Cambridge IGCSE0972

Sound

Physics 0972 Chapter Notes

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In this chapter

  1. 1The Nature of Sound Waves
  2. 2Pitch, Loudness and Oscilloscopes
  3. 3Speed of Sound and The Wave Equation
  4. 4Echoes and Distance Measurement
  5. 5Ultrasound and its Applications

1. The Nature of Sound Waves

Sound is a form of energy that travels as a mechanical wave, created by vibrations. When an object, like a guitar string or a loudspeaker cone, vibrates, it causes nearby particles of a medium (like air) to vibrate as well. This creates a chain reaction, passing the vibration along. The particles of the medium oscillate back and forth in the same direction that the wave is travelling. This makes sound a longitudinal wave. These oscillations create areas where the particles are bunched together, called compressions (high pressure), and areas where they are spread apart, called rarefactions (low pressure). Because sound relies on particle vibrations, it requires a medium (a solid, liquid, or gas) to travel through. It cannot travel in a vacuum, like outer space, because there are no particles to transmit the wave.

Key term

Longitudinal Wave: A wave in which the particles of the medium vibrate parallel to the direction of energy transfer.

Examiner insight

Examiners often test the understanding that sound requires a medium by asking to explain the 'bell-in-a-jar' experiment. Clearly state that a vacuum has no particles to transmit the vibrations.

Common pitfall

Confusing the movement of the wave with the movement of the particles. The wave travels from the source to the ear, but individual air particles just oscillate back and forth about their fixed positions.

Fun fact

Sound travels about four times faster in water than in air. This is why whales can communicate over hundreds of kilometres in the ocean.

Worked example 14 marks

An experiment is set up with an electric bell ringing inside a sealed glass jar connected to a vacuum pump. Explain what is heard, and why, as the air is slowly pumped out of the jar.

  1. 1

    Initially, with air in the jar, the sound of the bell is heard clearly. This is because the bell's vibrations create sound waves that travel through the air in the jar, then through the glass, and finally through the air to the observer's ear.

  2. 2

    As the air is pumped out, the sound becomes fainter and fainter.

  3. 3

    This happens because sound waves are mechanical waves and require a medium (particles) to travel. With fewer air particles in the jar, the transmission of the vibrations becomes less efficient.

  4. 4

    When a near-vacuum is achieved, the bell can no longer be heard, even though it is still visibly ringing. This is because there are virtually no particles left inside the jar to transmit the sound vibrations from the bell to the glass.

Recap

  • Sound is produced by vibrating objects.
  • Sound is a longitudinal wave, consisting of compressions and rarefactions.
  • Sound requires a medium (solid, liquid, or gas) to travel and cannot travel in a vacuum.
  • The particles of the medium oscillate parallel to the direction of the sound wave's travel.

Quick check

  1. What is the name for the regions in a sound wave where particles are bunched together at high pressure?1 mark
  2. Can sound travel from the Sun to the Earth? Explain your answer.2 marks

2. Pitch, Loudness and Oscilloscopes

The characteristics of a sound are determined by the properties of its wave. Pitch is how high or low a sound is, and it is determined by the wave's frequency. A higher frequency means a higher pitch. Frequency is the number of complete waves passing a point per second, measured in Hertz (Hz). Loudness is the intensity of the sound, and it is determined by the wave's amplitude. A larger amplitude means a louder sound. We can visualise sound waves using a microphone connected to an oscilloscope. The microphone converts the pressure variations of the sound wave into an electrical signal, which the oscilloscope displays as a waveform. On the screen, the horizontal axis represents time and the vertical axis represents amplitude. A sound with a higher pitch will have more waves squashed together on the screen (higher frequency), and a louder sound will have taller waves (larger amplitude).

Key term

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

Examiner insight

Students must be able to correctly interpret and draw oscilloscope traces. Be prepared to sketch or compare waveforms for sounds that are louder, quieter, higher-pitched, or lower-pitched for full marks.

