Cambridge IGCSE0625

General properties of waves

Physics 0625 Chapter Notes

What this chapter covers

General properties of waves
ShareWhatsAppPost
General properties of waves notes

Unable to load PDF

The notes viewer could not load. Please refresh the page.

Read online free. Download a watermarked copy with a free account.

Read the notes

The full General properties of waves notes as text: skim, search, and jump between subtopics.

~18 min read

1. Introduction to Waves

A wave is a disturbance that transfers energy from one place to another without transferring matter. Think of a Mexican wave in a stadium: the 'wave' moves around the stadium, but each person just stands up and sits down. The people don't move, but the energy does. There are two main types of waves. In transverse waves, the particles oscillate (vibrate) perpendicular (at 90 degrees) to the direction the wave's energy is travelling. Examples include light, all electromagnetic waves, and ripples on water. In longitudinal waves, the particles oscillate parallel to the direction of energy transfer. Sound waves are the key example of longitudinal waves.

Key term

Transverse Wave: A wave in which the oscillations of the particles are perpendicular to the direction of energy transfer.

Examiner insight

Examiners reward clear diagrams that correctly show the direction of oscillation relative to the direction of energy transfer for both wave types.

Common pitfall

Stating that waves transfer matter. A key definition is that waves transfer energy *without* a net transfer of matter.

Worked example 13 marks

A small cork is floating on a pond. Ripples from a dropped stone pass the cork.(a) Describe the motion of the cork.(b) Explain how this demonstrates that waves transfer energy but not matter.

  1. 1

    (a) The cork bobs up and down on the spot.

  2. 2

    (b) The cork moves, which shows that energy has been transferred to it. However, the cork does not travel across the pond with the wave, which shows that matter (the water) has not been transferred across the pond.

Recap

  • Waves transfer energy without transferring matter.
  • In transverse waves, oscillations are perpendicular to the direction of energy transfer.
  • In longitudinal waves, oscillations are parallel to the direction of energy transfer.
  • Light waves are transverse.
  • Sound waves are longitudinal.

Quick check

  1. Give one example of a transverse wave and one example of a longitudinal wave.2 marks

2. Describing Wave Properties

To understand waves, we use specific terms. The Amplitude is the maximum displacement of a particle from its undisturbed rest position. For a water wave, it's the height of a crest above the normal water level. The Wavelength (symbol λ, 'lambda') is the distance between one point on a wave and the same point on the next wave, for example, from crest to crest. The Frequency (symbolf) is the number of complete waves that pass a point per second, measured in Hertz (Hz). The Period (symbol T) is the time it takes for one complete wave to pass a point, measured in seconds (s). Frequency and period are related by the simple equation T = 1/f.

T = 1/f

Key term

Frequency (f): The number of complete waves passing a point per second, measured in Hertz (Hz).

Examiner insight

Candidates often lose marks by incorrectly labelling the amplitude on a wave diagram; it is the distance from the rest position to a crest, not from crest to trough.

Common pitfall

Confusing amplitude with the total crest-to-trough distance. Amplitude is measured from the centre line (rest position) to a peak.

Worked example 13 marks

The diagram shows a wave on a string. The wave is moving to the right. The frequency of the wave is 4 Hz.(a) What is the amplitude of the wave?(b) What is the wavelength of the wave?(c) What is the period of the wave? (A diagram shows a sine wave with peaks at +3 cm and troughs at -3 cm. One full wavelength is shown to be 8 cm long).

  1. 1

    (a) The amplitude is the maximum displacement from the rest position. From the diagram, this is 3 cm.

  2. 2

    (b) The wavelength is the length of one complete wave. From the diagram, this is 8 cm.

  3. 3

    (c) Use the formula T = 1/f. Given f = 4 Hz, T = 1/4 = 0.25 s.

Recap

  • Amplitude is the maximum displacement from the rest position.
  • Wavelength (λ) is the length of one complete wave cycle.
  • Frequency (f) is the number of waves per second, measured in Hertz (Hz).
  • Period (T) is the time for one wave to pass, and is equal to 1/f.

