverse wave drawn on a displacement-distance grid, crossing the centre line at 0, 4, 8 and 12 cm, peaking at +3 cm (x = 2 cm) and troughing at -3 cm (x = 6 cm).

(a)Use the description to state the wavelength of the wave.(1)

(b)Use the description to state the amplitude of the wave.(1)

(Total for Question 3 is 2 marks)

A water wave has a frequency of 2.5 Hz and a wavelength of 0.60 m. Use the Physics Equations Sheet: wave speed = frequency * wavelength (v = f * λ), to calculate the speed of the wave.

(Total for Question 4 is 2 marks)

A wave has a frequency of 50 Hz. Use the Physics Equations Sheet: period = 1 / frequency (T = 1/f), to calculate the period of the wave.

(Total for Question 5 is 2 marks)

A sound wave has a period of 0.0025 s (2.5 ms). Use the Physics Equations Sheet: frequency = 1 / period (f = 1/T), to calculate the frequency of the wave. Give your answer to 2 significant figures.

(Total for Question 6 is 2 marks)

A radio wave travels through air at the speed of light, 3.0 x 10⁸ m/s, and has a frequency of 9.6 x 10⁷ Hz (96 MHz). Rearrange the Physics Equations Sheet equation wave speed = frequency * wavelength (v = f * λ) to calculate the wavelength of the radio wave. Give your answer to 2 significant figures.

(Total for Question 7 is 3 marks)

Required practical: Meera uses a ripple tank to investigate water waves. She sets a motor-driven dipper vibrating at a constant rate to produce continuous straight waves across the water.

(a)Describe how Meera could measure the wavelength of the water waves produced in the ripple tank.(2)

(b)Meera uses a stopwatch to count the number of waves passing a fixed point in 10 s. She counts 24 waves. Calculate the frequency of the waves.(2)

(c)The wavelength Meera measures is 3.5 cm. Use the Physics Equations Sheet: wave speed = frequency * wavelength (v = f * λ), to calculate the speed of the waves. Give your answer in m/s.(3)

(d)Give one reason why repeating the wavelength measurement several times and calculating a mean improves the quality of Meera's result.(1)

(Total for Question 8 is 8 marks)

Required practical: Kwame investigates waves on a stretched string. He attaches one end of the string to a vibration transducer connected to a signal generator, and fixes the tension in the string.

(a)State the piece of equipment, in addition to the string and vibration transducer, that lets Kwame read off the frequency of the waves he produces.(1)

(b)The signal generator is set to 40 Hz. Kwame adjusts the length of the string until exactly 2 complete waves fit onto a 1.6 m length of string. Calculate the wavelength of the waves on the string.(2)

(c)Use the Physics Equations Sheet: wave speed = frequency * wavelength (v = f * λ), to calculate the speed of the waves on the string.(2)

(d)Kwame then increases the tension in the string but keeps the frequency at 40 Hz. He observes that the wavelength on the string increases. Suggest what this shows about the effect of increasing the tension in a string on the speed of waves travelling along it.(2)

(Total for Question 9 is 7 marks)

A microphone connected to an oscilloscope displays a displacement-time graph for a sound wave. The trace shows the wave completing exactly 5 full cycles in a time of 20 ms (0.020 s).

Figure (to be drawn): An oscilloscope displacement-time trace for a sound wave, showing exactly 5 complete cycles across a 20 ms (0.020 s) time base.

(a)Calculate the period of the sound wave.(2)

(b)Use your answer to part a and the Physics Equations Sheet equation frequency = 1 / period (f = 1/T) to calculate the frequency of the sound wave.(2)

(Total for Question 10 is 4 marks)

A ray of light is reflected from a plane (flat) mirror.

(a)State the law of reflection.(1)

(b)The angle between the incident ray and the mirror surface (not the normal) is 35 degrees. Calculate the angle of reflection, measured from the normal.(2)

(Total for Question 11 is 3 marks)

A ray of light travels from air into a glass block, crossing the boundary at an angle to the normal. The speed of light is lower in glass than in air. State and explain the direction in which the ray bends as it enters the glass.

(Total for Question 12 is 2 marks)

When a wave crosses a boundary between two different materials, its speed and wavelength change, but its frequency stays the same. Explain why the frequency does not change.

(Total for Question 13 is 2 marks)

Explain why sound waves cannot travel through a vacuum (for example, through outer space).

(Total for Question 14 is 1 mark)

State the approximate range of frequencies that a healthy young human ear can normally detect.

(Total for Question 15 is 1 mark)

Ultrasound is sound with a frequency above the upper limit of human hearing. State the approximate frequency above which a wave is classed as ultrasound.