Common pitfall

Mixing up the concepts: students often incorrectly state that a higher frequency means a louder sound. Remember: Frequency determines Pitch; Amplitude determines Loudness.

Worked example 14 marks

The diagram shows the waveform of a sound on an oscilloscope.a) On a copy of the axes, sketch the waveform of a sound that is quieter but has the same pitch.b) On a second copy, sketch the waveform of a sound that has a higher pitch but the same loudness.

  1. 1

    a) For a quieter sound, the amplitude must be smaller. For the same pitch, the frequency (and therefore the wavelength/period) must be the same. The new wave should be shorter in height but have the same number of peaks and troughs in the same time period.

  2. 2

    b) For a higher pitch, the frequency must be greater, meaning more waves in the same time period. For the same loudness, the amplitude must be the same. The new wave should have the same height but more waves packed into the same horizontal distance.

Recap

  • The pitch of a sound is determined by its frequency.
  • The loudness of a sound is determined by its amplitude.
  • An oscilloscope displays a sound wave as a graph of amplitude versus time.
  • On an oscilloscope trace, a higher frequency (more waves per second) corresponds to a higher pitch.
  • On an oscilloscope trace, a larger amplitude (taller waves) corresponds to a louder sound.

Quick check

  1. If a sound becomes lower in pitch, what property of its wave has decreased?1 mark
  2. Look at two oscilloscope traces. Trace A has waves that are twice as tall as Trace B, but has half the number of waves in the same time period. Describe the sound of A compared to B.2 marks

3. Speed of Sound and The Wave Equation

The speed of sound is the distance a sound wave travels per unit of time. It is not constant; it depends on the medium it is travelling through and the conditions of that medium, such as temperature and density. In general, sound travels fastest in solids, slower in liquids, and slowest in gases. This is because the particles in a solid are packed more tightly together than in a liquid or gas, allowing vibrations to be passed on more quickly. For example, the speed of sound in air is approximately 340 m/s, in water it's about 1500 m/s, and in steel it's around 5000 m/s. The relationship between the speed (v), frequency (f), and wavelength (λ) of any wave is given by the crucial wave equation: v = f × λ.

v = f × λ

Key term

Wavelength (λ): The distance between two consecutive corresponding points on a wave, for example, from the centre of one compression to the centre of the next.

Examiner insight

Marks are consistently awarded for showing clear working. Always write down the formula you are using, then show the substitution of values, and finally state the answer with the correct units.

Common pitfall

Forgetting to convert units before calculating. Frequency is often given in kHz (kilohertz) and must be converted to Hz. Wavelength might be in cm and must be converted to m.

Worked example 13 marks

A tuning fork produces a sound with a frequency of 256 Hz. The sound travels through the air at a speed of 340 m/s. Calculate the wavelength of the sound wave.

  1. 1

    State the wave equation: v = f × λ

  2. 2

    Rearrange the formula to make wavelength (λ) the subject: λ = v / f

  3. 3

    Substitute the given values into the formula: λ = 340 m/s / 256 Hz

  4. 4

    Calculate the result: λ ≈ 1.33 m

Worked example 23 marks

A sound wave in a solid has a wavelength of 1.2 m and a frequency of 4.0 kHz. Calculate the speed of the sound in the solid.

  1. 1

    First, convert the frequency from kilohertz (kHz) to hertz (Hz). 4.0 kHz = 4.0 × 1000 = 4000 Hz.

  2. 2

    State the wave equation: v = f × λ

  3. 3

    Substitute the values into the formula: v = 4000 Hz × 1.2 m

  4. 4

    Calculate the speed: v = 4800 m/s

Recap

  • The speed of sound depends on the medium's density and temperature.
  • Sound travels fastest in solids, slower in liquids, and slowest in gases.
  • The wave equation, v = f × λ, links speed, frequency, and wavelength.
  • Always ensure units are in metres (m), seconds (s), and Hertz (Hz) before calculating.
  • The speed of sound in air is approximately 330-340 m/s, a value often provided in exams.