Quick check

  1. If 20 waves pass a point in 5 seconds, what is the frequency?1 mark
  2. What is the period of the waves in the previous question?1 mark

3. The Wave Speed Equation

There is a crucial relationship that links the speed of a wave with its frequency and wavelength. The wave speed equation is one of the most important formulas in this topic. It states that the speed of a wave is the product of its frequency and its wavelength. The equation is: wave speed = frequency × wavelength, or in symbols, v = f × λ. For this equation to work correctly, you must use standard SI units: speed(v) in metres per second (m/s), frequency(f) in Hertz (Hz), and wavelength (λ) in metres (m).

wave speed = frequency × wavelength

v = f × λ

Key term

Wave Speed (v): The speed at which energy is transferred by a wave, calculated as frequency multiplied by wavelength.

Examiner insight

Examiners expect candidates to state the formula, show their substitution, and give the final answer with the correct units to gain full marks.

Common pitfall

Using inconsistent units, especially centimetres for wavelength with metres per second for speed. Always convert to metres first.

Worked example 13 marks

A radio station broadcasts at a frequency of 98.5 MHz. Radio waves travel at the speed of light, 3.0 × 10⁸ m/s. Calculate the wavelength of these radio waves.

  1. 1

    First, convert the frequency to Hertz. 1 MHz = 1,000,000 Hz. So, f = 98.5 × 10⁶ Hz.

  2. 2

    State the wave equation: v = f × λ.

  3. 3

    Rearrange the equation to find wavelength: λ = v / f.

  4. 4

    Substitute the values: λ = (3.0 × 10⁸ m/s) / (98.5 × 10⁶ Hz).

  5. 5

    Calculate the result: λ = 3.046 m. The wavelength is approximately 3.05 m.

Worked example 24 marks

Water waves in a swimming pool have a wavelength of 40 cm. A wave passes a point every 0.5 s. Calculate the speed of the waves in m/s.

  1. 1

    First, find the frequency. The period T = 0.5 s. Frequency f = 1/T = 1/0.5s = 2 Hz.

  2. 2

    Next, convert the wavelength to metres: λ = 40 cm = 0.40 m.

  3. 3

    Use the wave equation: v = f × λ.

  4. 4

    Substitute the values: v = 2 Hz × 0.40 m.

  5. 5

    Calculate the speed: v = 0.8 m/s.

Recap

  • The wave speed equation is v = f × λ.
  • Wave speed (v) is measured in m/s.
  • Frequency (f) is measured in Hz.
  • Wavelength (λ) is measured in m.
  • Always convert units to standard SI units before calculating.

Quick check

  1. A wave has a speed of 20 m/s and a frequency of 5 Hz. What is its wavelength?2 marks

4. Reflection, Refraction and Diffraction

Waves can interact with barriers and boundaries in three important ways, often demonstrated in a ripple tank.

  1. Reflection: This is when a wave bounces off a surface. The angle at which the wave hits the surface (angle of incidence) is equal to the angle at which it reflects (angle of reflection).
  2. Refraction: This is the change in direction of a wave as it passes from one medium to another, caused by a change in speed. For example, water waves slow down and bend as they move from deep to shallow water. When a wave slows down, its wavelength decreases, but its frequency stays the same.
  3. Diffraction: This is the spreading out of waves as they pass through a gap or around an obstacle. The effect is most significant when the size of the gap is similar to the wavelength of the waves. Longer wavelengths diffract more than shorter wavelengths.

angle of incidence = angle of reflection

Key term

Diffraction: The spreading out of waves as they pass through an aperture or around an obstacle.

Fun fact

You can hear someone talking from around a corner because sound waves have a long enough wavelength to diffract around it. You can't see them because light's wavelength is tiny and barely diffracts at all around everyday objects.

Worked example 13 marks

Plane waves in a ripple tank travel from a deep region to a shallow region, as shown. The frequency of the waves is 5 Hz.(a) Name the effect occurring at the boundary.(b) In the shallow region, are the waves faster or slower?(c) How does the wavelength in the shallow region compare to the wavelength in the deep region?

  1. 1

    (a) The effect is refraction.

  2. 2

    (b) Waves travel more slowly in shallower water.

  3. 3

    (c) Since the waves slow down and frequency remains constant (v=fλ), the wavelength must decrease. The waves are closer together in the shallow region.

