General properties of waves

1. Overview

Waves are oscillations that transfer energy from one place to another without transferring matter. Understanding wave properties is essential because it explains how we see (light waves), how we hear (sound waves), and how modern communication technology functions.

Key Definitions

• Amplitude: The maximum displacement of a point on a wave from its undisturbed (rest) position.
• Wavelength ($\lambda$): The distance between a point on one wave and the equivalent point on the next adjacent wave (e.g., crest to crest).
• Frequency ($f$): The number of waves passing a point per second (measured in Hertz, Hz).
• Period ($T$): The time taken for one complete wave to pass a point.
• Wavefront: A line representing all the points on a wave that are in the same phase (e.g., a line connecting all the crests).
• Crest (Peak): The highest point of a transverse wave.
• Trough: The lowest point of a transverse wave.
• Wave Speed ($v$): The speed at which energy is transferred through a medium.

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Core Content

Wave Motion and Energy Transfer

Waves transfer energy without transferring matter.

• In a rope: If you shake one end, the energy travels to the other end, but the rope fibers only move up and down.
• In water: A buoy or a leaf on the surface will bob up and down as a wave passes, but it will not move forward with the wave.

Transverse vs. Longitudinal Waves

1. Transverse Waves:

   • Vibrations are at right angles (90°) to the direction of energy travel.
   • Examples: Light (all electromagnetic radiation), water waves, and seismic S-waves.
   • 📊A wave showing peaks and troughs with an arrow pointing right for 'Direction of Travel' and a vertical double-headed arrow for 'Direction of Vibration'

2. Longitudinal Waves:

   • Vibrations are parallel to the direction of energy travel.
   • Consist of compressions (points bunched together) and rarefactions (points stretched apart).
   • Examples: Sound waves and seismic P-waves.
   • 📊A spring/slinky showing compressed coils and stretched coils

Describing Waves

• Wave Speed ($v$): Calculated by multiplying frequency by wavelength.

Worked Example 1: Finding wave speed A wave has a frequency of 10 Hz and a wavelength of 2 m. Calculate the speed.

1. $v = f \times \lambda = 10 \times 2 = 20\text{ m/s}$

Worked Example 2: Finding wavelength (most common exam rearrangement) A sound wave in water has a speed of $1400\text{ m/s}$ and a frequency of $7{,}000\text{ Hz}$. Calculate the wavelength.

1. Rearrange: $\lambda = v / f$
2. $\lambda = 1400 / 7{,}000 = 0.20\text{ m}$

The exam almost always gives you two values and asks for the third. Practise all three rearrangements: $v = f\lambda$, $\lambda = v/f$, $f = v/\lambda$.

Wave Behaviors & The Ripple Tank

A ripple tank uses a vibrating bar to create water waves.

1. Reflection: Waves hit a flat (plane) barrier and bounce off. The angle of incidence equals the angle of reflection.
2. Refraction: Waves change speed and direction when crossing a boundary. In a ripple tank, this is done by changing the depth.
   • Deep water = Faster waves = Longer wavelength.
   • Shallow water = Slower waves = Shorter wavelength.
   • Note: The frequency never changes during refraction.
3. Diffraction: The spreading out of waves as they pass through a gap or around an edge.

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Extended Content (Extended Only)

Factors Affecting Diffraction

Diffraction is most noticeable when the size of the gap is similar to the wavelength of the wave. This is a key exam concept.

1. Gap Size:

   • If the gap is much wider than the wavelength, the waves pass through with very little spreading (mostly just at the edges).
   • If the gap is narrow (close to the size of the wavelength), the waves spread out in wide semi-circles.
   • 📊Two panels; one shows waves passing through a wide gap with slight curving at edges; the second shows waves passing through a narrow gap becoming circular

2. Wavelength:

   • Longer wavelengths diffract (spread) more than shorter wavelengths when passing through the same gap or around an edge.
   • This explains why you can hear someone talking around a corner (long sound waves diffract) but you cannot see them (short light waves do not diffract significantly).

Why can you hear bass through a doorway but not treble? Imagine someone playing a violin in the next room with the door open. You can hear the deep, low notes clearly, but the high-pitched notes seem to disappear when you move to the side of the doorway. This is diffraction in action.

Low-frequency sounds have long wavelengths (up to several metres), which are comparable to the width of a typical doorway (~1 m). When the wavelength is close to the gap size, the wave spreads out widely after passing through — it bends around the edges and fills the room beyond. High-frequency sounds have much shorter wavelengths (a few centimetres), so the doorway is enormous compared to the wave. These waves pass straight through without much spreading, like shining a torch through a wide opening.

