Wave Properties

Wave Fundamentals

Waves on a String

Investigate how wave speed, frequency, wavelength, and amplitude are related. Experiment with different end conditions (fixed, loose, open) to observe standing waves and resonance.

What is a Wave?

A wave is a disturbance that transfers energy through a medium or space without transferring matter.

Types of Waves

Property	Transverse	Longitudinal
Oscillation direction	Perpendicular to propagation	Parallel to propagation
Examples	Light, water surface waves, strings	Sound, pressure waves
Crests and troughs	Yes	Compressions and rarefactions
Polarisation	Can be polarised	Cannot be polarised

Wave Terminology

Term	Symbol	Definition	Unit
Displacement	$x$ or $y$	Distance from equilibrium	m
Amplitude	$A$	Maximum displacement	m
Wavelength	$\lambda$	Distance between two consecutive identical points	m
Frequency	$f$	Number of complete oscillations per second	Hz
Period	$T$	Time for one complete oscillation	s
Wave speed	$v$	Speed at which the wave propagates	m/s
Phase	$\phi$	Position in the cycle of oscillation	rad

Relationships

$$f = \frac{1}{T}$$ $$v = f\lambda$$

Example

A sound wave has frequency $440\mathrm{ Hz}$ and travels at $343\mathrm{ m/s}$. Find its wavelength.

$$\lambda = \frac{v}{f} = \frac{343}{440} = 0.780\mathrm{ m}$$

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The Wave Equation

General Form

For a travelling wave:

$$y(x, t) = A\sin(\omega t - kx + \phi)$$

where:

• $\omega = 2\pi f$ is the angular frequency
• $k = \dfrac{2\pi}{\lambda}$ is the wave number
• $\phi$ is the phase constant

Key Relations

$$v = \frac{\omega}{k} = f\lambda$$

Intensity

The intensity of a wave is the power per unit area:

$$I = \frac{P}{A}$$

For a point source radiating equally in all directions:

$$I = \frac{P}{4\pi r^2}$$

Intensity is proportional to amplitude squared:

$$I \propto A^2$$

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Sound Waves

Nature of Sound

Sound is a longitudinal mechanical wave that requires a medium. It consists of compressions (high pressure) and rarefactions (low pressure).

Speed of Sound

Medium	Speed (m/s)
Air at $20\degree\mathrm{C}$	343
Water at $20\degree\mathrm{C}$	1482
Steel	5960
Glass	5640

The speed of sound in air depends on temperature:

$$v \approx 331 + 0.6T_C \mathrm{ m/s}$$

Inverse Square Law

For a point source of sound:

$$I \propto \frac{1}{r^2}$$

Doubling the distance from a source reduces the intensity to one quarter.

Sound Intensity Level

Measured in decibels (dB):

$$\beta = 10\log_{10}\!\left(\frac{I}{I_0}\right)$$

where $I_0 = 10^{-12}\mathrm{ W/m}^2$ is the threshold of hearing.

Source	Level (dB)
Threshold of hearing	0
Whisper	30
Normal conversation	60
Busy traffic	80
Rock concert	120
Jet engine at 30 m	150

Exam Tip

A 10 dB increase represents a tenfold increase in intensity. A 3 dB increase approximately doubles the intensity. Decibels are logarithmic, so you cannot simply add them.

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Electromagnetic Spectrum

Region	Wavelength Range	Frequency Range
Radio waves	$\gt 0.1\mathrm{ m}$	$\lt 3 \times 10^9\mathrm{ Hz}$
Microwaves	$0.1\mathrm{ mm}$ to $0.1\mathrm{ m}$	$3 \times 10^9$ to $3 \times 10^{12}\mathrm{ Hz}$
Infrared	$700\mathrm{ nm}$ to $0.1\mathrm{ mm}$	$3 \times 10^{12}$ to $4.3 \times 10^{14}\mathrm{ Hz}$
Visible light	$400\mathrm{ nm}$ to $700\mathrm{ nm}$	$4.3 \times 10^{14}$ to $7.5 \times 10^{14}\mathrm{ Hz}$
Ultraviolet	$10\mathrm{ nm}$ to $400\mathrm{ nm}$	$7.5 \times 10^{14}$ to $3 \times 10^{16}\mathrm{ Hz}$
X-rays	$0.01\mathrm{ nm}$ to $10\mathrm{ nm}$	$3 \times 10^{16}$ to $3 \times 10^{19}\mathrm{ Hz}$
Gamma rays	$\lt 0.01\mathrm{ nm}$	$\gt 3 \times 10^{19}\mathrm{ Hz}$

Key Properties

• All EM waves travel at $c = 3.0 \times 10^8\mathrm{ m/s}$ in a vacuum.
• They are all transverse waves.
• They can all travel through a vacuum.
• They can all be polarised.

