Waves — IB Physics SL Study Guide

For: IB Physics SL candidates sitting IB Physics SL — Theme C (Wave behaviour).

Covers: All 8 core Theme C subtopics including simple harmonic motion, wave fundamentals, transverse/longitudinal waves, wave properties, superposition, standing waves, Doppler effect, and polarisation.

You should already know: Trigonometry (sine, cosine), basic mechanics (oscillations).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Are Waves?

Waves are propagating disturbances that transfer energy from one point to another without net transfer of matter. They are a core Theme C concept in the 2024 IB Physics SL syllabus, appearing in both multiple-choice and structured response questions, with ~15% of Theme C marks allocated to wave fundamentals and applications. Synonyms include wave motion and oscillatory propagation; you will encounter both mechanical waves (require a propagation medium, e.g. sound, string waves) and electromagnetic waves (no medium required, e.g. light, radio waves) across the syllabus.

2. Simple harmonic motion (SHM) — defining equation $a=-{\omega }^{2}x$

SHM is a type of periodic oscillation where the restoring force (and thus acceleration) is directly proportional to displacement from equilibrium, and acts in the opposite direction to displacement. The defining equation is $a=-{\omega }^{2}x$, where:

• $a$ = acceleration of the oscillating object ($m{s}^{-2}$)
• $\omega$ = angular frequency ($rad{s}^{-1}$, related to period $T$ by $\omega =\frac{2\pi }{T}$ and to frequency $f$ by $\omega =2\pi f$)
• $x$ = displacement from the equilibrium position ($m$) The negative sign confirms acceleration always points back to the equilibrium position, opposing displacement. Key SHM terms include amplitude $x_0$ (maximum displacement from equilibrium) and period $T$ (time for one full oscillation).

Worked Example: A mass on a spring oscillates with a period of 0.4 s. Calculate its acceleration when displaced 0.02 m from equilibrium.

1. Calculate angular frequency: $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.4}=5\pi \approx 15.71rad{s}^{-1}$
2. Substitute into the SHM equation: $a=-{\omega }^{2}x=-(15.71{)}^2\ast 0.02\approx -4.93m{s}^{-2}$ The negative sign confirms acceleration is directed toward equilibrium, as expected for SHM. Exam tip: Examiners often ask you to justify if an oscillation is SHM; you must explicitly state acceleration is proportional to displacement and opposite in direction, not just write the equation.

3. Wave fundamentals — frequency, period, wavelength, $v=f\lambda$

When SHM propagates through a medium (or vacuum for electromagnetic waves), it forms a travelling wave. Four core parameters describe all waves:

• Frequency $f$: Number of complete wave cycles passing a fixed point per second, unit hertz (Hz)
• Period $T$: Time for one full wave cycle to pass a fixed point, $T=\frac{1}{f}$, unit seconds (s)
• Wavelength $\lambda$: Distance between two consecutive identical points on a wave (e.g. two crests, two compressions), unit meters (m)
• Wave speed $v$: Speed at which the wave propagates through its medium, unit $m{s}^{-1}$

The core wave equation comes from the fact that a wave travels one wavelength in one period: $v=\frac{\lambda }{T}=f\lambda$. Note that wave speed depends only on the properties of the medium, not on frequency or amplitude.

Worked Example: A 440 Hz concert A sound wave travels through air at $340m{s}^{-1}$. Calculate its wavelength.

1. Rearrange the wave equation to solve for $\lambda$: $\lambda =\frac{v}{f}$
2. Substitute values: $\lambda =\frac{340}{440}\approx 0.77m$ Exam tip: If you increase the frequency of a wave in a fixed medium, its wavelength will decrease proportionally to keep wave speed constant — this is a common multiple-choice trap.

4. Transverse vs longitudinal waves

Mechanical waves are classified by the direction of particle oscillation relative to wave propagation direction:

• Transverse waves: Particles oscillate perpendicular to the direction of wave travel. Examples include light, string waves, and seismic S-waves. They have distinct crests (maximum positive displacement) and troughs (maximum negative displacement).
• Longitudinal waves: Particles oscillate parallel to the direction of wave travel. Examples include sound waves and seismic P-waves. They have distinct compressions (regions of high particle density/pressure) and rarefactions (regions of low particle density/pressure).

Worked Example: Identify the wave type for each scenario: (a) Particles move up and down as the wave travels right; (b) A sound wave travelling through water. Solution: (a) Transverse, as oscillation direction is perpendicular to propagation. (b) Longitudinal, as sound propagates via pressure oscillations parallel to travel direction. Exam trap: Mechanical transverse waves cannot propagate through fluids (gases/liquids) because fluids have no shear strength to support perpendicular oscillations.