(Total for Question 16 is 1 mark)

A fishing boat sailing off the coast of Cornwall uses ultrasound to find the depth of the water below it. The ultrasound pulse travels from the boat down to the seabed and reflects back up to the boat. Use the Physics Equations Sheet: distance = speed * time, to calculate the depth of the water, given that the pulse takes 0.084 s to travel down to the seabed and back again, and the speed of sound in seawater is 1500 m/s.

(Total for Question 17 is 3 marks)

Give one medical use and one industrial (non-medical) use of ultrasound waves.

(Total for Question 18 is 2 marks)

A student wants to find out how the depth of water in a ripple tank affects the speed of the water waves produced. Describe a method the student could use to investigate this, making sure the method gives valid, repeatable and reliable results. In your answer you should refer to: how the depth of water is varied and measured; how the frequency and wavelength of the waves are measured; how the wave speed is calculated; and how repeatable results are obtained.

(Total for Question 19 is 6 marks)

A loudspeaker emits a sound wave that travels through air at 330 m/s and has a wavelength of 1.65 m. A dog whistle emits an ultrasound wave that travels through the same air at the same speed, but its frequency is 100 times higher than the frequency of the sound wave from the loudspeaker. Use the Physics Equations Sheet: wave speed = frequency * wavelength (v = f * λ), to calculate the wavelength of the ultrasound wave from the dog whistle.

(Total for Question 20 is 4 marks)

Mark scheme · P6a Wave Properties and Behaviour

Question 1

B1 (i) longitudinal
B1 (ii) transverse
B1 (iii) transverse
Answer: (i) longitudinal (ii) transverse (iii) transverse

Question 2

B1 transverse wave: the oscillations/vibrations are at right angles (perpendicular) to the direction of energy transfer
B1 longitudinal wave: the oscillations/vibrations are parallel to (along the same line as) the direction of energy transfer
Answer: Transverse: vibrations perpendicular to energy transfer. Longitudinal: vibrations parallel to energy transfer.

Question 3

(a) B1 8 cm (0.08 m), oe
(a) Answer: 8 cm (0.08 m)
(b) B1 3 cm (0.03 m), oe - not 6 cm (the peak-to-peak distance)
(b) Answer: 3 cm (0.03 m)

Question 4

M1 substitutes correctly: v = 2.5 * 0.60
A1 v = 1.5 m/s, cao
Answer: 1.5 m/s

Question 5

M1 substitutes correctly: T = 1/50
A1 T = 0.02 s, cao
Answer: 0.02 s

Question 6

M1 substitutes correctly: f = 1/0.0025
A1 f = 400 Hz (2 s.f.), cao
Answer: 400 Hz

Question 7

M1 rearranges correctly: λ = v/f
M1 substitutes correctly: λ = (3.0 x 10⁸)/(9.6 x 10⁷)
A1 λ = 3.1 m (2 s.f.), cao
Answer: 3.1 m

Question 8

(a) B1 view the wave pattern on a screen below the tank (or use a lamp/stroboscope to freeze the pattern) and use a ruler/metre rule to measure across several complete waves
(a) B1 divide the measured distance by the number of waves measured across, to find a mean (average) wavelength
(a) Answer: Measure the distance across several waves with a ruler and divide by the number of waves to get a mean wavelength.
(b) M1 substitutes correctly: f = 24/10
(b) A1 f = 2.4 Hz, cao
(b) Answer: 2.4 Hz
(c) M1 converts 3.5 cm to 0.035 m
(c) M1 substitutes correctly: v = 2.4 * 0.035
(c) A1 v = 0.084 m/s, cao (ft from their answer to part b)
(c) Answer: 0.084 m/s
(d) B1 reduces the effect of random error in an individual reading, giving a more accurate/reliable mean value, oe
(d) Answer: It reduces the effect of random error, giving a more accurate mean value.

Question 9

(a) B1 the signal generator (with a frequency display/dial)
(a) Answer: The signal generator.
(b) M1 substitutes correctly: λ = 1.6/2
(b) A1 λ = 0.8 m, cao
(b) Answer: 0.8 m
(c) M1 substitutes correctly: v = 40 * 0.8
(c) A1 v = 32 m/s, cao (ft from their answer to part b)
(c) Answer: 32 m/s
(d) M1 applies v = f * λ: since f is constant and λ has increased, v must have increased
(d) A1 concludes that increasing the tension in the string increases the speed of waves on it, oe
(d) Answer: Increasing the tension increases the wave speed on the string.