Quick check

  1. In which of these does sound travel slowest: concrete, sea water, or helium gas?1 mark
  2. A sound wave has a wavelength of 0.5 m and travels at 340 m/s. What is its frequency?2 marks

4. Echoes and Distance Measurement

An echo is a reflection of sound. When a sound wave hits a hard, flat surface like a wall, cliff, or the seabed, it bounces back. If you hear the original sound and then its reflection a short time later, you have heard an echo. This principle is extremely useful for measuring distances. The key is to remember that the sound has to travel a total distance of 'there and back'. If the distance to the reflecting surface is 'd', the total distance the sound travels is 2d. By measuring the time delay(t) between sending the sound and receiving the echo, and knowing the speed of sound (v), we can calculate the distance 'd' using a rearranged speed formula: speed = total distance / time. This becomes v = 2d / t, which can be rearranged to find the distance: d = (v ×t) / 2. This is the basis for SONAR (SOund Navigation And Ranging) used by ships and submarines.

speed = (2 × distance) / time

d = (v × t) / 2

Key term

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

Examiner insight

Examiners specifically look for the '2d' term or the final division by 2 in the calculation to award method marks. Always show this step clearly in your working.

Common pitfall

Forgetting that the sound travels a round trip. Students often calculate the total distance (v × t) and forget to divide by two to find the one-way distance to the object.

Worked example 13 marks

A person stands some distance from a large cliff and claps their hands. They hear an echo 1.5 seconds later. Taking the speed of sound in air as 330 m/s, calculate the distance from the person to the cliff.

  1. 1

    The sound travels to the cliff and back, so the total distance travelled is 2d.

  2. 2

    State the formula relating speed, distance and time for an echo: v = 2d / t.

  3. 3

    Rearrange the formula to find the one-way distance (d): d = (v ×t) / 2.

  4. 4

    Substitute the values: d = (330 m/s × 1.5s) / 2.

  5. 5

    Calculate the result: d = 495 / 2 = 247.5 m.

Worked example 23 marks

A fishing boat uses an echo-sounder to measure the depth of the sea. It sends a pulse of sound and detects the echo after 0.12 s. The speed of sound in seawater is 1500 m/s. What is the depth of the sea below the boat?

  1. 1

    Identify the knowns: time t = 0.12 s, speed v = 1500 m/s.

  2. 2

    Use the echo formula: distance = (speed × time) / 2.

  3. 3

    Substitute the values: depth = (1500 m/s × 0.12s) / 2.

  4. 4

    Calculate the result: depth = 180 / 2 = 90 m.

Recap

  • An echo is a reflected sound wave.
  • Echoes can be used to measure distances to objects.
  • The sound travels to the object and back again, a total distance of 2d.
  • The key formula for echo calculations is d = (v × t) / 2.
  • This principle is used in sonar for mapping the seabed and locating submarines.

Quick check

  1. A bat is 17 m away from a wall. How long will it take for its ultrasonic squeak to return as an echo? (Speed of sound = 340 m/s)2 marks

5. Ultrasound and its Applications

Ultrasound is sound with a frequency above the upper limit of human hearing, which is approximately 20,000 Hz (or 20 kHz). Although we cannot hear it, ultrasound has many important uses in medicine and industry. It works on the same principle as echoes. A device called a transducer emits short pulses of ultrasound waves. These waves travel into a material (like the human body). When the waves hit a boundary between different types of tissue (e.g., between fluid and bone, or muscle and organ), some are reflected back. The transducer detects these returning echoes. A computer measures the time it takes for the echoes to return and uses the speed of sound in that tissue to calculate the depth of the boundary. By processing thousands of these echoes from different directions, the computer can build up a detailed, real-time image. This is called ultrasonography.

distance to boundary = (speed of ultrasound × time delay) / 2

Key term

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

Fun fact

Some high-tech security alarms use ultrasound. They fill a room with a pattern of ultrasonic waves, and if an intruder moves in the room, they disturb the pattern of echoes, triggering the alarm.