5. Wave properties — reflection, refraction, diffraction

All waves exhibit three core properties when interacting with boundaries or obstacles:

1. Reflection: Waves bounce off a boundary they cannot pass through, following the law of reflection: the angle of incidence equals the angle of reflection, both measured relative to the normal (perpendicular line) to the boundary. This effect causes echoes for sound and specular reflection for light.
2. Refraction: Waves change speed and direction when passing between two media with different propagation properties. For light, Snell's Law describes this: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$, where $n$ is the refractive index of the medium, and $\theta$ is the angle between the wave and the normal.
3. Diffraction: Waves spread out when passing through a gap or around an obstacle. Maximum diffraction occurs when the gap size is approximately equal to the wavelength of the wave. Diffraction does not change the frequency, wavelength, or speed of the wave, only its direction of spread.

Worked Example: A water wave of wavelength 1.2 m passes through a 1 m wide gap in a harbour wall. Will significant diffraction occur? Solution: Yes, because the gap size (1 m) is very close to the wavelength (1.2 m), so the wave will spread widely into the harbour behind the wall.

6. Superposition and interference (constructive / destructive)

The superposition principle states that when two or more waves meet at a point, the resultant displacement at that point is equal to the vector sum of the displacements of each individual wave at that point. Interference is the stable net effect of superposition of two coherent waves (same frequency, same amplitude, constant phase difference):

• Constructive interference: Occurs when waves arrive in phase (phase difference of $0,2\pi ,4\pi ...$ or path difference of $n\lambda$, where $n$ is an integer). The resultant amplitude is the sum of individual amplitudes, producing maximum intensity.
• Destructive interference: Occurs when waves arrive exactly out of phase (phase difference of $\pi ,3\pi ...$ or path difference of $(n+\frac{1}{2})\lambda$). The resultant amplitude is the difference of individual amplitudes, producing minimum (zero, if amplitudes are equal) intensity.

Worked Example: Two coherent microwaves of wavelength 3 cm meet at a point with a path difference of 4.5 cm. State the type of interference occurring, justifying your answer. Solution: The path difference ratio is $\frac{4.5cm}{3cm}=1.5=(1+\frac{1}{2})$, so path difference equals $(n+\frac{1}{2})\lambda$ for $n=1$. This meets the condition for destructive interference. Exam tip: Coherence is a required condition for stable, observable interference — examiners frequently test this requirement in short-answer questions.

7. Standing waves and harmonics on strings and in pipes

Standing waves form when two identical waves travelling in opposite directions superpose, producing a stationary pattern that does not propagate. Key features include nodes (points of zero displacement, permanent destructive interference) and antinodes (points of maximum displacement, permanent constructive interference). Harmonic rules vary by system:

• Strings fixed at both ends / open pipes (both ends open): Fundamental frequency (1st harmonic, lowest possible frequency) is $f_1=\frac{v}{2L}$, where $L$ = length of the string/pipe, $v$ = wave speed. All integer multiples of the fundamental are allowed: $f_n=n{f}_1$, $n=1,2,\mathrm{3...}$
• Closed pipes (one end closed, one end open): Fundamental frequency is $f_1=\frac{v}{4L}$. Only odd multiples of the fundamental are allowed: $f_n=(2n-1)f_1$, $n=1,2,\mathrm{3...}$

Worked Example: A guitar string of length 0.6 m has a wave speed of $360m{s}^{-1}$. Calculate the frequency of its 3rd harmonic.

1. Calculate the fundamental frequency: $f_1=\frac{360}{2\ast 0.6}=300Hz$
2. 3rd harmonic = $3\ast f_1=3\ast 300=900Hz$ Exam trap: Closed pipes do not support even harmonics; if a question asks for the 2nd harmonic of a closed pipe, it does not exist, a common mark-losing error.

8. Doppler effect — source vs observer motion

The Doppler effect is the apparent change in frequency of a wave observed when the source of the wave or the observer moves relative to one another. For mechanical waves (e.g. sound), the observed frequency $f^{\prime }$ is calculated as:

• Source moving towards stationary observer: $f^{\prime }=f\frac{v}{v-v_s}$
• Source moving away from stationary observer: $f^{\prime }=f\frac{v}{v+v_s}$
• Observer moving towards stationary source: $f^{\prime }=f\frac{v+v_o}{v}$
• Observer moving away from stationary source: $f^{\prime }=f\frac{v-v_o}{v}$ Where $v$ = wave speed in the medium, $v_s$ = speed of the source, $v_o$ = speed of the observer, $f$ = actual frequency of the source. A simple mnemonic: motion towards the other party increases observed frequency, motion away decreases it. For electromagnetic waves, the simplified formula $\frac{\Delta f}{f}\approx \frac{v}{c}$ applies, where $v$ is relative speed, $c$ = speed of light, as EM waves do not require a medium.

Worked Example: A fire engine siren emits an 800 Hz tone, moving towards a stationary observer at $20m{s}^{-1}$. Speed of sound in air is $340m{s}^{-1}$. Calculate the observed frequency.