Worked example 23 marks

A harbour wall has two gaps. Gap A is 50 m wide and Gap B is 5 m wide. Water waves with a wavelength of 4 m approach the harbour. Through which gap will the waves diffract more, and why?

  1. 1

    The waves will diffract more when passing through Gap B.

  2. 2

    This is because diffraction is most significant when the size of the gap is close to the wavelength of the waves.

  3. 3

    The width of Gap B (5m) is much closer to the wavelength (4m) than the width of Gap A (50 m).

Recap

  • Reflection is the bouncing of waves from a surface.
  • Refraction is the bending of waves due to a change in speed when entering a new medium.
  • Diffraction is the spreading of waves after passing through a gap or around an edge.
  • In refraction, frequency remains constant while speed and wavelength change.
  • Maximum diffraction occurs when the gap size is similar to the wavelength.

Quick check

  1. When a water wave enters a shallower region of a ripple tank, what happens to its frequency?1 mark

5. The Nature of Sound

Sound is produced by vibrating objects, such as a guitar string, a loudspeaker cone, or your vocal cords. These vibrations are passed through a medium (a substance like a solid, liquid, or gas) as a longitudinal wave. As the object vibrates forwards, it pushes particles together, creating a region of high pressure called a compression. As it vibrates backwards, it leaves a region of low pressure called a rarefaction. This series of compressions and rarefactions travels outwards, carrying sound energy. Because sound relies on particle vibrations, it cannot travel through a vacuum where there are no particles. Sound travels at different speeds in different media: it is fastest in solids, slower in liquids, and slowest in gases, because the particles are closer together in solids, allowing vibrations to be passed on more quickly.

Key term

Compression: A region in a longitudinal wave where the particles are closest together (high pressure).

Examiner insight

Students must be explicit that sound needs a medium because it requires particles to oscillate and pass on the vibration; a vacuum has no particles, so the wave cannot propagate.

Worked example 14 marks

An experiment is set up with an electric bell inside a sealed glass jar connected to a vacuum pump.(a) Why can the bell be heard when the pump is off?(b) What is observed as the air is pumped out of the jar?(c) Explain this observation.

  1. 1

    (a) When the pump is off, the jar is full of air. The sound waves travel through the air to the glass and then to the observer's ear.

  2. 2

    (b) As air is pumped out, the sound of the bell becomes fainter and eventually cannot be heard, even though the hammer can still be seen hitting the bell.

  3. 3

    (c) This shows that sound requires a medium (particles) to travel. A vacuum is an empty space with no particles, so the sound vibrations cannot be transmitted.

Recap

  • Sound is a longitudinal wave produced by vibrating objects.
  • Sound waves consist of a series of compressions and rarefactions.
  • Sound requires a medium (solid, liquid, or gas) to travel and cannot travel in a vacuum.
  • Sound travels fastest in solids and slowest in gases.

Quick check

  1. What is the name for the low-pressure region of a sound wave?1 mark
  2. Why would an explosion in space be silent to a nearby observer?1 mark

6. Pitch, Loudness and Oscilloscopes

Sounds have two main characteristics: pitch and loudness. Pitch describes how high or low a sound is, and it is determined by the wave's frequency. A high frequency wave produces a high-pitched sound, while a low frequency wave produces a low-pitched sound. Loudness is determined by the wave's amplitude. A wave with a large amplitude carries more energy and sounds louder. A wave with a small amplitude is quieter. We can visualise sound waves using a Cathode-Ray Oscilloscope (CRO). The CRO displays a graph of the wave, with the vertical axis representing amplitude (loudness) and the horizontal axis representing time. A louder sound will have a taller trace on the CRO, and a higher-pitched sound will have more waves packed together on the screen (shorter wavelength). The typical range of human hearing is from 20 Hz to 20,000 Hz (20 kHz).

Key term

Pitch: The subjective perception of a sound's frequency; high frequency corresponds to high pitch.

Examiner insight

When asked to draw a new waveform on an oscilloscope trace, candidates must correctly modify both the amplitude (for loudness) and the number of waves shown (for pitch/frequency).

Common pitfall

Mixing up the relationships. Remember: Amplitude -> Loudness; Frequency -> Pitch.

