Sound

1. Overview

Sound is a form of energy transfer produced by vibrating sources. Understanding sound waves is crucial as they are the primary way we communicate and navigate our environment, ranging from simple speech to advanced medical imaging and underwater exploration.

Key Definitions

• Longitudinal Wave: A wave in which the vibration of the medium is parallel to the direction the wave travels.
• Frequency: The number of vibrations per second, measured in Hertz (Hz).
• Amplitude: The maximum displacement of a point on a wave from its rest position (determines volume).
• Pitch: How "high" or "low" a sound is, determined by its frequency.
• Echo: The reflection of a sound wave from a surface.
• Ultrasound: Sound waves with a frequency higher than the upper limit of human hearing (above 20,000 Hz).
• Medium: The substance (solid, liquid, or gas) through which a sound wave travels.

---

Core Content

Production and Nature of Sound

• Vibrating Sources: Sound is produced whenever an object vibrates. These vibrations push and pull on surrounding air molecules, creating a wave.
• Longitudinal Nature: Sound travels as a longitudinal wave. The particles of the medium oscillate back and forth in the same direction that the wave is moving.
• Medium Requirement: Sound cannot travel through a vacuum. It requires a medium (particles) to transmit the vibrations.
  • 📊A bell jar experiment showing a ringing bell inside a vacuum where no sound is heard once the air is removed.

Human Hearing

• The approximate range of audible frequencies for a healthy human ear is 20 Hz to 20,000 Hz (or 20 kHz).

Speed of Sound

• The speed of sound in air is approximately 330 – 350 m/s.
• Determining Speed (Experimental Method):
  1. Two people stand a measured distance apart (e.g., 500 meters) in a large open field.
  2. Person A fires a starting pistol or clashes two cymbals together.
  3. Person B starts a stopwatch when they see the flash/action and stops it when they hear the sound.
  4. Speed is calculated using: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.

Characteristics of Sound

• Pitch: Controlled by frequency. Higher frequency = higher pitch (a "squeaky" sound).
• Loudness: Controlled by amplitude. Larger amplitude = louder sound.
  • 📊Two wave traces; one with short peaks (quiet) and one with tall peaks (loud), both having the same spacing between peaks.

Echoes

• An echo is simply a reflected sound wave. When sound hits a hard, flat surface, it bounces back toward the source.

---

Extended Content (Extended Curriculum Only)

Compression and Rarefaction

Longitudinal waves consist of:

• Compressions: Regions of high pressure where particles are bunched together.
• Rarefactions: Regions of low pressure where particles are spread apart.

Speed in Different Media

The speed of sound depends on how close the particles are to one another.

• Solids (Fastest): Particles are tightly packed, allowing vibrations to pass quickly.
• Liquids: Slower than solids, but faster than gases.
• Gases (Slowest): Particles are far apart, so it takes longer for vibrations to transfer.
• Order: $v_{\text{solid}} > v_{\text{liquid}} > v_{\text{gas}}$

Uses of Ultrasound

1. Non-Destructive Testing (NDT): Ultrasound is beamed into metal pipes or moving parts. If there is a crack or flaw, the sound reflects back earlier than expected.
2. Medical Scanning: Ultrasound reflects off different boundaries of soft tissue (e.g., scanning an unborn baby). It is safer than X-rays as it is non-ionizing.
3. Sonar (Sound Navigation and Ranging): Used by ships to find the depth of the sea or locate fish.
   • Calculation Note: In Sonar and Echo-location, the sound travels to the object and back — so you must halve the total distance.

Worked Example: Sonar Depth Calculation A fishing boat sends an ultrasound pulse to the seabed. The pulse returns after $0.080\text{ s}$. The speed of sound in seawater is $1400\text{ m/s}$. Calculate the depth.

1. Total distance = speed × time = $1400 \times 0.080 = 112\text{ m}$
2. Depth = total distance / 2 = $112 / 2 = 56\text{ m}$

The ÷2 step is where most marks are lost. If there is a fish between the ship and the seabed, the echo from the fish arrives sooner because the fish is closer — this is also a common exam question.

---

Key Equations

• Wave Equation: $v = f\lambda$
  • $v$ = speed (m/s)
  • $f$ = frequency (Hz)
  • $\lambda$ = wavelength (m)
• Speed Equation: $v = \frac{d}{t}$
  • $d$ = distance (m)
  • $t$ = time (s)
• Echo Distance: $d = \frac{v \times t}{2}$

---

Common Mistakes to Avoid

• ❌ Wrong: Thinking sound travels faster in air than in water or solids.
• ✅ Right: Sound requires a medium and travels fastest in solids because the particles are closer together.
• ❌ Wrong: Forgetting to divide by two in echo/sonar calculations.
• ✅ Right: Always check if the question asks for the total distance traveled (there and back) or the distance to the object (divide by 2).
• ❌ Wrong: Confusing pitch and loudness.
• ✅ Right: Remember that Amplitude = Loudness and Frequency = Pitch. Changing one does not affect the other.
• ❌ Wrong: Assuming sound can travel through a vacuum like light does.
• ✅ Right: Sound is a mechanical wave; it cannot travel through a vacuum.