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Key Equations

Equation	Symbols	Units
$v = f \lambda$	$v$ = speed, $f$ = frequency, $\lambda$ = wavelength	$v$ (m/s), $f$ (Hz), $\lambda$ (m)
$f = 1 / T$	$f$ = frequency, $T$ = period	$f$ (Hz), $T$ (s)

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Common Mistakes to Avoid

• ❌ Wrong: Measuring amplitude from the trough to the crest.
• ✅ Right: Amplitude is measured from the center (equilibrium) line to the crest. If the total vertical height is 6 cm, the amplitude is 3 cm.
• ❌ Wrong: Thinking the frequency changes when waves enter shallow water or a new medium.
• ✅ Right: Frequency stays constant. Only speed and wavelength change during refraction.
• ❌ Wrong: Assuming a longer wave must be taller.
• ✅ Right: Wavelength (length) and Amplitude (height) are independent properties. A long wave can have a small amplitude.
• ❌ Wrong: Calculating wave speed from a displacement-distance graph alone.
• ✅ Right: You can find wavelength from a distance graph, but you need frequency or a displacement-time graph to find the speed.

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Exam Tips

1. Check your units: Ensure wavelength is in meters (m) and not centimeters (cm) before using the $v = f \lambda$ equation.
2. The "Frequency Constant" Rule: In any question about refraction (waves changing speed), always remember that the frequency is determined by the source and does not change.
3. Diffraction Drawing: When drawing diffracted waves, ensure the wavelength stays the same before and after the gap. Only the shape changes (it becomes circular), not the distance between the lines.

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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 [6 marks]

Question:

A student sets up a ripple tank to investigate water waves.

(a) State what is meant by the wavelength of a wave. [1]

(b) The student generates waves with a frequency of 2.0 Hz. The distance between five successive crests is measured to be 12 cm.

(i) Calculate the wavelength of the waves. [2]

(ii) Calculate the speed of the water waves. [3]

Worked Solution:

(a)

1. The wavelength is the distance between two successive crests or troughs. $\boxed{\text{The distance between two successive identical points on a wave.}}$ Definition of wavelength.

How to earn full marks:

• State "distance between two successive points" – 1 mark

(b) (i)

1. There are 4 wavelengths between 5 successive crests. $4 \lambda = 12 \text{ cm}$ Recognize that there are 4 wavelengths

2. Divide the total distance by the number of wavelengths. $\lambda = \frac{12 \text{ cm}}{4} = 3 \text{ cm}$ $\boxed{\lambda = 3.0 \text{ cm}}$ Correctly calculates the wavelength

How to earn full marks:

• Recognize 4 wavelengths - 1 mark
• Correct final answer with unit - 1 mark

(ii)

1. Recall the wave equation $v = f\lambda$ State the wave equation

2. Convert wavelength to meters and substitute values into the wave equation. $v = 2.0 \text{ Hz} \times 0.03 \text{ m} = 0.06 \text{ m/s}$ Correctly substitutes frequency and wavelength into the wave equation.

3. State the final answer with correct unit. $\boxed{v = 0.06 \text{ m/s}}$ Correct final answer with unit

How to earn full marks:

• Recall the wave equation $v = f\lambda$ - 1 mark
• Correct substitution with wavelength in meters - 1 mark
• Correct final answer with unit - 1 mark

Common Pitfall: Many students forget to convert the wavelength to meters before using the wave equation. Always double-check your units and make sure they are consistent (SI units) before performing calculations. Also, remember that the distance between n crests is n-1 wavelengths.

Exam-Style Question 2 — Short Answer [5 marks]

Question:

A loudspeaker emits sound waves.

(a) State whether sound waves are transverse or longitudinal. [1]

(b) Describe the difference between transverse and longitudinal waves, in terms of the direction of vibration and the direction of propagation. [4]

Worked Solution:

(a)

1. State the type of wave. $\boxed{\text{Longitudinal}}$ Correctly identifies the type of wave

How to earn full marks:

• State "Longitudinal" – 1 mark

(b)

1. Describe the direction of vibration and propagation in transverse waves. In transverse waves, the direction of vibration is perpendicular to the direction of propagation.

2. Describe the direction of vibration and propagation in longitudinal waves. In longitudinal waves, the direction of vibration is parallel to the direction of propagation.

How to earn full marks:

• Transverse waves: Vibration is perpendicular to propagation - 2 marks
• Longitudinal waves: Vibration is parallel to propagation - 2 marks

Common Pitfall: It's easy to mix up the definitions of transverse and longitudinal waves. A helpful way to remember is that "longitudinal" sounds like "parallel," which is the relationship between vibration and propagation in longitudinal waves.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A student investigates the diffraction of water waves through a gap using a ripple tank.