Worked example 12 marks

The diagram shows the trace of a sound wave on an oscilloscope. Copy the diagram and on the same axes, draw the trace for a sound that is quieter but has a higher pitch.

  1. 1
    1. 'Quieter' means the amplitude must be smaller. The new wave's peaks and troughs should be closer to the centre line than the original wave.
  2. 2
    1. 'Higher pitch' means the frequency must be higher. The new wave must have more complete cycles in the same amount of time, meaning its wavelength on the screen will be shorter.
  3. 3
    1. The resulting drawing should show a wave that is less tall and more squashed together horizontally than the original.

Recap

  • The pitch of a sound is determined by its frequency.
  • The loudness of a sound is determined by its amplitude.
  • High frequency means high pitch; large amplitude means loud sound.
  • An oscilloscope displays amplitude on the y-axis and time on the x-axis.
  • The range of human hearing is approximately 20 Hz to 20,000 Hz.

Quick check

  1. If you tighten a guitar string, it produces a higher pitch. Has its frequency of vibration increased or decreased?1 mark
  2. When you turn down the volume on a TV, are you changing the amplitude or the frequency of the sound waves?1 mark

7. Speed of Sound and Echoes

An echo is simply a reflected sound wave. When a sound wave hits a hard, flat surface like a wall or a cliff, it bounces back. If the surface is far enough away, you will hear the reflected sound a short time after the original sound. We can use echoes to measure distances or the speed of sound itself. The key is to remember that the sound has to travel to the reflecting surface AND back again. Therefore, the total distance travelled by the sound is twice the distance to the object. The standard `speed = distance / time` formula is used, but you must be careful with the distance value. For example, if 'd' is the distance to a wall, the echo travels a total distance of '2d'.

speed = total distance / time taken

distance to object (d) = (speed of sound × time for echo) / 2

Key term

Echo: A sound wave that has been reflected off a surface and is heard after the original sound.

Common pitfall

Forgetting that the sound in an echo problem travels a distance of 2d (there and back). Many students use just 'd' and get the wrong answer by a factor of two.

Worked example 13 marks

A student stands in front of a tall cliff and claps her hands. She hears the echo 1.5 seconds later. If the speed of sound in air is 340 m/s, how far away is the cliff?

  1. 1
    1. First, calculate the total distance the sound travelled (to the cliff and back).
  2. 2

    Use the formula: distance = speed × time.

  3. 3

    Total distance = 340 m/s × 1.5 s = 510 m.

  4. 4
    1. This is the distance for the round trip. The distance to the cliff is half of this.
  5. 5

    Distance to cliff = 510 m / 2 = 255 m.

Worked example 23 marks

To measure the speed of sound, two students stand 200 m apart. One student bangs two wooden blocks together. The second student starts a stopwatch when they see the blocks hit and stops it when they hear the sound. The measured time is 0.59 s. Calculate the speed of sound.

  1. 1
    1. State the formula: speed = distance / time.
  2. 2
    1. The distance the sound travelled is 200 m.
  3. 3
    1. The time taken is 0.59 s.
  4. 4
    1. Substitute values: speed = 200 m / 0.59 s.
  5. 5
    1. Calculate the result: speed = 338.98... m/s. The speed of sound is approximately 339 m/s.

Recap

  • An echo is a reflected sound wave.
  • In echo calculations, the sound travels to the object and back again.
  • The total distance travelled by an echo is twice the distance to the reflecting surface (2d).
  • Use the formula: distance = speed × time, being careful with the distance value.
  • Measuring the time for an echo to return from a known distance is a way to find the speed of sound.