---

Exam Tips

1. Check Units: Ensure frequency is in Hz (not kHz) and time is in seconds before using equations.
2. Echo Questions: Read carefully! If the question says "The sound took 2 seconds to return to the ship," you must divide that time by 2 to find the depth, or calculate the total distance and then divide by 2.
3. Drawing Waves: If asked to draw a "louder sound of a lower pitch," make the waves taller (amplitude) and the peaks further apart (frequency).

---

Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) Define the term 'ultrasound'. [1]

(b) State the approximate range of audible frequencies for a healthy young person. [1]

(c) Explain why ultrasound is used in medical imaging instead of audible sound. [3]

Worked Solution:

(a)

1. Ultrasound is sound with a frequency above the upper limit of human hearing, specifically above 20 kHz. [Definition of ultrasound]

How to earn full marks:

• State that ultrasound is sound with a frequency above 20 kHz.

(b)

1. Approximately 20 Hz to 20,000 Hz. [Audible frequency range]

How to earn full marks:

• State the correct range of frequencies, with both lower and upper limits.
• Include units (Hz).

(c)

1. Ultrasound has a much shorter wavelength than audible sound. [Relating frequency to wavelength]
2. Shorter wavelengths allow for better resolution in imaging because they diffract less. [Explaining the advantage of shorter wavelength]
3. Ultrasound is non-ionising and therefore safer than X-rays, posing less risk to the patient. [Stating safety advantage]

How to earn full marks:

• Mention shorter wavelength leading to better resolution.
• Explain that ultrasound is safer than X-rays because it is non-ionising.
• Simply stating "better image" or "safer" is not sufficient; explain why.

Common Pitfall: Many students only give one reason why ultrasound is preferred. Remember to discuss both the improved resolution due to shorter wavelengths and the safety aspect of being non-ionising. Also, don't just say "better resolution" – explain why shorter wavelengths give better resolution.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) Describe how sound waves are produced by a vibrating object. [2]

(b) State what is meant by the terms 'compression' and 'rarefaction' in the context of sound waves. [2]

(c) Explain why sound cannot travel through a vacuum. [2]

Worked Solution:

(a)

1. A vibrating object causes the air molecules (or particles of another medium) around it to vibrate. [Vibrating object and air molecules]
2. These vibrations are passed on to neighboring air molecules, creating a series of compressions (high pressure) and rarefactions (low pressure) that propagate outwards as a longitudinal wave. [Explaining propagation of sound waves]

How to earn full marks:

• Mention that the vibrating object makes the air (or other medium) vibrate.
• Describe how the vibrations are passed on, creating compressions and rarefactions as a longitudinal wave.

(b)

1. Compression: a region in a longitudinal wave where the particles are closer together than normal, resulting in higher pressure. [Definition of compression]
2. Rarefaction: a region in a longitudinal wave where the particles are further apart than normal, resulting in lower pressure. [Definition of rarefaction]

How to earn full marks:

• Define both compression and rarefaction.
• Mention high/low pressure or close/spread air molecules for each.

(c)

1. Sound waves are longitudinal waves, which means they require a medium to travel. [Sound wave type]
2. A vacuum is a space devoid of matter (no particles), so there is nothing to vibrate and transmit the energy. [Explanation requiring a medium]

How to earn full marks:

• State that sound is a longitudinal wave.
• Explain that a medium (particles) is needed to transmit the vibrations, and a vacuum has no particles.

Common Pitfall: Students often forget to mention that sound is a longitudinal wave. This is important because it directly relates to why a medium is needed – longitudinal waves are vibrations of the medium itself. Also, be specific about what a vacuum is: not just "empty space," but space without particles.

Exam-Style Question 3 — Extended Response [9 marks]

Question:

A student performs an experiment to determine the speed of sound in air. They stand a distance away from a large wall and clap their hands. They measure the time it takes for the echo to return.

(a) Describe how the student can use this method to determine the speed of sound in air. Include the measurements they need to take and the calculation they need to perform. [4]

(b) The student measures the distance to the wall to be 85 m and the time for the echo to return to be 0.50 s. Calculate the speed of sound in air. [3]

(c) Suggest one way the student could improve the accuracy of their experiment. Explain why this improvement would increase accuracy. [2]

Worked Solution:

(a)

1. The student needs to measure the distance, $d$, from themselves to the wall using a tape measure or a laser distance measurer. [Stating distance measurement]
2. The student needs to clap their hands and measure the time, $t$, it takes for the echo to return using a stopwatch or timing app. [Stating time measurement]
3. The total distance travelled by the sound is $2d$ (to the wall and back), because it's an echo. [Total distance clarification]
4. The speed of sound, $v$, can then be calculated using the formula $v = \frac{2d}{t}$. [Stating the speed equation]

How to earn full marks:

• State that the student needs to measure the distance to the wall.
• State that the student needs to measure the time for the echo to return.
• Clearly state the formula $v = \frac{2d}{t}$ or equivalent, and explain that $2d$ is the total distance.