(a) Describe what is meant by diffraction. [2]

(b)

Draw a diagram to show the pattern of the water waves (i) before they reach the gap and (ii) after they pass through the gap. [3]

(c) The student observes that when the gap is made much wider than the wavelength of the water waves, the amount of diffraction is reduced. Explain why this happens. [3]

Worked Solution:

(a)

1. Diffraction is the spreading of waves. $\boxed{\text{Diffraction is the spreading of waves when they pass through a gap or around an obstacle.}}$ Clear and complete definition of diffraction.

How to earn full marks:

• State "spreading of waves" – 1 mark
• State "passing through a gap or around an obstacle" – 1 mark

(b)

1. 📊(i) Before the gap: straight, parallel wavefronts approaching the barrier. Draw at least 3 wavefronts. The wavefronts should be continuous and evenly spaced. (ii) After the gap: circular wavefronts spreading out from the gap. The wavefronts should be continuous and approximately circular. The wavelength should be the same before and after the gap.

How to earn full marks:

• (i) Straight, parallel wavefronts before the gap – 1 mark
• (ii) Circular wavefronts after the gap – 1 mark
• Same wavelength before and after the gap – 1 mark

(c)

1. Relate diffraction to the size of the gap and the wavelength. When the gap is much wider than the wavelength, the waves pass through the gap with little or no bending.

2. Explain the effect on the wave pattern. The waves continue to travel in approximately straight lines, with minimal spreading at the edges of the gap.

3. Explain how the gap size affects the amount of diffraction. Significant diffraction only occurs when the gap size is approximately equal to or smaller than the wavelength of the wave.

How to earn full marks:

• Gap much wider than wavelength implies little bending - 1 mark
• Waves travel in straight lines with minimal spreading - 1 mark
• Diffraction best when gap is about the same size as wavelength - 1 mark

Common Pitfall: Students often forget that the wavelength of the wave remains constant during diffraction. The only thing that changes is the direction of the wave's propagation. Also, remember that significant diffraction only occurs when the gap size is comparable to the wavelength.

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

Question:

A transverse wave travels along a rope. Figure 1 shows the displacement of the rope at one instant in time.

(a) State what is meant by a transverse wave. [2]

(b) Use Figure 1 to determine:

(i) the wavelength of the wave. [2]

(ii) the amplitude of the wave. [1]

(c) The frequency of the wave is 2.5 Hz.

(i) Calculate the speed of the wave. [2]

(d) The wave now passes into a different section of rope where the wave speed is higher. State and explain how this change in speed affects the wavelength of the wave, assuming the frequency remains constant. [3]

Worked Solution:

(a)

1. Direction of vibration and propagation. $\boxed{\text{A transverse wave is a wave in which the direction of vibration is perpendicular to the direction of propagation.}}$ Clear definition relating vibration and propagation

How to earn full marks:

• State "vibration is perpendicular to propagation" – 2 marks

(b) (i)

1. Measure the distance between two successive crests or troughs. $\lambda = 3.0 \text{ m} - 1.0 \text{ m} = 2.0 \text{ m}$ $\boxed{\lambda = 2.0 \text{ m}}$ Correctly determines the wavelength from the graph.

How to earn full marks:

• Correctly reading values from the graph (3.0 m and 1.0 m) - 1 mark
• Correct final answer with unit - 1 mark

(ii)

1. Measure the maximum displacement from the equilibrium position. $\boxed{\text{Amplitude} = 2.0 \text{ cm}}$ Correctly determines the amplitude from the graph.

How to earn full marks:

• Correct final answer with unit - 1 mark

(c) (i)

1. Recall the wave equation. $v = f\lambda$ State the wave equation

2. Substitute the frequency and wavelength into the wave equation. $v = 2.5 \text{ Hz} \times 2.0 \text{ m} = 5.0 \text{ m/s}$ $\boxed{v = 5.0 \text{ m/s}}$ Correctly substitutes frequency and wavelength into the wave equation and gets the correct answer.

How to earn full marks:

• Recall the wave equation $v = f\lambda$ - 1 mark
• Correct substitution and final answer with unit - 1 mark

(d)

1. State the effect on wavelength. The wavelength will increase.

2. Relate the change in wavelength to the change in wave speed. Since the wave speed increases, and the frequency remains constant, the wavelength must also increase.

3. Explain the relationship using the wave equation. From the equation $v = f\lambda$, if $v$ increases and $f$ is constant, then $\lambda$ must increase to maintain the equality.

How to earn full marks:

• State "wavelength increases" - 1 mark
• Wave speed increases implies wavelength increases (frequency constant) - 1 mark
• Reference the wave equation $v = f\lambda$ to support the explanation - 1 mark

Common Pitfall: When reading amplitude from a graph, many students use the total vertical distance (from crest to trough) instead of measuring from the equilibrium position. Also, remember that when a wave enters a new medium, its speed and wavelength can change, but its frequency remains constant.