Quick check

  1. A ship's sonar sends a pulse to the seabed, which is 3000 m deep. The sound travels at 1500 m/s in water. How long does the pulse take to return to the ship?2 marks

8. Ultrasound and Its Uses

Ultrasound is sound with a frequency above the upper limit of human hearing, which is about 20,000 Hz (or 20 kHz). Although we can't hear it, ultrasound has many important applications. Its high frequency means it has a short wavelength, which allows it to produce detailed images and be focused into narrow beams. Medical Uses: The most common medical use is in prenatal scanning to create images of a fetus in the womb. Pulses of ultrasound are sent into the body, and they reflect off boundaries between different tissues. A computer analyses the returning echoes to build a real-time image. It is used because it is non-ionising, meaning it doesn't carry enough energy to damage cells, making it much safer than X-rays for delicate tissues. It is also used to break up kidney stones. Industrial Uses: Ultrasound is used for industrial cleaning of delicate items like jewellery, and for quality control to check for flaws inside metal castings. Echo-sounding (Sonar): Ships use ultrasound to measure the depth of the sea or locate shoals of fish. A pulse is sent downwards, and the time taken for the echo to return from the seabed is used to calculate the depth.

Key term

Ultrasound: Sound waves with a frequency higher than the upper limit of human hearing (above 20 kHz).

Fun fact

Bats and dolphins are masters of echolocation, using ultrasound to navigate and hunt in complete darkness with incredible accuracy.

Worked example 15 marks

(a) What is ultrasound?(b) Explain why ultrasound is preferred over X-rays for producing an image of a fetus.(c) A fishing boat uses an echo-sounder which sends out an ultrasound pulse. The echo from a shoal of fish is detected 0.08 s later. The speed of sound in seawater is 1500 m/s. Calculate the depth of the shoal of fish.

  1. 1

    (a) Ultrasound is sound with a frequency greater than 20,000 Hz.

  2. 2

    (b) Ultrasound is non-ionising radiation, which means it does not damage living cells. X-rays are ionising and can be harmful to a developing fetus.

  3. 3

    (c) Total distance travelled = speed × time = 1500 m/s × 0.08 s = 120 m. This is the distance to the fish and back, so the depth is half this value. Depth = 120 m / 2 = 60 m.

Recap

  • Ultrasound is high-frequency sound (> 20 kHz) that is inaudible to humans.
  • Ultrasound is used for medical imaging (e.g. prenatal scans) because it is non-ionising and safe.
  • It is also used for industrial cleaning and flaw detection.
  • Sonar uses ultrasound pulses and echoes to measure the depth of the sea.

Quick check

  1. State one non-medical use of ultrasound.1 mark
  2. Is ultrasound a transverse or a longitudinal wave?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State two differences between a light wave and a sound wave.2 marks
  2. A microwave oven uses electromagnetic waves with a frequency of 2450 MHz. The speed of these waves is 3.0 × 10⁸ m/s. Calculate their wavelength in metres.3 marks
  3. The diagram shows a wave on an oscilloscope. The time for the whole trace to be produced is 0.04 s. (a) How many complete waves are shown? (b) Calculate the frequency of the wave.3 marks
  4. A girl shouts towards a cliff 495 m away. The speed of sound in air is 330 m/s. Calculate the time it takes for her to hear the echo.3 marks
  5. Plane water waves travel from a region of deep water to a region of shallow water. Their speed decreases from 0.4 m/s to 0.3 m/s. If the wavelength in the deep water is 8 cm, calculate the wavelength in the shallow water.4 marks
  6. Sound waves with a wavelength of 0.5 m approach a doorway that is 0.7 m wide. Describe what happens to the waves as they pass through the doorway and explain why someone in the next room can hear the sound even if they cannot see the source.4 marks
  7. A loudspeaker cone vibrates with a frequency of 250 Hz. (a) Describe the motion of the air particles in front of the loudspeaker. (b) What is the name for the regions where the air particles are bunched together? (c) Calculate the wavelength of the sound produced, if the speed of sound is 340 m/s.4 marks
  8. A boat's sonar system is used for echo-sounding. It sends out a pulse of ultrasound and receives the echo 0.4 seconds later. (a) Define ultrasound. (b) Explain why ultrasound is used for this purpose. (c) If the speed of sound in water is 1500 m/s, calculate the depth of the sea.5 marks
  9. A wave has a period of 0.002 s and a wavelength of 0.7 m. (a) Calculate the frequency of the wave. (b) Calculate the speed of the wave.4 marks
  10. Explain, in terms of particles, why sound travels faster in a solid like steel than in a gas like air, and why it cannot travel through a vacuum at all.4 marks

Go deeper

Practise and revise with member-only material for this chapter.

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters