Cambridge IGCSE0625

Sound

Physics 0625 Chapter Notes

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In this chapter

  1. 1The Nature of Sound Waves
  2. 2Pitch, Loudness and Waveforms
  3. 3Speed of Sound and Echoes
  4. 4Ultrasound and Its Applications

1. The Nature of Sound Waves

Sound is a form of energy that travels as a wave, produced by vibrating objects. For example, a guitar string vibrates, pushing and pulling on nearby air particles. These vibrations pass from particle to particle, transferring energy through the medium. Sound waves are longitudinal, which means the particles of the medium oscillate back and forth parallel to the direction the wave is travelling. This creates 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 particles to be transmitted, it requires a medium (a solid, liquid, or gas) and cannot travel through a vacuum, where there are no particles to vibrate.

Key term

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

Examiner insight

Examiners reward clear descriptions of particle motion in a longitudinal wave, specifically mentioning oscillations parallel to the wave's direction of energy transfer.

Common pitfall

Thinking that air particles travel from the source to the ear. They only vibrate about fixed positions; it is the energy that is transferred.

Fun fact

The crack of a whip is a mini sonic boom. The tip of the whip moves faster than the speed of sound, creating a shockwave we hear as a 'crack'.

Worked example 13 marks

A bell is placed inside a glass jar connected to a vacuum pump. The bell is rung, and then the air is pumped out of the jar. Explain what an observer would hear.

  1. 1

    Step 1: Initially, with air in the jar, the sound of the bell can be heard. This is because the sound waves travel through the air in the jar, then the glass, and finally the air outside to the observer's ears.

  2. 2

    Step 2: As the air is pumped out, the sound becomes fainter. This is because there are fewer air particles inside the jar to transmit the vibrations.

  3. 3

    Step 3: When a vacuum is created, the sound cannot be heard at all, even though the bell's hammer can be seen striking the bell. This is because sound requires a medium (particles) to travel, and a vacuum is a space with no particles.

Worked example 22 marks

The diagram shows air particles in front of a loudspeaker. Describe the motion of a single air particle as the sound wave passes.

  1. 1

    Step 1: The sound wave is longitudinal, meaning the particles oscillate parallel to the direction of wave travel.

  2. 2

    Step 2: A single air particle will oscillate (vibrate) back and forth about its fixed equilibrium position. It does not travel along with the wave.

Recap

  • Sound is produced by vibrating sources.
  • Sound waves are longitudinal, consisting of compressions and rarefactions.
  • The particles in the medium vibrate parallel to the direction of energy transfer.
  • Sound requires a medium (solid, liquid, or gas) to travel.
  • Sound cannot travel through a vacuum.

Quick check

  1. What is the name for the regions of low pressure in a sound wave?1 mark
  2. Why is there no sound in space?1 mark

2. Pitch, Loudness and Waveforms

We can visualise a sound wave by connecting a microphone to an oscilloscope. The oscilloscope displays the sound as a voltage-time graph, which looks like a transverse wave. The properties of this displayed waveform are directly related to what we hear. The pitch of a sound is determined by its frequency (how many waves pass a point per second, measured in Hertz, Hz). A high frequency means a high pitch. The loudness of a sound is determined by its amplitude (the maximum displacement or height of the wave from its equilibrium position). A large amplitude means a loud sound. The relationship between a wave's speed (v), frequency (f), and wavelength (λ) is given by the wave equation: v = f × λ.

v = f × λ

Key term

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

Examiner insight

Marks are often awarded for correctly sketching or interpreting oscilloscope traces. Be precise: to show a louder sound, draw a taller wave; for a higher pitch, draw more waves in the same space.

Common pitfall

Confusing the oscilloscope trace with the actual wave. Sound is a longitudinal wave, but the oscilloscope displays it as a transverse graph of pressure or voltage against time.

Fun fact

The standard 'A' note that orchestras tune to is 440 Hz. This means the air particles are oscillating back and forth 440 times every second.

Worked example 14 marks

The diagram shows the oscilloscope traces for two sounds, A and B. Compare the pitch and loudness of the two sounds.

  1. 1

    Step 1 (Pitch/Frequency): Sound B has more waves on the screen in the same amount of time as sound A. This means sound B has a higher frequency.

  2. 2

    Step 2 (Pitch Conclusion): Since pitch is related to frequency, sound B has a higher pitch than sound A.

  3. 3

    Step 3 (Loudness/Amplitude): Both sound A and sound B have the same height (amplitude).

  4. 4

    Step 4 (Loudness Conclusion): Since loudness is related to amplitude, both sounds have the same loudness.