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Superposition

Principle of Superposition

When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the individual displacements:

$$y_{\mathrm{total}} = y_1 + y_2 + y_3 + \cdots$$

Constructive Interference

Waves arrive in phase (path difference $= n\lambda$, where $n$ is an integer):

$$\mathrm{Path difference} = n\lambda$$

The resultant amplitude is $A_1 + A_2$ (maximum).

Destructive Interference

Waves arrive out of phase (path difference $= (n + \frac{1}{2})\lambda$):

$$\mathrm{Path difference} = \left(n + \frac{1}{2}\right)\lambda$$

The resultant amplitude is $|A_1 - A_2|$ (minimum).

Two-Source Interference

For coherent sources (same frequency, constant phase relationship), interference produces a pattern of bright and dark fringes (for light) or loud and quiet regions (for sound).

For double-slit interference with slit separation $d$ and distance to screen $D$:

$$d\sin\theta = n\lambda \quad (\mathrm{bright fringes})$$ $$d\sin\theta = \left(n + \frac{1}{2}\right)\lambda \quad (\mathrm{dark fringes})$$

For small angles ($\sin\theta \approx \tan\theta \approx \dfrac{x}{D}$):

$$x_n = \frac{n\lambda D}{d}$$

Fringe spacing:

$$\Delta x = \frac{\lambda D}{d}$$

Example

Light of wavelength $600\mathrm{ nm}$ passes through a double slit with separation $0.2\mathrm{ mm}$. The screen is $1.5\mathrm{ m}$ away. Find the fringe spacing.

$$\Delta x = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.5}{0.2 \times 10^{-3}} = \frac{9 \times 10^{-7}}{2 \times 10^{-4}} = 4.5 \times 10^{-3}\mathrm{ m} = 4.5\mathrm{ mm}$$

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Standing Waves

Formation

Standing waves form when two identical waves travelling in opposite directions superpose. This occurs due to reflection at a boundary.

Nodes and Antinodes

• Node: point of zero displacement (destructive interference)
• Antinode: point of maximum displacement (constructive interference)

Standing Waves on Strings

For a string of length $L$ fixed at both ends:

Harmonics:

Harmonic	Wavelength	Frequency
1st (fundamental)	$\lambda_1 = 2L$	$f_1 = \dfrac{v}{2L}$
2nd	$\lambda_2 = L$	$f_2 = \dfrac{v}{L} = 2f_1$
3rd	$\lambda_3 = \dfrac{2L}{3}$	$f_3 = \dfrac{3v}{2L} = 3f_1$
$n$th	$\lambda_n = \dfrac{2L}{n}$	$f_n = \dfrac{nv}{2L} = nf_1$

The wave speed on a string under tension $T$ with mass per unit length $\mu$:

$$v = \sqrt{\frac{T}{\mu}}$$

Standing Waves in Pipes

Open pipe (open at both ends):

f_n = \frac`\{nv}`{2L}, \quad n = 1, 2, 3, \ldots

Both ends are antinodes.

Closed pipe (closed at one end):

f_n = \frac`\{nv}`{4L}, \quad n = 1, 3, 5, \ldots \mathrm{ (odd harmonics only)}

The closed end is a node, the open end is an antinode.

Example

A string of length $0.75\mathrm{ m}$ has a fundamental frequency of $220\mathrm{ Hz}$. Find the speed of waves on the string.

$$f_1 = \frac{v}{2L} \implies v = 2Lf_1 = 2(0.75)(220) = 330\mathrm{ m/s}$$

Example

An open pipe has a fundamental frequency of $440\mathrm{ Hz}$. Find the frequency of the third harmonic.

$$f_3 = 3f_1 = 3(440) = 1320\mathrm{ Hz}$$

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The Doppler Effect

Definition

The Doppler effect is the change in observed frequency of a wave when there is relative motion between the source and the observer.

Moving Source, Stationary Observer

$$f' = \frac{f}{1 \pm \frac{v_s}{v_w}}$$

where $v_s$ is the speed of the source and $v_w$ is the wave speed.

