Cambridge AS & A Level9702

Waves

Physics 9702 Chapter Notes

What this chapter covers

Waves - Progressive wavesWaves - Transverse and longitudinal wavesWaves - Doppler effect for sound wavesWaves - Electromagnetic spectrumWaves - Polarisation
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1. Wave Fundamentals and Key Terms

A progressive wave is a disturbance that transfers energy from one place to another without any net transfer of matter. We can represent waves using graphs. A displacement-distance graph shows the displacement of all particles along the wave at a single instant in time. A displacement-time graph shows the oscillation of a single particle over a period of time. From these, we can define key properties. Amplitude (A) is the maximum displacement from the equilibrium position. Wavelength (λ) is the distance between two consecutive points in phase, like two crests. Period (T) is the time for one complete oscillation of a particle. Frequency(f) is the number of complete oscillations per second, measured in Hertz (Hz). Frequency and period are reciprocals: f = 1/T.

f = 1/T

Key term

Amplitude (A): The maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position.

Common pitfall

Confusing a displacement-distance graph with a displacement-time graph. Always check the label on the x-axis: 'distance' gives you wavelength, while 'time' gives you the period.

Worked example 12 marks

The diagram shows a snapshot of a wave on a rope. Determine(a) the amplitude and(b) the wavelength of the wave.

  1. 1

    a) Amplitude is the maximum displacement from the equilibrium (zero) position. From the graph, the peak is at 0.5 m.

  2. 2

    Therefore, Amplitude A = 0.5 m.

  3. 3

    b) Wavelength is the length of one full cycle. From the graph, one complete wave cycle finishes at the 4 m mark.

  4. 4

    Therefore, Wavelength λ = 4.0 m.

Worked example 22 marks

A single point on a wave takes 0.02 seconds to complete one full oscillation. Calculate the frequency of the wave.

  1. 1

    The time for one full oscillation is the period, T. So, T = 0.02 s.

  2. 2

    The relationship between frequency(f) and period (T) is f = 1/T.

  3. 3

    f = 1 / 0.02 s

  4. 4

    f = 50 Hz

Recap

  • Waves transfer energy, not matter.
  • Amplitude is the maximum displacement from the equilibrium position.
  • Wavelength is the length of one complete wave.
  • Period is the time for one complete oscillation.
  • Frequency is the number of oscillations per second (f = 1/T).

Quick check

  1. Define the term 'period' of a wave.1 mark
  2. A wave has a frequency of 200 Hz. What is its period?1 mark

2. Transverse and Longitudinal Waves

Waves can be classified based on the direction of oscillation relative to the direction of energy transfer. In a transverse wave, the particles of the medium oscillate perpendicular (at 90°) to the direction of energy transfer. Examples include all electromagnetic waves (like light), waves on a guitar string, and ripples on water. In a longitudinal wave, the particles of the medium oscillate parallel to the direction of energy transfer. These waves travel as a series of compressions (regions of high pressure) and rarefactions (regions of low pressure). The most common example is a sound wave.

Key term

Longitudinal Wave: A wave in which the oscillations of the medium's particles are parallel to the direction of energy transfer.

Examiner insight

Examiners often ask for examples of each wave type; be sure to know that all electromagnetic waves are transverse and sound is longitudinal.

Worked example 12 marks

For a sound wave travelling through air, describe the motion of an air particle relative to the direction of the wave's travel. What type of wave is this?

  1. 1

    A sound wave causes air particles to oscillate back and forth.

  2. 2

    This oscillation is parallel to the direction in which the sound energy is travelling.

  3. 3

    Because the oscillations are parallel to the direction of energy transfer, a sound wave is a longitudinal wave.

Worked example 22 marks

Give two examples of transverse waves and one example of a longitudinal wave.

  1. 1

    Transverse wave examples: Light (or any EM wave), a wave on a string.

  2. 2

    Longitudinal wave example: Sound.

Recap

  • In transverse waves, oscillations are perpendicular to energy transfer.
  • In longitudinal waves, oscillations are parallel to energy transfer.
  • All electromagnetic waves are transverse.
  • Sound waves are longitudinal.
  • Longitudinal waves consist of compressions and rarefactions.