Worked example 23 marks

A sound wave has a frequency of 220 Hz and travels through air at a speed of 330 m/s. Calculate its wavelength.

  1. 1

    Step 1: Write down the wave equation: v = f × λ

  2. 2

    Step 2: Rearrange the formula to make wavelength (λ) the subject: λ = v / f

  3. 3

    Step 3: Substitute the given values into the formula: λ = 330 m/s / 220 Hz

  4. 4

    Step 4: Calculate the result: λ = 1.5 m

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's frequency as how compressed the waves are horizontally.
  • An oscilloscope displays a sound wave's amplitude as the vertical height of the waves.
  • The wave equation, v = f × λ, links speed, frequency, and wavelength.

Quick check

  1. If a guitarist plays a note, then plays it again but louder, what has changed about the sound wave?1 mark
  2. How would the trace on an oscilloscope change if a sound's pitch was lowered?1 mark

3. Speed of Sound and Echoes

The speed of sound depends on the medium it travels through. It is fastest in solids, slower in liquids, and slowest in gases. This is because particles are closest together and more rigidly bonded in solids, allowing vibrations to pass more efficiently. In air at room temperature, the speed of sound is approximately 340 m/s. An echo is a reflection of a sound wave from a hard surface. By measuring the time it takes for an echo to return, we can calculate the distance to the reflecting surface. The sound has to travel to the surface and back, so the total distance travelled is twice the distance to the surface. This leads to the echo formula: speed = (2 × distance) / time.

speed = distance / time

speed = (2 × distance to reflector) / echo time

Key term

Echo: A reflection of a sound wave that arrives at the listener with a delay after the direct sound.

Examiner insight

In questions about measuring the speed of sound, examiners look for an understanding of how to handle reaction time errors. Using a long distance or measuring an echo makes the time interval larger and reduces the percentage error from human reaction.

Common pitfall

Forgetting to double the distance (or halve the time) in echo calculations. The sound travels to the object AND back.

Fun fact

Because sound travels faster in water (about 1500 m/s) than in air, whales can communicate over hundreds or even thousands of kilometres.

Worked example 13 marks

A person stands 100 m from a large cliff and shouts. They hear an echo a short time later. If the speed of sound in air is 340 m/s, calculate the time it takes for the echo to be heard.

  1. 1

    Step 1: Identify the total distance the sound travels. It goes to the cliff and back, so distance = 2 × 100 m = 200 m.

  2. 2

    Step 2: Write down the formula relating speed, distance, and time: speed = distance / time.

  3. 3

    Step 3: Rearrange the formula to find time: time = distance / speed.

  4. 4

    Step 4: Substitute the values: time = 200 m / 340 m/s.

  5. 5

    Step 5: Calculate the result: time ≈ 0.59 s.

Worked example 23 marks

Two students want to measure the speed of sound. They stand 200 m apart. One student bangs two wooden blocks together, and the second student starts a stopwatch when they see the blocks hit and stops it when they hear the sound. The measured time is 0.60 s. Calculate the speed of sound.

  1. 1

    Step 1: Write down the relevant formula: speed = distance / time.

  2. 2

    Step 2: Identify the values: distance = 200 m, time = 0.60 s.

  3. 3

    Step 3: Substitute the values into the formula: speed = 200 m / 0.60 s.

  4. 4

    Step 4: Calculate the result: speed ≈ 333 m/s.

Recap

  • Sound travels fastest in solids, slower in liquids, and slowest in gases.
  • The speed of sound in air is approximately 340 m/s.
  • An echo is a reflected sound wave.
  • For echo calculations, the total distance travelled is twice the distance to the reflector.
  • The formula for echoes is speed = 2d / t.

Quick check

  1. In which material would sound travel faster: steel or water?1 mark
  2. A ship is 450 m from a cliff. How long does it take for an echo from its horn to return? (Speed of sound = 330 m/s)2 marks

4. Ultrasound and Its Applications

Ultrasound is sound with a frequency above the range of human hearing, which is typically considered to be anything over 20,000 Hz (or 20 kHz). Although we can't hear it, ultrasound has many important uses. It works on the principle of sending out pulses of ultrasound and detecting the echoes that return. The time taken for the echoes to return can be used to build up an image or measure a distance. Because ultrasound is non-ionising (unlike X-rays), it is very safe for medical imaging, most notably for prenatal scans to check the health and development of a fetus. In industry, it is used for quality control to detect cracks in metal castings and for ultrasonic cleaning of delicate items like jewellery. Ships and submarines use a similar principle, called SONAR (Sound Navigation and Ranging), to map the seabed or detect other underwater objects.

depth = (speed of ultrasound × time delay) / 2

Key term

Ultrasound: Sound waves with a frequency higher than the upper audible limit of human hearing, typically above 20,000 Hz.