1. Use the source-moving-towards formula: $f^{\prime }=800\ast \frac{340}{340-20}=800\ast \frac{340}{320}=850Hz$
2. Verify: The source is moving towards the observer, so $f^{\prime }>f$, which matches the result.

9. Polarisation (transverse waves only)

Polarisation is the restriction of wave oscillations to a single plane perpendicular to the direction of propagation. Only transverse waves can be polarised, as longitudinal waves oscillate parallel to propagation direction, so there is no perpendicular plane to restrict. Common applications of polarisation include polaroid sunglasses (block horizontally polarised glare from water/roads) and liquid crystal displays (LCDs). Unpolarised light (e.g. sunlight) consists of waves oscillating in all perpendicular planes; passing it through a polariser filters out all oscillations except those aligned with the polariser axis.

Exam tip: A standard exam question asks you to explain why sound cannot be polarised — your answer must state that sound is a longitudinal wave, so oscillations are parallel to propagation, so polarisation is impossible.

10. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the negative sign in the SHM defining equation, or justifying SHM only by referencing the equation without explaining the proportionality and direction of acceleration. Why students do it: They memorize the formula but not its physical meaning. Correct move: Always include the negative sign, and explicitly state "acceleration is proportional to displacement from equilibrium and acts in the opposite direction to displacement" when justifying SHM.
• Wrong move: Assuming wave speed changes with frequency. Why students do it: They rearrange $v=f\lambda$ and incorrectly assume changing frequency changes speed, rather than wavelength. Correct move: For a fixed medium, wave speed is constant; increasing frequency decreases wavelength proportionally.
• Wrong move: Calculating even harmonics for closed pipes. Why students do it: They mix up closed and open pipe harmonic rules. Correct move: Always note closed pipes only support odd harmonics, so 2nd, 4th etc. harmonics do not exist for closed systems.
• Wrong move: Using the wrong sign in Doppler effect formulas, leading to an observed frequency that contradicts the direction of motion. Why students do it: They memorize formulas without using the intuitive frequency check. Correct move: After calculation, verify that motion towards increases observed frequency, and motion away decreases it, to catch sign errors.
• Wrong move: Stating longitudinal waves can be polarised. Why students do it: They forget the link between oscillation direction and polarisation eligibility. Correct move: Explicitly state only transverse waves can be polarised, no exceptions.

11. Practice Questions (IB Physics SL — Theme C Style)

Question 1 (3 marks)

A mass oscillating on a spring follows SHM with a period of 0.2 s. Calculate its displacement from equilibrium when its acceleration is $-15m{s}^{-2}$. Solution:

1. Calculate angular frequency: $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.2}=10\pi \approx 31.42rad{s}^{-1}$ (1 mark)
2. Rearrange the SHM equation to solve for x: $x=-\frac{a}{{\omega }^2}$ (1 mark)
3. Substitute values: $x=-\frac{(-15)}{(31.42{)}^2}\approx 0.015m=1.5cm$ (1 mark)

Question 2 (2 marks)

A closed air column of length 0.25 m produces sound waves, with speed of sound in air equal to $340m{s}^{-1}$. Calculate the frequency of the 3rd harmonic of the pipe. Solution:

1. Calculate fundamental frequency for closed pipe: $f_1=\frac{v}{4L}=\frac{340}{4\ast 0.25}=340Hz$ (1 mark)
2. 3rd harmonic for closed pipe = $3\ast f_1=1020Hz$ (1 mark)

Question 3 (2 marks)

Two coherent light waves of wavelength 500 nm meet at a point with a path difference of 1250 nm. State the type of interference occurring, justifying your answer. Solution:

1. Calculate path difference ratio: $\frac{1250nm}{500nm}=2.5=(2+\frac{1}{2})$, so path difference equals $(n+\frac{1}{2})\lambda$ for $n=2$ (1 mark)
2. This meets the condition for destructive interference, so destructive interference occurs (1 mark)

12. Quick Reference Cheatsheet

13. What's Next

This waves topic is the foundation for all subsequent Theme C content in the IB Physics SL 2024 syllabus, including electromagnetic wave propagation, diffraction grating applications, and optical instruments like lenses and telescopes. It also connects to core Topic 4 content, and you will encounter wave principles in later topics such as quantum physics, where wave-particle duality relies on core wave properties like interference and diffraction. Mastering these fundamentals will make higher-difficulty Theme C questions significantly easier to solve, and will reduce the time you need to spend revising related topics later in your course.

To reinforce your understanding, practice solving more structured and multiple-choice questions on these subtopics, and make sure you can apply each formula correctly to novel scenarios presented in exams. If you have any questions about specific concepts, tricky exam questions, or need extra practice tailored to your weak areas, you can ask Ollie directly on the homepage, where you will also find flashcards, past paper walkthroughs, and personalized study plans for IB Physics SL.