(b)

1. $v = \frac{2d}{t} = \frac{2 \times 85 \text{ m}}{0.50 \text{ s}}$ [Substituting values into the speed equation]
2. $v = 340 \text{ m/s}$ [Calculating the speed of sound]
3. $\boxed{v = 340 \text{ m/s}}$

How to earn full marks:

• Substitute the correct values into the speed equation.
• Calculate the correct speed of sound.
• Include the correct unit (m/s).

(c)

1. The student could repeat the experiment multiple times and calculate the average time. [Suggesting repetition]
2. This would reduce the effect of random errors in timing, leading to a more reliable average value for the time and therefore a more accurate result. [Explaining the benefit of repetition]

How to earn full marks:

• Suggest repeating the experiment multiple times and averaging the results.
• Explain that this reduces random errors in timing, leading to a more accurate average.

Common Pitfall: A very common mistake is forgetting to double the distance when calculating the speed of sound, as the sound travels to the wall and back. Also, be specific about why repeating the experiment improves accuracy – it's not just about "getting a better result," but about reducing the impact of random timing errors.

Exam-Style Question 4 — Extended Response [10 marks]

Question:

A ship uses sonar to detect the depth of the ocean. The sonar emits a pulse of ultrasound and measures the time it takes for the pulse to reflect off the seabed and return to the ship.

(a) Define the term 'ultrasound'. [1]

(b) Explain why ultrasound is used in sonar rather than audible sound. [2]

(c) The speed of sound in seawater is 1500 m/s. The time taken for the ultrasound pulse to return to the ship is 0.80 s. Calculate the depth of the ocean. [3]

(d) The ship also uses radar to detect other ships. Compare and contrast the properties of ultrasound waves and radio waves, including how they travel and their typical speeds. [4]

Worked Solution:

(a)

1. Ultrasound is sound with a frequency above 20 kHz, which is the upper limit of human hearing. [Definition of ultrasound]

How to earn full marks:

• State that ultrasound is sound with a frequency above 20 kHz.

(b)

1. Ultrasound has a shorter wavelength than audible sound. [Relating frequency to wavelength]
2. Shorter wavelengths diffract less, allowing for more precise detection of objects and the seabed because the signal is more directional. [Explaining the advantage of shorter wavelength and reduced diffraction]

How to earn full marks:

• Mention shorter wavelength.
• Explain that this leads to less diffraction and a more directional signal, resulting in more precise detection.

(c)

1. The total distance travelled by the ultrasound pulse is $2d$, where $d$ is the depth of the ocean. [Total distance clarification]
2. $v = \frac{2d}{t}$, therefore $d = \frac{vt}{2} = \frac{1500 \text{ m/s} \times 0.80 \text{ s}}{2}$ [Rearranging the speed equation and substituting values]
3. $d = 600 \text{ m}$ [Calculating the depth]
4. $\boxed{d = 600 \text{ m}}$

How to earn full marks:

• State the relationship $d = \frac{vt}{2}$ or explain that the distance is half of $vt$.
• Substitute the correct values into the equation.
• Calculate the correct depth.
• Include the correct unit (m).

(d)

1. Ultrasound is a longitudinal wave; radio waves are transverse waves. [Wave type comparison]
2. Ultrasound requires a medium to travel (e.g., water); radio waves can travel through a vacuum because they are electromagnetic waves. [Medium comparison]
3. The speed of ultrasound in water is approximately 1500 m/s; the speed of radio waves is $3.0 \times 10^8$ m/s (the speed of light). Radio waves are significantly faster. [Speed comparison]
4. Both can be reflected, but ultrasound's reflection is more affected by the density of the medium, while radio wave reflection depends on the electrical conductivity of the reflecting surface. [Reflection and diffraction comparison]

How to earn full marks:

• State that ultrasound is longitudinal and radio waves are transverse.
• State that ultrasound needs a medium, but radio waves do not, and explain why radio waves don't need a medium.
• Compare the speeds of the two waves, giving approximate values and stating that radio waves are much faster.
• Mention that both can be reflected, and briefly describe the different factors affecting their reflection.

Common Pitfall: When calculating the depth using sonar, remember to divide by two, as the time given is for the sound to travel down and back. Also, when comparing ultrasound and radio waves, many students forget to mention the type of wave (longitudinal vs. transverse) and the vast difference in their speeds. Don't just state the speeds; emphasize how much faster radio waves are.