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As Physics Topic 7: Waves

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AS Level CIE Physics: Waves

Note: An understanding of colour from Cambridge IGCSE/O Level Physics or equivalent is assumed.

7.1 Progressive Waves

1. Description of Wave Motion

Wave motion is the propagation of a disturbance (energy) from one point to another through a medium or a vacuum, without the net transfer of the medium itself.

  • Ropes & Springs: A flick on a rope creates a pulse. The rope particles move up and down (transverse to the direction of energy travel), transferring the pulse along its length.
  • Ripple Tanks: Dropping a dipper into water creates circular ripples. These are transverse waves where the water surface moves up and down while the wave travels outwards.

2. Key Wave Terms

Term Definition Symbol/Unit
Displacement The distance and direction of a point on the wave from its equilibrium position x (m)
Amplitude The maximum magnitude of displacement from equilibrium A (m)
Phase Difference How much one point on a wave is "behind" or "ahead" of another Radians or degrees
Period Time for one complete wave to pass a fixed point T (s)
Frequency Number of complete waves passing a point per second f (Hz)
Wavelength Distance between two consecutive points in phase λ (m)
Wave Speed Speed at which the wave profile travels v (m/s)

3. Cathode-Ray Oscilloscope (CRO)

A CRO displays a voltage signal as a function of time.

  • Time-base: Controls horizontal scaling (s/div)
  • Y-gain: Controls vertical scaling (V/div)
Example 1 (CRO):

A CRO screen shows a sinusoidal wave. The time-base is set to 5.0 ms/div. Two consecutive peaks are 4.0 divisions apart. The y-gain is set to 2.0 V/div, and the peak-to-peak height is 6.0 divisions.

a) Calculate the frequency of the wave.

b) Calculate the amplitude of the wave.

Solution:

a) Period, T = (4.0 div) × (5.0 ms/div) = 20 ms = 0.020 s
Frequency, f = 1T = 10.020 = 50 Hz

b) Peak-to-peak voltage = (6.0 div) × (2.0 V/div) = 12.0 V
Amplitude, A = 12.02 = 6.0 V

4. & 5. The Wave Equation

v = f × λ

Derivation: In one period T, the wave travels one wavelength λ.
Therefore, wave speed v = distancetime = λT.
Since f = 1T, we get v = f × λ.

Wave Equation Calculator

Calculate any one variable given the other two:

Example 2 (Wave Equation):

A sound wave has a frequency of 2.0 kHz and a wavelength of 17 cm. Calculate its speed.

Solution:

f = 2000 Hz, λ = 0.17 m
v = f × λ = 2000 × 0.17 = 340 m/s

6. Energy Transfer

A progressive wave transfers energy from one place to another. The medium itself does not travel; energy is transferred through particle oscillations.

7. Intensity

I = PA

Where I is intensity (W/m²), P is power (W), and A is area (m²).

I ∝ A²

Intensity is proportional to the square of amplitude.

Example 3 (Intensity):

A wave has an amplitude of 2.0 cm and an intensity of 6.0 × 10⁻⁴ W/m².

a) Calculate the intensity when the amplitude is 5.0 cm.

b) Calculate the amplitude when the intensity is 1.5 × 10⁻³ W/m².

Solution:

a) I ∝ A², so I₂I₁ = A₂A₁²
I₂ = I₁ × A₂A₁² = (6.0×10⁻⁴) × 5.02.0² = 3.75×10⁻³ W/m²

b) A ∝ √I, so A₂A₁ = √I₂I₁
A₂ = A₁ × √I₂I₁ = 2.0 × √1.5×10⁻³6.0×10⁻⁴ = 3.16 cm

7.2 Transverse and Longitudinal Waves

1. Comparison

Feature Transverse Waves Longitudinal Waves
Particle Oscillation Perpendicular to energy direction Parallel to energy direction
Wave Profile Crests and troughs Compressions and rarefactions
Examples Waves on string, EM waves, water ripples Sound waves, seismic P-waves
Can be Polarised? Yes No

2. Graphical Representations

  • Displacement-Distance Graph: A "snapshot" of the wave at a single instant
  • Displacement-Time Graph: Shows how one particle's displacement varies with time
Example 4 (Graphs):

A displacement-distance graph shows a wave with amplitude 5 cm and wavelength 8 cm.

a) State the amplitude and wavelength.

b) If the wave speed is 16 cm/s and moving right, draw a displacement-time graph for the particle at x = 2 cm.

Solution:

a) Amplitude = 5 cm, Wavelength = 8 cm

b) v = fλ → 16 = f×8 → f = 2 Hz → T = 1f = 0.5 s
The particle at x=2 cm is at equilibrium and will start moving upward.
The displacement-time graph is a sine wave starting from (0,0) with amplitude 5 cm and period 0.5 s.

7.3 Doppler Effect for Sound Waves

1. Understanding the Effect

The Doppler effect is the change in observed frequency due to relative motion between source and observer.

  • Source moving towards observer: Wavefronts bunch up → wavelength decreases → frequency increases
  • Source moving away from observer: Wavefronts stretch out → wavelength increases → frequency decreases

2. The Equation

fo = fs ×
v v ± vs

Where:

  • fo = observed frequency (Hz)
  • fs = source frequency (Hz)
  • v = speed of sound (m/s)
  • vs = speed of source (m/s)

Sign Convention:

  • Use minus (-) when source moves towards observer
  • Use plus (+) when source moves away from observer

Doppler Effect Calculator

Example 5 (Doppler Effect):

A police car siren emits sound at 1000 Hz. The car moves towards a stationary observer at 30 m/s. Speed of sound is 340 m/s.

a) Calculate the frequency heard by the observer.

b) What frequency is heard when the car moves away at the same speed?

Solution:

a) Source moving towards: fo = fs × v v - vs = 1000 × 340 340 - 30 = 1097 Hz

b) Source moving away: fo = fs × v v + vs = 1000 × 340 340 + 30 = 919 Hz

7.4 Electromagnetic Spectrum

1. General Properties

All electromagnetic waves:

  • Are transverse waves
  • Consist of oscillating electric and magnetic fields
  • Travel at c = 3.00 × 10⁸ m/s in vacuum
  • Can travel through vacuum
  • Obey the wave equation c = fλ

2. Principal Regions

Region Approximate Wavelength Range
Radio Waves > 10 cm
Microwaves 10 cm - 1 mm
Infrared (IR) 1 mm - 700 nm
Visible Light 700 nm - 400 nm
Ultraviolet (UV) 400 nm - 10 nm
X-rays 10 nm - 10⁻³ nm
Gamma Rays (γ) < 10⁻³ nm

3. Visible Light

The range 400-700 nm is visible to the human eye:

  • ~700 nm = Red light
  • ~400 nm = Violet light
Example 6 (EM Spectrum):

Calculate the frequency of blue light with wavelength 450 nm in vacuum.

Solution:

λ = 450 × 10⁻⁹ m, c = 3.00 × 10⁸ m/s
f = cλ = 3.00×10⁸450×10⁻⁹ = 6.67 × 10¹⁴ Hz

7.5 Polarisation

1. Phenomenon of Polarisation

Polarisation proves EM waves are transverse.

  • Unpolarised Wave: Oscillations in all directions perpendicular to travel
  • Plane-Polarised Wave: Oscillations confined to one plane

Longitudinal waves cannot be polarised because their oscillations are parallel to the direction of travel.

2. Malus's Law

I = I₀ × cos²θ

Where:

  • I = transmitted intensity
  • I₀ = incident intensity of plane-polarised wave
  • θ = angle between polarization direction and filter's transmission axis

Malus's Law Calculator

Example 7 (Polarisation):

Unpolarised light of intensity I₀ passes through polarising filter P.

a) What is the intensity I₁ after P?

The light then passes through a second filter A at 60° to P.

b) Calculate the final intensity I₂ in terms of I₀.

Solution:

a) First filter polarises the light: I₁ = I₀2

b) Malus's Law: I₂ = I₁ × cos²θ = I₀2 × cos²(60°) = I₀2 × 14 = I₀8

Effect of Time Base on a CRO — Notes & Example

Effect of the Time Base on a CRO

Definition: The time base controls how much time each horizontal division represents on the cathode ray oscilloscope (CRO).

Key effects when the time base is increased

  • The waveform becomes compressed horizontally (each cycle occupies fewer divisions).
  • More cycles appear on the screen.
  • The trace sweeps across the screen faster.
  • Individual cycles appear smaller, so fine time details are harder to measure.

Key effects when the time base is decreased

  • The waveform becomes stretched horizontally (each cycle occupies more divisions).
  • Fewer cycles appear on the screen.
  • The trace sweeps across the screen slower.
  • Individual cycles are larger, making fine time measurements easier.

Example scenario (numerical)

Signal: 100 Hz sine wave

QuantityValue
Period, T1 / 100 = 0.01 s = 10 ms
Initial time base1 ms/div → one cycle occupies 10 divisions
Increased time base5 ms/div → one cycle occupies 2 divisions

Result: increasing the time base compresses the waveform horizontally (from 10 div to 2 div), so the trace looks smaller and more cycles fit on screen.

Abel Masitsa

Wave Intensity and Power

Wave Intensity and Power: A Detailed Explanation

Understanding the concepts of wave intensity and power is fundamental in physics, with applications ranging from acoustics to optics. Let's explore these concepts in detail, including their mathematical relationships and practical implications.

1. Power (P)

What is Wave Power?

Power is a fundamental concept in physics that refers to the rate at which energy is transferred or converted. In the context of a wave, it is the amount of energy the wave carries past a given point per unit of time.

Simple Analogy

Imagine a wave on a string. The power would be the total energy (kinetic + potential) that flows past a specific point on the string every second.

SI Unit: Watt (W), where 1 Watt = 1 Joule per second (1 J/s)

For a wave, the power is proportional to the square of its amplitude. For example, doubling the amplitude of a wave on a string will quadruple the power it transmits.

2. Intensity (I)

What is Wave Intensity?

While power tells us the total energy transferred, Intensity tells us how concentrated that power is. It is defined as the power per unit area through which the wave travels.

Practical Example

Think of a 100-watt light bulb:

  • The Power is always 100 W, regardless of where you are.
  • The Intensity changes with distance. If you hold your hand very close to the bulb, the light is intense and feels hot because the 100 W is concentrated on a small area of your skin. If you move far away, the same 100 W is now spread over a much larger area, so the intensity is much lower.

SI Unit: Watt per square meter (W/m²)

Intensity is what our senses (like our ears for sound or our eyes for light) typically perceive. A louder sound or a brighter light has a higher intensity.

The Key Relationship: Power vs. Intensity

The fundamental relationship is:

I =
PA
Intensity = Power / Area

This formula highlights that for a given power, intensity decreases as the area over which it is spread increases.

3. Intensity for a Spherical Wave

Many waves, like sound from a speaker or light from a star, radiate outward equally in all directions. This is called a spherical wave.

  • The "area" we are interested in is the surface area of the sphere that the wave has reached.
  • The surface area of a sphere is A = 4πr², where r is the distance from the source.

If a source emits a wave with a total power P, this power is spread over the entire spherical surface. Therefore, the intensity at a distance r from the source is:

I =
P4πr²
Intensity of a spherical wave

Consequences of I ∝
1r²
(The Inverse-Square Law)

This relationship shows that the intensity of a wave is inversely proportional to the square of the distance from the source.

Examples of the Inverse-Square Law:

  • If you double the distance (r becomes 2r), the intensity drops to one-fourth of its original value.
  • If you triple the distance, the intensity drops to one-ninth.

This is why stars that are far away appear very dim, and why you can barely hear a loudspeaker from a long distance.

Visualizing the Inverse-Square Law

Distance r
Intensity I
Distance 2r
Intensity
14
I
Distance 3r
Intensity
19
I

As distance increases, the same power spreads over a larger area, reducing intensity.

4. Intensity in Terms of Wave Properties

For a traveling wave (like sound or a wave on a string), the intensity can also be expressed in terms of the wave's fundamental properties: its amplitude and frequency.

For a sinusoidal wave, the intensity I is proportional to:

  1. The square of the amplitude (A²)
  2. The square of the angular frequency (ω²)
  3. The speed of the wave (v)

The general formula for the average intensity of a mechanical wave is:

I =
12
ρ v ω² A²
Intensity in terms of wave properties

Where:

  • I is the average intensity (W/m²)
  • ρ (rho) is the density of the medium (kg/m³)
  • v is the speed of the wave in the medium (m/s)
  • ω (omega) is the angular frequency (rad/s) = 2πf, where f is the frequency in Hz
  • A is the amplitude of the wave (m)

Why is this important?

This formula explains why waves behave the way they do:

High Frequency, High Intensity

A high-pitched sound (high ω) of the same amplitude as a low-pitched sound carries more energy. This is why a piccolo can be heard over an orchestra.

Amplitude is Crucial

Doubling the amplitude quadruples the intensity. This is why plucking a guitar string harder (increasing amplitude) makes a much louder sound.

Summary and Key Formulas

Concept
Definition
Key Formula(s)
SI Unit
Power (P)
The rate of energy transfer by the wave
(Base definition)
Watt (W)
Intensity (I)
The power per unit area carried by a wave
1. I =
PA
2. I =
P4πr²
(spherical waves)
3. I =
12
ρvω²A² (sinusoidal wave)
W/m²

Practical Example: A Sound Wave

Imagine a small speaker emitting a 1 kHz sound wave with a total power of 0.1 Watts.

At a distance of 1 meter:

  • The area is A = 4π(1)² ≈ 12.6 m²
  • The intensity is I =
    0.1 W12.6 m²
    ≈ 0.008 W/m²

At a distance of 10 meters:

  • The area is A = 4π(10)² ≈ 1257 m²
  • The intensity is I =
    0.1 W1257 m²
    ≈ 0.00008 W/m²

The intensity at 10 meters is

1100
of the intensity at 1 meter, perfectly demonstrating the inverse-square law (10² = 100).

Tags: Physics Waves Intensity Power Inverse Square Law

Understanding Wave Phase Difference: A Complete Guide

Published: November 10, 2025 Physics > Waves

What is Phase Difference?

Phase difference describes how "out of sync" two points on a wave are with each other. It tells us how much one point lags behind or leads ahead of another point in their oscillation cycle.

The Phase Difference Formula

φ =
xλ
× 360°

Where:

  • φ = Phase difference in degrees
  • x = Distance between the two points
  • λ = Wavelength of the wave

Key Phase Relationships

0° Phase Difference

Separation: 0, λ, 2λ, 3λ...

Relationship: IN PHASE

Points move together in perfect sync

90° Phase Difference

Separation: λ/4

Relationship: QUARTER CYCLE APART

When one is at crest, other is at equilibrium

180° Phase Difference

Separation: λ/2

Relationship: COMPLETELY OUT OF PHASE

Points move in opposite directions

Practical Examples

Example 1: Points A and B

If points A and B are separated by one wavelength (x = λ):

φ =
λλ
× 360° = 360° ≡ 0°

Result: Points A and B are IN PHASE

Example 2: Points C and D

If points C and D are separated by quarter wavelength (x = λ/4):

φ =
λ/4λ
× 360° = 90°

Result: Points C and D have 90° PHASE DIFFERENCE

Quick Reference Table

Separation
Phase Difference
Relationship
0
0°
In Phase
λ/4
90°
Quarter Cycle Apart
λ/2
180°
Completely Out of Phase
3λ/4
270°
Three Quarters Apart
λ
360° (0°)
In Phase

Why Phase Difference Matters

Interference Patterns

Phase difference determines whether waves interfere constructively (in phase) or destructively (out of phase)

Standing Waves

Nodes and antinodes in standing waves are created by specific phase relationships

Sound & Music

Phase affects sound quality and how different frequencies interact

Electronics

AC circuits rely on phase relationships between voltage and current

Polarization of Light Using Filters

1. Light as a Wave

Light is an electromagnetic (EM) wave — it has two fields:

  • Electric field (E)
  • Magnetic field (B)

These fields vibrate at right angles to each other and to the direction of wave travel.

2. Unpolarized Light

From sources like the Sun or a bulb, the E-field vibrates in all directions perpendicular to the wave's motion.

Imagine many arrows spinning around — that's unpolarized light.

3. Polarizing Filter

A polarizer only allows the E-field in one direction to pass through (say, vertical).

All other directions of the E-field are blocked or absorbed.

After passing through one polarizer → the light becomes linearly polarized.

4. Using Two Polarizers

When you use a second polarizer (analyzer), the amount of light that passes depends on the angle (θ) between their transmission axes:

I = I₀ cos²θ

(This is called Malus' Law)

θ = 0°

I = I₀ (maximum light passes)

θ = 45°

I = I₀/2

θ = 90°

I = 0 (completely dark)

5. Why the Polarizer Only Affects the E-Field

Polarizers work on the electric field because the E-field interacts with the molecules in the polarizing material.

The B-field adjusts automatically (since it's always at right angles to E), so it doesn't need separate filtering.

🧭 Summary

Stage Description Effect on E-field Light Intensity
Before polarizer Unpolarized light Many directions Full
After 1st polarizer Linearly polarized One direction Half
After 2nd polarizer (angle θ) Depends on angle E = E₀ cosθ I = I₀ cos²θ

Why a Polariser Reduces Unpolarised Light to ½

Unpolarised light vibrates in all directions. A polarising filter only allows vibrations that are parallel to its transmission axis to pass.

For each vibration direction at angle θ, the transmitted electric field becomes E cos θ. Since intensity depends on the square of the electric field, the transmitted intensity becomes E² cos² θ.

Unpolarised light contains all vibration angles equally. When all the values of cos² θ are averaged, the result is:

⟨cos² θ⟩ = ½

This means that, on average, only half of the original energy is aligned with the filter’s axis.

Therefore, the first polariser reduces the intensity to ½ I₀.

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