Worked example 13 marks

Explain why ultrasound is used for prenatal scanning to check the health of a foetus, and state one advantage it has over using X-rays for this purpose.

  1. 1

    Ultrasound is used by sending high-frequency sound pulses into the mother's abdomen from a transducer.

  2. 2

    These pulses reflect from the surfaces and organs of the foetus.

  3. 3

    A computer detects the returning echoes and uses the time delays to build a real-time image of the foetus.

  4. 4

    Advantage: Ultrasound is non-ionising radiation, meaning it does not carry enough energy to damage cells or DNA. This makes it much safer for a developing foetus than X-rays, which are ionising and can be harmful.

Worked example 23 marks

An industrial ultrasound probe is used to check for flaws in a steel pipe. A pulse of ultrasound returns an echo from a small crack in 0.00002 s. If the speed of sound in steel is 5000 m/s, how deep is the crack inside the pipe wall?

  1. 1

    State the formula for echo location: depth(d) = (speed × time) / 2.

  2. 2

    Substitute the given values: d = (5000 m/s × 0.00002s) / 2.

  3. 3

    Calculate the total distance travelled by the pulse: 5000 × 0.00002 = 0.1 m.

  4. 4

    Divide by two to find the one-way depth: d = 0.1 m / 2 = 0.05 m.

  5. 5

    The crack is 0.05 m (or 5 cm) deep.

Recap

  • Ultrasound is high-frequency sound, above 20,000 Hz.
  • It is used in medical imaging (e.g., prenatal scans) and industrial quality control (e.g., finding flaws).
  • Ultrasound imaging works by emitting pulses and detecting the echoes from internal boundaries.
  • The time delay of the echo is used to calculate the distance to the reflecting surface.
  • A key advantage of ultrasound is that it is non-ionising and therefore safe for imaging sensitive tissues.

Quick check

  1. State one non-medical use of ultrasound.1 mark
  2. Why is a gel used between an ultrasound transducer and the skin?1 mark

End-of-chapter exercise

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

  1. A loudspeaker cone vibrates to produce a sound wave. Describe the motion of the air particles in front of the cone and explain what is meant by the terms 'compression' and 'wavelength' in this context.4 marks
  2. A sound has a frequency of 400 Hz. The speed of sound in air is 320 m/s. a) Calculate the wavelength of the sound. b) What would be the wavelength of a sound with a frequency of 800 Hz (one octave higher)?4 marks
  3. Explain why you see a flash of lightning several seconds before you hear the thunder, even though they are produced at the same time. If the time delay is 5 seconds, calculate the distance to the lightning strike. (Speed of sound = 330 m/s).3 marks
  4. A submarine emits a sonar pulse which returns from the seabed in 1.2 s. The speed of sound in seawater is 1500 m/s. Calculate the depth of the sea.3 marks
  5. Two sounds, A and B, are displayed on an oscilloscope. The trace for sound A has a larger amplitude and a lower frequency than the trace for sound B. Compare the loudness and pitch of sound A to sound B.2 marks
  6. Describe how ultrasound is used in medicine to produce an image of an internal organ, such as the kidney. You should mention the role of the transducer, reflection and a computer.4 marks
  7. The range of human hearing is 20 Hz to 20,000 Hz. Calculate the longest and shortest wavelength of sound that a human can hear in air, assuming the speed of sound is 340 m/s.4 marks
  8. A student shouts towards a wall 85 m away. a) Calculate the total distance the sound must travel to produce an echo. b) Calculate the time it takes for the echo to be heard. (Speed of sound = 340 m/s).3 marks
  9. Sound can be diffracted. Explain what diffraction is and give an everyday example of sound diffraction.2 marks
  10. A metal rod of length 2.0 m is struck at one end. The sound travels through the metal and is detected at the other end 0.0004 s later. a) Calculate the speed of sound in the metal. b) Why is this value much greater than the speed of sound in air?4 marks

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