Quick check

  1. Is a wave on a rope transverse or longitudinal?1 mark
  2. What is the name for a region of low pressure in a sound wave?1 mark

3. The Wave Equation and Oscilloscopes

The speed of a wave(v) is related to its frequency(f) and wavelength (λ) by the wave equation: v = fλ. This is one of the most fundamental equations in wave physics. It tells us that for a given speed, a higher frequency means a shorter wavelength, and vice versa. We can measure wave properties using a Cathode-Ray Oscilloscope (CRO). A CRO displays a voltage-time graph. The 'Y-gain' or 'volts/div' setting controls the vertical scale, allowing us to determine the amplitude (peak voltage). The 'time-base' or 's/div' setting controls the horizontal scale, allowing us to measure the period (T) of the wave. From the period, we can calculate the frequency using f = 1/T.

v = fλ

f = 1/T

Key term

Wave Speed (v): The speed at which energy is transferred by the wave, calculated by multiplying its frequency and wavelength.

Common pitfall

Forgetting to convert units, such as milliseconds (ms) to seconds (s) or megahertz (MHz) to hertz (Hz), before using them in equations.

Worked example 13 marks

A radio station broadcasts at a frequency of 98.4 MHz. Radio waves travel at 3.0 x 10⁸ m/s. Calculate the wavelength of these radio waves.

  1. 1

    First, convert the frequency to Hz: f = 98.4 MHz = 98.4 x 10⁶ Hz.

  2. 2

    State the wave equation: v = fλ.

  3. 3

    Rearrange for wavelength: λ = v / f.

  4. 4

    Substitute the values: λ = (3.0 x 10⁸ m/s) / (98.4 x 10⁶ Hz).

  5. 5

    λ = 3.05 m (to 3 significant figures).

Worked example 24 marks

The trace on a CRO screen shows a wave. The time-base is set to 5 ms/div and the Y-gain is 2 V/div. The wave occupies 4 divisions horizontally for one cycle and has a peak 1.5 divisions above the centre. Find(a) the frequency and(b) the amplitude of the signal.

  1. 1

    a) First find the period, T. T = (number of horizontal divisions for one cycle) × (time-base setting).

  2. 2

    T = 4 div × 5 ms/div = 20 ms = 0.020 s.

  3. 3

    Now find the frequency: f = 1/T = 1 / 0.020 s = 50 Hz.

  4. 4

    b) Find the amplitude (peak voltage). Amplitude = (peak vertical divisions) × (Y-gain setting).

  5. 5

    Amplitude = 1.5 div × 2 V/div = 3.0 V.

Recap

  • The wave equation is v = fλ.
  • Wave speed is constant for a given medium.
  • A CRO displays a voltage-time graph of a signal.
  • The CRO time-base is used to find the period and frequency.
  • The CRO Y-gain is used to find the amplitude (peak voltage).

Quick check

  1. A wave has a speed of 10 m/s and a wavelength of 0.5 m. What is its frequency?2 marks

4. Wave Intensity and Amplitude

The intensity of a wave is a measure of the energy it carries per unit time, per unit area. Formally, intensity (I) is defined as the power (P) transmitted per unit area (A) perpendicular to the wave's velocity. This gives the equation: Intensity = Power / Area. The unit of intensity is watts per square metre (W m⁻²). For any wave, the intensity is directly proportional to the square of its amplitude (A). This is written as I ∝ A². This is a crucial relationship. It means that if you double the amplitude of a wave, you multiply its intensity by 2², which is 4. If you triple the amplitude, the intensity increases by a factor of 3², which is 9. This is why a slightly louder sound (small increase in pressure amplitude) can carry significantly more energy.

Intensity = Power / Area

I ∝ A²

Key term

Intensity (I): The power transferred per unit area by a wave, measured in watts per square metre (W m⁻²).

Examiner insight

Students who can explain the consequence of the I ∝ A² relationship (e.g., 'doubling amplitude means quadrupling intensity') score higher than those who just state the formula.

Worked example 13 marks

A laser beam has a power of 5.0 mW and a cross-sectional area of 1.2 x 10⁻⁶ m². Calculate the intensity of the laser beam.

  1. 1

    First, convert power to watts: P = 5.0 mW = 5.0 x 10⁻³ W.

  2. 2

    State the formula for intensity: I = P / A.

  3. 3

    Substitute the values: I = (5.0 x 10⁻³ W) / (1.2 x 10⁻⁶ m²).

  4. 4

    I = 4167 W m⁻².

  5. 5

    To 2 significant figures, I = 4200 W m⁻².

Worked example 23 marks

Wave X has an amplitude of 12 cm and an intensity I. Wave Y has the same frequency but an amplitude of 4 cm. What is the intensity of Wave Y in terms of I?