Examiner insight

When describing uses of ultrasound, specific and detailed answers gain more marks. For example, 'prenatal scanning to monitor fetal development' is better than 'seeing babies', and 'detecting flaws in metal castings' is better than 'checking things in factories'.

Common pitfall

Confusing ultrasound with other imaging techniques like X-rays or MRI. Remember ultrasound uses sound waves, is non-ionising, and is particularly good for imaging soft tissues.

Fun fact

Some high-tech pest repellents emit high-frequency ultrasound, which is silent to humans but intensely annoying to rodents and insects.

Worked example 13 marks

A fishing boat uses sonar equipment to find the depth of the sea. It sends a pulse of ultrasound downwards and detects the echo 0.12 seconds later. If the speed of sound in seawater is 1500 m/s, calculate the depth of the sea.

  1. 1

    Step 1: Write down the echo formula, adapted for this situation: speed = (2 × depth) / time.

  2. 2

    Step 2: Rearrange the formula to find the depth: depth = (speed × time) / 2.

  3. 3

    Step 3: Substitute the known values: depth = (1500 m/s × 0.12s) / 2.

  4. 4

    Step 4: Calculate the intermediate product: 1500 × 0.12 = 180 m. This is the total distance travelled (down and back up).

  5. 5

    Step 5: Calculate the final depth: depth = 180 m / 2 = 90 m.

Worked example 24 marks

Describe one medical and one non-medical use of ultrasound.

  1. 1

    Step 1 (Medical): A key medical use is prenatal scanning. A probe sends ultrasound waves into the womb, and the reflections from the fetus are used by a computer to create a real-time image. This allows doctors to monitor the baby's growth and health without using harmful radiation.

  2. 2

    Step 2 (Non-medical): A non-medical use is industrial quality control. Ultrasound can be used to check for internal cracks or flaws in materials like metal pipes or welds. A flaw will cause the ultrasound to reflect unexpectedly, revealing its location.

Recap

  • Ultrasound is sound with a frequency above 20,000 Hz.
  • Humans cannot hear ultrasound, but many animals like bats and dolphins can.
  • Ultrasound is used for safe medical imaging, such as prenatal scans.
  • Sonar uses ultrasound echoes to measure depth and locate underwater objects.
  • Industrial uses of ultrasound include cleaning and detecting flaws in materials.

Quick check

  1. What is the typical upper frequency limit of human hearing, in kHz?1 mark
  2. State one reason why ultrasound is preferred over X-rays for observing a fetus.1 mark

End-of-chapter exercise

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

  1. Explain why sound is described as a longitudinal wave. You may draw a diagram to help your explanation.3 marks
  2. A sound wave has a wavelength of 0.50 m and travels at 340 m/s in air. Calculate the frequency of the sound. State the formula you use.3 marks
  3. Describe an experiment to measure the speed of sound in air using a starting pistol, a stopwatch, and a long measuring tape. State the measurements you would take and the calculation needed.5 marks
  4. A boy shouts at a cliff wall and hears an echo 1.2 seconds later. If the speed of sound is 330 m/s, how far away is the cliff wall?3 marks
  5. The diagram shows two oscilloscope traces, P and Q. Describe the differences between the sound that produces trace P and the sound that produces trace Q in terms of their loudness and pitch.4 marks
  6. Sound travels at about 340 m/s in air, but at about 5000 m/s in steel. Explain why sound travels so much faster in a solid like steel compared to a gas like air. Refer to the particles in your answer.3 marks
  7. A ship uses sonar to measure the depth of the sea. It sends out an ultrasound pulse and receives an echo 0.9 seconds later. The speed of ultrasound in seawater is 1500 m/s. (a) Calculate the depth of the sea. (b) Why is ultrasound used for this purpose instead of audible sound?4 marks
  8. The frequency range for human hearing is 20 Hz to 20,000 Hz. Calculate the wavelength of the sound with the highest frequency a human can hear. (Take the speed of sound in air as 340 m/s).3 marks
  9. Describe, in detail, one medical application and one industrial application of ultrasound.4 marks
  10. A loudspeaker produces a note of frequency 500 Hz. An observer stands in front of a wall, 85 m away from it. The loudspeaker is placed between the observer and the wall. The observer hears the sound directly from the speaker and also hears the echo from the wall. Calculate the time delay between hearing the direct sound and the echo. (Speed of sound = 340 m/s).4 marks

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