Question 10

(a) M1 substitutes correctly: T = 0.020/5
(a) A1 T = 0.004 s (4 ms), cao
(a) Answer: 0.004 s (4 ms)
(b) M1 substitutes correctly: f = 1/0.004
(b) A1 f = 250 Hz, cao (ft from their answer to part a)
(b) Answer: 250 Hz

Question 11

(a) B1 the angle of incidence equals the angle of reflection (both measured from the normal)
(a) Answer: Angle of incidence = angle of reflection.
(b) M1 finds the angle of incidence from the normal: 90 - 35 = 55 degrees
(b) A1 angle of reflection = 55 degrees, cao (equal to the angle of incidence, by the law of reflection)
(b) Answer: 55 degrees

Question 12

B1 the ray bends towards the normal
B1 because the wave slows down as it enters the optically denser medium (glass), which changes its direction, oe
Answer: The ray bends towards the normal, because it slows down entering the denser medium (glass).

Question 13

B1 the frequency of a wave is set by the source producing it (the rate at which the source vibrates)
B1 the number of wave cycles arriving at the boundary each second must equal the number leaving it, as waves are not created or destroyed there, so the frequency cannot change, oe
Answer: Frequency is fixed by the source; the number of waves arriving at and leaving the boundary each second must match, so the frequency is unchanged.

Question 14

B1 sound is a longitudinal wave that travels by making particles of a medium vibrate; a vacuum contains no particles, so there is nothing for the wave to travel through, oe
Answer: Sound needs a medium (particles) to travel through, and a vacuum has no particles.

Question 15

B1 20 Hz to 20 000 Hz (20 kHz)
Answer: 20 Hz to 20,000 Hz (20 kHz)

Question 16

B1 above about 20 000 Hz (20 kHz)
Answer: Above about 20,000 Hz (20 kHz)

Question 17

M1 substitutes correctly to find the total distance travelled: distance = 1500 * 0.084 = 126 m
dM1 divides the total distance by 2, because the pulse travels down to the seabed and back again (2 x depth = 126 m)
A1 depth = 63 m, cao
Answer: 63 m

Question 18

B1 medical use, e.g. pre-natal (antenatal) scanning to image a fetus in the womb, or breaking up kidney stones (any suitable medical use)
B1 industrial use, e.g. non-destructive testing to detect cracks/flaws inside metal objects, or measuring the depth of the sea/a lake, or cleaning delicate objects (any suitable industrial use)
Answer: Medical: antenatal scanning of a fetus. Industrial: detecting flaws inside metal objects (non-destructive testing).

Question 19

Level 1 (1-2): Simple, isolated statements are made about the method, for example mentioning changing the water depth and/or measuring wave speed, but with little detail, poor organisation, and key steps such as controlling other variables or taking repeats are missing.
Level 2 (3-4): Relevant steps of the method are given in a broadly logical order, showing how the depth is varied and how wave speed is found from measured frequency and wavelength, but some detail is missing, for example on repeats, control of other variables, or how measurements are made accurately.
Level 3 (5-6): A logical, detailed and coherent method is described. It clearly explains how the depth is varied and measured, how frequency and wavelength are measured accurately, how wave speed is calculated using v = f * lambda, and how repeatable and reliable results are obtained, for example repeats, mean values, controlling other variables. The steps are clearly linked and in a sensible sequence.
Indicative content:
- Vary the depth of water in the ripple tank, e.g. by placing rigid sheets/blocks of known thickness under part of the tank, or adding measured volumes of water, and record each depth with a ruler.
- Keep the frequency (and amplitude) of the dipper the same for every depth tested, so depth is the only variable being changed (control variable).
- Keep the liquid used (water) and its temperature the same for every test (control variables).
- Measure the frequency of the waves, e.g. by counting the number of waves passing a fixed point in a measured time using a stopwatch, or by reading the value set on the signal generator/motor.
- Measure the wavelength by viewing the wave pattern, e.g. on a screen, or using a lamp or stroboscope, and using a ruler to measure the distance across several complete waves, then dividing by the number of waves to find a mean wavelength.
- Calculate the wave speed at each depth using the equation v = f * lambda.
- Repeat the frequency and wavelength measurements at each depth at least three times and calculate mean values, to improve the reliability of the results.
- Test a range of at least five different depths, spread evenly, to reveal a clear pattern.
- Record results in a table and/or plot a graph of wave speed against depth to identify the relationship between them.

Question 20

M1 finds the frequency of the loudspeaker's sound wave: f = v/λ = 330/1.65 = 200 Hz
M1 finds the frequency of the ultrasound wave: 200 * 100 = 20 000 Hz
M1 rearranges and substitutes to find the new wavelength: λ = v/f = 330/20 000
A1 λ = 0.0165 m (1.65 cm or 1.65 x 10^-2 m), cao
Answer: 0.0165 m (1.65 cm)

Mark my answers