  1. 1

    The relationship between intensity and amplitude is I ∝ A².

  2. 2

    We can write this as a ratio: I_Y / I_X = (A_Y / A_X)².

  3. 3

    The ratio of the amplitudes is A_Y / A_X = 4 cm / 12 cm = 1/3.

  4. 4

    Substitute this into the intensity ratio: I_Y / I = (1/3)² = 1/9.

  5. 5

    Therefore, the intensity of Wave Y is I/9.

Recap

  • Intensity is power per unit area (I = P/A).
  • The unit of intensity is W m⁻².
  • Intensity is proportional to the square of the amplitude (I ∝ A²).
  • Doubling a wave's amplitude quadruples its intensity.

Quick check

  1. If a wave's amplitude is halved, what happens to its intensity?1 mark

5. The Doppler Effect for Sound

The Doppler effect is the change in the observed frequency of a wave when there is relative motion between the source of the wave and the observer. For sound, this results in a change in pitch. As a sound source (like an ambulance siren) moves towards you, the wave crests arrive more frequently, so you hear a higher frequency (higher pitch). As it moves away, the wave crests are more spread out and arrive less frequently, so you hear a lower frequency (lower pitch). The observed frequency (f_o) can be calculated using the formula f_o = f_s * v / (v ± v_s), where f_s is the source frequency, v is the speed of sound, and v_s is the speed of the source. Use the minus sign (-) in the denominator when the source is moving towards the observer, and the plus sign (+) when it is moving away.

f_o = f_s * v / (v ± v_s)

Key term

Doppler Effect: The apparent change in the frequency of a wave in relation to an observer who is moving relative to the wave source.

Common pitfall

Mixing up the plus and minus signs in the Doppler equation. Remember: moving 'towards' leads to a higher frequency, which requires a smaller denominator (so use the minus sign).

Fun fact

The 'redshift' of light from distant galaxies, a key piece of evidence for the expansion of the universe, is an example of the Doppler effect applied to light waves.

Worked example 13 marks

An ambulance siren emits a sound of frequency 550 Hz. The ambulance is travelling towards a stationary observer at 30 m/s. The speed of sound in air is 340 m/s. Calculate the frequency heard by the observer.

  1. 1

    Identify the variables: f_s = 550 Hz, v_s = 30 m/s, v = 340 m/s.

  2. 2

    The source is moving towards the observer, so we use the minus sign in the denominator.

  3. 3

    Formula: f_o = f_s * v / (v - v_s).

  4. 4

    Substitute values: f_o = 550 * 340 / (340 - 30).

  5. 5

    f_o = 550 * 340 / 310.

  6. 6

    f_o = 603.2 Hz. The observed frequency is approximately 603 Hz.

Worked example 22 marks

For the same ambulance in the previous question, what frequency does the observer hear after it has passed them and is moving away at the same speed?

  1. 1

    The variables are the same, but now the source is moving away.

  2. 2

    We must use the plus sign in the denominator.

  3. 3

    Formula: f_o = f_s * v / (v + v_s).

  4. 4

    Substitute values: f_o = 550 * 340 / (340 + 30).

  5. 5

    f_o = 550 * 340 / 370.

  6. 6

    f_o = 505.4 Hz. The observed frequency is approximately 505 Hz.

Recap

  • The Doppler effect is the observed frequency change due to relative motion.
  • Observed frequency is higher for an approaching source.
  • Observed frequency is lower for a receding source.
  • Use f_o = f_s * v / (v - v_s) for an approaching source.
  • Use f_o = f_s * v / (v + v_s) for a receding source.

Quick check

  1. As a train blowing its horn passes you, does the pitch you hear increase or decrease?1 mark

6. The Electromagnetic Spectrum

Electromagnetic (EM) waves are transverse waves made of oscillating electric and magnetic fields. They are unique because they do not require a medium and can travel through a vacuum. In a vacuum, all EM waves travel at the same speed, the speed of light, c = 3.00 x 10⁸ m/s. The EM spectrum is the continuous range of all EM waves, organised by frequency and wavelength. The wave equation c = fλ applies to all of them. In order of increasing wavelength (and decreasing frequency), the spectrum is: Gamma rays, X-rays, Ultraviolet (UV), Visible light, Infrared (IR), Microwaves, Radio waves. The small part of this spectrum that our eyes can detect is visible light, with wavelengths from approximately 400 nm (violet) to 700 nm (red).

c = fλ

Key term

Electromagnetic Spectrum: The range of all types of electromagnetic radiation, ordered by frequency or wavelength, from gamma rays to radio waves.

Examiner insight

Marks are often awarded for correctly recalling the order of the EM spectrum. A mnemonic like 'Rich Men In Vegas Use X-pensive Gadgets' (Radio, Microwave, Infrared, Visible, UV, X-ray, Gamma) is very helpful.

Fun fact

Your body emits infrared radiation, which is how thermal imaging cameras can 'see' you in the dark.

Worked example 13 marks

A Wi-Fi router operates using microwaves with a frequency of 2.4 GHz. Calculate the wavelength of these microwaves. (c = 3.00 x 10⁸ m/s)

  1. 1

    First, convert the frequency to Hz: f = 2.4 GHz = 2.4 x 10⁹ Hz.

  2. 2

    State the wave equation for EM waves: c = fλ.

  3. 3

    Rearrange for wavelength: λ = c / f.

  4. 4

    Substitute the values: λ = (3.00 x 10⁸ m/s) / (2.4 x 10⁹ Hz).

  5. 5

    λ = 0.125 m (or 12.5 cm).

Recap

  • EM waves are transverse and can travel in a vacuum.
  • All EM waves travel at the speed of light (c) in a vacuum.
  • The EM spectrum ranges from gamma rays (shortest λ) to radio waves (longest λ).
  • Visible light has wavelengths between approximately 400 nm and 700 nm.
  • The equation c = fλ relates speed, frequency, and wavelength for all EM waves.

Quick check

  1. Which has a higher frequency: X-rays or Microwaves?1 mark
  2. What is the approximate speed of ultraviolet light in a vacuum?1 mark

7. Polarisation of Transverse Waves

In a normal, unpolarised transverse wave (like light from the sun or a bulb), the oscillations occur in all possible planes perpendicular to the direction of energy transfer. Polarisation is the process of restricting these oscillations to a single plane. A polarising filter (or 'polariser') achieves this by acting like a set of microscopic parallel slits, only allowing oscillations aligned with its 'transmission axis' to pass through. The resulting wave is plane-polarised. A key fact is that only transverse waves can be polarised. Longitudinal waves, like sound, cannot be polarised because their oscillations are already fixed in one direction: parallel to the direction of energy transfer. This property is considered definitive proof that light is a transverse wave. Polarised sunglasses work by blocking horizontally-polarised light, which is the primary component of glare reflected from surfaces like water or roads.

Key term

Polarisation: A phenomenon in which the oscillations of a transverse wave are restricted to a single plane.

Examiner insight

A classic question is 'Explain why light can be polarised but sound cannot'. The key is to state that light is a transverse wave while sound is a longitudinal wave, and then define both wave types in your explanation.

Worked example 14 marks

Explain how two polarising filters can be used to block almost all light from a lamp.

  1. 1

    Place the first polarising filter in front of the lamp. The unpolarised light from the lamp passes through it and becomes vertically (or horizontally, depending on orientation) plane-polarised.

  2. 2

    The intensity of the light is reduced to 50% of the original.

  3. 3

    Place the second polarising filter (the analyser) after the first one.

  4. 4

    Rotate the second filter so its transmission axis is at 90° to the axis of the first filter.

  5. 5

    The vertically polarised light from the first filter cannot pass through the second filter, which now only allows horizontal oscillations.

  6. 6

    As a result, almost no light is transmitted through the second filter.

Recap

  • Polarisation restricts wave oscillations to a single plane.
  • Only transverse waves can be polarised.
  • Longitudinal waves (like sound) cannot be polarised.
  • A polarising filter produces plane-polarised light from an unpolarised source.
  • Two polarisers with axes at 90° to each other are 'crossed' and block light.

Quick check

  1. Explain briefly why sound waves cannot be polarised.2 marks

8. Analysing Polarised Light with Malus's Law

When unpolarised light passes through a polarising filter, its intensity is halved (I = I₀/2). If this now plane-polarised light then passes through a second filter (called an 'analyser'), the transmitted intensity depends on the angle between the two filters. Malus's Law describes this relationship: I = I_max cos²θ. Here, I_max is the intensity of the light incident on the analyser, I is the intensity transmitted through the analyser, and θ is the angle between the transmission axis of the polariser and the analyser. If the analyser is aligned with the polariser (θ=0°), cos²(0°)=1, and all the light gets through (I = I_max). If the analyser is 'crossed' with the polariser (θ=90°), cos²(90°)=0, and no light gets through (I = 0).

I = I_max cos²θ

Key term

Malus's Law: A law stating that the intensity of plane-polarised light that passes through an analyser is proportional to the square of the cosine of the angle between the light's polarisation plane and the analyser's transmission axis.

Common pitfall

Forgetting to square the cosine term in Malus's Law. It's cos²θ, not cos(2θ) or just cos(θ). Also, remember that your calculator must be in degrees mode.

Worked example 13 marks

Plane-polarised light of intensity 20 W m⁻² is incident on a polarising filter. The transmission axis of the filter is at an angle of 60° to the plane of polarisation of the light. Calculate the intensity of the transmitted light.

  1. 1

    State Malus's Law: I = I_max cos²θ.

  2. 2

    Identify the given values: I_max = 20 W m⁻², θ = 60°.

  3. 3

    Substitute the values into the equation: I = 20 × (cos 60°)².

  4. 4

    Calculate cos 60° = 0.5.

  5. 5

    I = 20 × (0.5)² = 20 × 0.25.

  6. 6

    I = 5.0 W m⁻².

Worked example 24 marks

Unpolarised light of intensity I₀ is passed through two polarising filters. The second filter is oriented at 45° to the first. What is the final intensity of the light in terms of I₀?

  1. 1

    Step 1: The unpolarised light passes through the first filter. The intensity is halved.

  2. 2

    Intensity after first filter, I₁ = I₀ / 2.

  3. 3

    Step 2: This polarised light (I₁) now acts as the incident light for the second filter. Here, I_max = I₁ = I₀ / 2 and θ = 45°.

  4. 4

    Using Malus's Law: I₂ = I_max cos²θ = (I₀ / 2) × (cos 45°)².

  5. 5

    cos 45° = 1/√2 or ≈0.707. So (cos 45°)² = 1/2 = 0.5.

  6. 6

    I₂ = (I₀ / 2) × (1/2) = I₀ / 4.

  7. 7

    The final intensity is one quarter of the original intensity.

Recap

  • Malus's Law is I = I_max cos²θ.
  • I_max is the intensity of polarised light incident on the second filter (analyser).
  • θ is the angle between the polarisation plane and the analyser's axis.
  • Maximum transmission occurs at θ = 0°.
  • Zero transmission occurs at θ = 90°.
  • Remember that the first polariser halves the intensity of unpolarised light.

Quick check

  1. What is the value of cos²(90°)?1 mark
  2. Polarised light passes through an analyser with its axis at 30° to the plane of polarisation. What fraction of the intensity is transmitted?2 marks

End-of-chapter exercise

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

  1. A sound wave has a frequency of 440 Hz and travels through air at 330 m/s. Calculate its wavelength.2 marks
  2. Describe the difference between a transverse wave and a longitudinal wave, giving one example of each.4 marks
  3. A signal is displayed on a CRO. One complete cycle occupies 5 horizontal divisions and the peak is 2.5 vertical divisions from the centre line. The time-base is 100 μs/div and the Y-gain is 50 mV/div. Calculate the frequency and the peak voltage of the signal.4 marks
  4. A car horn sounds with a frequency of 400 Hz. The car travels towards a stationary observer at 25 m/s. Calculate the frequency heard by the observer. (Speed of sound = 340 m/s)3 marks
  5. Wave P has an amplitude of 3A and wave Q has an amplitude of A. Both waves have the same frequency. What is the ratio of the intensity of wave P to the intensity of wave Q?3 marks
  6. State the seven main regions of the electromagnetic spectrum in order of increasing wavelength. For which region is the wavelength approximately 5 x 10⁻⁷ m?3 marks
  7. Explain why light is described as a transverse wave, using the phenomenon of polarisation in your answer.3 marks
  8. Unpolarised light with an initial intensity of 80 W m⁻² passes through a polarising filter. The transmitted light then passes through a second polarising filter, whose transmission axis is at an angle of 30° to the first. Calculate the final intensity of the light.4 marks
  9. A speedboat's engine produces a sound with a frequency of 150 Hz. The boat moves directly away from a stationary observer on a jetty at a speed of 20 m/s. It then turns around and moves directly towards the observer at the same speed. Calculate the highest and lowest frequencies heard by the observer. (Speed of sound = 340 m/s)5 marks
  10. A loudspeaker radiates 10 W of power uniformly in all directions. Calculate the intensity of the sound at a distance of 5.0 m from the speaker. (Hint: The area of a sphere is 4πr²).3 marks

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