• Source approaching: $f' = \dfrac{f}{1 - v_s/v_w}$ (frequency increases)
• Source receding: $f' = \dfrac{f}{1 + v_s/v_w}$ (frequency decreases)

Moving Observer, Stationary Source

$$f' = f\!\left(1 \pm \frac{v_o}{v_w}\right)$$

• Observer approaching: $f' = f(1 + v_o/v_w)$ (frequency increases)
• Observer receding: $f' = f(1 - v_o/v_w)$ (frequency decreases)

General Doppler Formula

$$f' = f\!\left(\frac{v_w \pm v_o}{v_w \mp v_s}\right)$$

Upper signs when approaching, lower signs when receding.

Electromagnetic Doppler Effect

For light:

$$f' = f\sqrt{\frac{1 + \beta}{1 - \beta}}$$

where $\beta = \dfrac{v}{c}$. For $v \ll c$:

$$\frac{\Delta f}{f} \approx \frac{v}{c}$$

• Redshift: source receding, observed wavelength increases
• Blueshift: source approaching, observed wavelength decreases

Applications

Application	Description
Radar guns	Measure speed of vehicles
Medical ultrasound	Measure blood flow velocity
Astronomy	Measure speed of stars/galaxies (redshift)
Weather radar	Track storm systems

Example

A fire engine with siren at $700\mathrm{ Hz}$ approaches at $30\mathrm{ m/s}$. What frequency does a stationary observer hear? ($v_{\mathrm{sound}} = 343\mathrm{ m/s}$)

$$f' = \frac{f}{1 - v_s/v_w} = \frac{700}{1 - 30/343} = \frac{700}{0.9125} = 767\mathrm{ Hz}$$

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Diffraction

Definition

Diffraction is the spreading of waves when they pass through an aperture or around an obstacle.

Conditions

• Diffraction is most significant when the wavelength is comparable to the size of the aperture or obstacle.
• For light (very small $\lambda$), diffraction requires very narrow slits.
• For sound (larger $\lambda$), diffraction is more noticeable around everyday objects.

Single Slit Diffraction

For light of wavelength $\lambda$ passing through a slit of width $a$:

Minima occur at:

$$a\sin\theta = n\lambda, \quad n = \pm 1, \pm 2, \ldots$$

The central maximum is twice as wide as the secondary maxima.

Rayleigh Criterion

Two sources are just resolvable when the central maximum of one diffraction pattern coincides with the first minimum of the other:

$$\theta_{\min} = 1.22\frac{\lambda}{D}$$

where $D$ is the diameter of the circular aperture.

Resolution

The ability to distinguish between two closely spaced objects depends on:

• Wavelength (shorter $\lambda$ gives better resolution)
• Aperture diameter (larger $D$ gives better resolution)

This is why astronomical telescopes use large mirrors and electron microscopes use short wavelengths (electrons).

Example

A telescope with a mirror of diameter $0.1\mathrm{ m}$ observes light of wavelength $550\mathrm{ nm}$. Find the minimum angular separation it can resolve.

$$\theta_{\min} = 1.22 \times \frac{550 \times 10^{-9}}{0.1} = 6.71 \times 10^{-6}\mathrm{ rad}$$

This is approximately $1.38$ arcseconds.

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Polarisation

Definition

Polarisation is the restriction of the oscillation direction of a transverse wave to one plane.

Methods

Method	Description
Polarising filter	Allows only one plane of oscillation to pass
Reflection	Light reflected from a surface is partially polarised
Birefringence	Certain crystals split light into two polarised beams
Scattering	Light scattered by the atmosphere is partially polarised

Malus's Law

When polarised light of intensity $I_0$ passes through an analyser at angle $\theta$ to the polarisation direction:

$$I = I_0\cos^2\theta$$

Brewster's Angle

When light hits a surface at Brewster's angle, the reflected light is completely polarised:

$$\tan\theta_B = \frac{n_2}{n_1}$$

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IB Exam-Style Questions

Question 1 (Paper 1 style)

Light of wavelength $590\mathrm{ nm}$ is incident on a double slit with separation $0.5\mathrm{ mm}$. The screen is $2\mathrm{ m}$ away. Find the distance from the central maximum to the third bright fringe.

$$x_3 = \frac{3\lambda D}{d} = \frac{3 \times 590 \times 10^{-9} \times 2}{0.5 \times 10^{-3}} = \frac{3.54 \times 10^{-6}}{5 \times 10^{-4}} = 7.08 \times 10^{-3}\mathrm{ m} = 7.08\mathrm{ mm}$$

Question 2 (Paper 2 style)

A string of length $0.8\mathrm{ m}$ and mass $4\mathrm{ g}$ is under tension $50\mathrm{ N}$.

(a) Find the speed of waves on the string.

$$\mu = \frac{0.004}{0.8} = 0.005\mathrm{ kg/m}$$ $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{50}{0.005}} = \sqrt{10000} = 100\mathrm{ m/s}$$

(b) Find the fundamental frequency and the first three harmonic frequencies.

$$f_1 = \frac{v}{2L} = \frac{100}{1.6} = 62.5\mathrm{ Hz}$$ $$f_2 = 125\mathrm{ Hz}, \quad f_3 = 187.5\mathrm{ Hz}$$

(c) Draw the standing wave pattern for the second harmonic.

The second harmonic has one node at the centre and antinodes at each quarter point. There are 3 nodes (including both ends) and 2 antinodes.

Question 3 (Paper 1 style)

An ambulance with siren at $800\mathrm{ Hz}$ moves away from a stationary observer at $20\mathrm{ m/s}$. What frequency does the observer hear? ($v_{\mathrm{sound}} = 340\mathrm{ m/s}$)

$$f' = \frac{f}{1 + v_s/v_w} = \frac{800}{1 + 20/340} = \frac{800}{1.0588} = 756\mathrm{ Hz}$$

Question 4 (Paper 2 style)

Unpolarised light of intensity $I_0$ passes through two polarising filters. The axis of the second filter is at $60\degree$ to the first.

(a) Find the intensity after the first filter.

$$I_1 = \frac{I_0}{2}$$

(b) Find the intensity after the second filter.

$$I_2 = I_1\cos^2 60\degree = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$$

Question 5 (Paper 2 style)

A car horn produces sound at $400\mathrm{ Hz}$. The car approaches a stationary observer at $25\mathrm{ m/s}$, then passes and moves away at the same speed.

(a) Find the frequency heard by the observer as the car approaches.

$$f' = \frac{400}{1 - 25/343} = \frac{400}{0.927} = 431\mathrm{ Hz}$$

(b) Find the frequency heard as the car moves away.

$$f' = \frac{400}{1 + 25/343} = \frac{400}{1.073} = 373\mathrm{ Hz}$$

(c) Calculate the change in frequency.

$$\Delta f = 431 - 373 = 58\mathrm{ Hz}$$

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Summary

Quantity	Formula
Wave speed	$v = f\lambda$
Intensity	$I = \dfrac{P}{4\pi r^2}$
Sound level	$\beta = 10\log_{10}(I/I_0)$
Double-slit maxima	$d\sin\theta = n\lambda$
Single-slit minima	$a\sin\theta = n\lambda$
String harmonics	$f_n = \dfrac{nv}{2L}$
Doppler (source moving)	$f' = \dfrac{f}{1 \mp v_s/v_w}$
Malus's law	$I = I_0\cos^2\theta$
Rayleigh criterion	$\theta = 1.22\dfrac{\lambda}{D}$

Exam Strategy

For wave problems, always identify the type of wave and the relevant equations. For interference problems, determine whether you need path difference or phase difference. For standing waves, clearly identify whether the system is a string, open pipe, or closed pipe. For Doppler problems, identify what is moving (source, observer, or both).

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Wave Intensity and Amplitude

Relationship Between Intensity and Amplitude

For a wave, intensity is proportional to the square of the amplitude:

$$I \propto A^2$$

If the amplitude doubles, the intensity quadruples.

Intensity at a Distance from a Point Source

$$I = \frac{P}{4\pi r^2}$$

This means:

$$I_1 r_1^2 = I_2 r_2^2$$

Example

At $10\mathrm{ m}$ from a source, the intensity is $0.5\mathrm{ W/m}^2$. Find the intensity at $25\mathrm{ m}$.

$$I_2 = I_1 \times \frac{r_1^2}{r_2^2} = 0.5 \times \frac{100}{625} = 0.08\mathrm{ W/m}^2$$

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Phase and Phase Difference

Phase Difference

Phase difference $\Delta\phi$ between two waves at a point:

$$\Delta\phi = \frac{2\pi \Delta x}{\lambda}$$

where $\Delta x$ is the path difference.

Phase Difference	Description
$0, 2\pi, 4\pi, \ldots$	In phase (constructive)
$\pi, 3\pi, 5\pi, \ldots$	Anti-phase (destructive)
$\pi/2$	$90\degree$ out of phase

Coherence

Two sources are coherent if they have:

• The same frequency.
• A constant phase relationship.

Only coherent sources produce a stable interference pattern.

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Diffraction Gratings

Equation

For a diffraction grating with $N$ slits per metre (slit separation $d = 1/N$):

$$d\sin\theta = n\lambda$$

The maximum number of orders visible:

$$n_{\max} = \frac{d}{\lambda}$$

(rounded down to the nearest integer).

Advantages Over Double Slit

• Sharper, brighter fringes.
• Larger angular separation.
• More accurate measurement of wavelength.

Example

A diffraction grating has $500\mathrm{ lines/mm}$. Light of wavelength $600\mathrm{ nm}$ is incident normally. Find the angles of the first and second-order maxima.

$$d = \frac{1}{500000} = 2 \times 10^{-6}\mathrm{ m}$$$$\sin\theta_1 = \frac{\lambda}{d} = \frac{600 \times 10^{-9}}{2 \times 10^{-6}} = 0.3 \implies \theta_1 = 17.5\degree$$$$\sin\theta_2 = \frac{2\lambda}{d} = 0.6 \implies \theta_2 = 36.9\degree$$

Maximum order: $n_{\max} = \dfrac{2 \times 10^{-6}}{600 \times 10^{-9}} = 3.33$, so 3 orders are visible.

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Additional IB Exam-Style Questions

Question 6 (Paper 2 style)

A string of length $0.6\mathrm{ m}$ is fixed at both ends. The speed of waves on the string is $240\mathrm{ m/s}$.

(a) Calculate the fundamental frequency.

$$f_1 = \frac{v}{2L} = \frac{240}{1.2} = 200\mathrm{ Hz}$$

(b) Draw the standing wave pattern for the third harmonic and state its frequency.

The third harmonic has 3 half-wavelengths fitting on the string, with 4 nodes and 3 antinodes.

$$f_3 = 3f_1 = 600\mathrm{ Hz}$$

(c) The tension in the string is doubled. Find the new fundamental frequency.

$$v = \sqrt{\frac{T}{\mu}} \implies v' = \sqrt{2}v$$ $$f_1' = \sqrt{2} \times 200 = 283\mathrm{ Hz}$$

Question 7 (Paper 2 style)

Two loudspeakers are $3\mathrm{ m}$ apart and emit sound of frequency $686\mathrm{ Hz}$ in phase. The speed of sound is $343\mathrm{ m/s}$.

(a) Calculate the wavelength.

$$\lambda = \frac{v}{f} = \frac{343}{686} = 0.5\mathrm{ m}$$

(b) A listener walks along a line parallel to the speakers, $4\mathrm{ m}$ away. Find the positions of the first two points of constructive interference.

For constructive interference: path difference $= n\lambda$.

Using geometry, the path difference $\Delta = d\sin\theta$ where $\theta$ is the angle from the perpendicular bisector.

$$3\sin\theta = n \times 0.5$$

For $n = 1$: $\sin\theta = 1/6$, $\theta = 9.6\degree$. Distance from centre: $4\tan 9.6\degree = 0.68\mathrm{ m}$.

For $n = 2$: $\sin\theta = 1/3$, $\theta = 19.5\degree$. Distance from centre: $4\tan 19.5\degree = 1.41\mathrm{ m}$.

Question 8 (Paper 1 style)

Light of wavelength $\lambda$ passes through a single slit of width $a$ and produces a diffraction pattern. If the slit width is halved, what happens to the width of the central maximum?

The first minimum occurs at $a\sin\theta = \lambda$. If $a$ is halved, $\sin\theta$ doubles, so the angular width of the central maximum approximately doubles. The width of the central maximum is inversely proportional to the slit width.

Question 9 (Paper 2 style)

Unpolarised light of intensity $200\mathrm{ W/m}^2$ passes through three polarising filters. The first has its axis vertical. The second is at $30\degree$ to the vertical. The third is at $60\degree$ to the vertical.

Find the intensity after each filter.

After filter 1: $I_1 = \dfrac{200}{2} = 100\mathrm{ W/m}^2$.

After filter 2: $I_2 = 100\cos^2 30\degree = 100 \times 0.75 = 75\mathrm{ W/m}^2$.

After filter 3: $I_3 = 75\cos^2(60\degree - 30\degree) = 75\cos^2 30\degree = 75 \times 0.75 = 56.25\mathrm{ W/m}^2$.

For the A-Level treatment of this topic, see Wave Properties.

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tip

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