Wave Types and Basic Properties — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Mechanical vs electromagnetic wave classification, transverse vs longitudinal classification, definitions of wavelength, frequency, period, amplitude, the universal wave speed formula $v=f\lambda$, and wave energy-amplitude proportionality.

You should already know: Definitions of period and frequency for simple harmonic motion, basic kinematics of constant speed motion, core energy relationships for oscillating systems.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Wave Types and Basic Properties?

A wave is a disturbance that propagates (travels) through space or a medium, transferring energy from one location to another without any net transfer of matter. This topic, the first foundational topic of Unit 7 (Simple Harmonic Motion and Waves) in the AP Physics 1 CED, makes up roughly 2-4% of the total exam score, with concepts underlying all other wave topics that contribute to the unit’s 12-16% total exam weight. Questions on this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as standalone MCQ or as the first part of a multi-part FRQ on standing waves or sound. Standard notation used throughout this chapter and the exam matches AP conventions: $v$ for wave speed (m/s), $f$ for frequency (Hz = s⁻¹), $T$ for period (s), $\lambda$ for wavelength (m), and $A$ for amplitude (m). Common synonyms you may see on the exam include: periodic wave = continuous wave, wave pulse = single non-repeating disturbance, oscillation = cycle of wave motion.

2. Wave Classifications

Waves are sorted into two main classification schemes on the AP exam, based on what they require to travel and how the medium moves relative to the wave’s direction of travel. First classification: mechanical vs electromagnetic. Mechanical waves require a physical medium (material) to propagate; they cannot travel through a vacuum. Examples include sound waves, water waves, and waves on a stretched string. Electromagnetic waves do not require a medium and can travel through vacuum; all electromagnetic waves travel at the speed of light $c$ in vacuum, and AP Physics 1 only requires you to classify them, not analyze their properties in depth. The second classification, more commonly tested, is transverse vs longitudinal. Transverse waves have particle displacement of the medium perpendicular to the direction of wave propagation. For example, if you shake a string side-to-side to send a wave along its length, the string moves side-to-side (perpendicular) while the wave moves along the string (parallel to the string’s length). Longitudinal waves have particle displacement parallel to the direction of wave propagation. Sound is a classic example: air molecules oscillate back and forth along the direction the sound is traveling, creating regions of high pressure (compressions) and low pressure (rarefactions). Wavelength for longitudinal waves is measured as the distance between two consecutive compressions or two consecutive rarefactions, just like wavelength for transverse waves is the distance between two consecutive crests or troughs.

Worked Example

Problem: A student investigates wave types using a stretched horizontal slinky fixed at one end. The student holds the free end and creates two separate waves that travel from the free end toward the fixed end (along the length of the slinky): Wave 1: The student moves the free end back and forth perpendicular to the length of the slinky. Wave 2: The student moves the free end forward and back parallel to the length of the slinky. Which correctly classifies Wave 1 and Wave 2? A. Wave 1: Longitudinal, Wave 2: Longitudinal B. Wave 1: Transverse, Wave 2: Longitudinal C. Wave 1: Longitudinal, Wave 2: Transverse D. Wave 1: Transverse, Wave 2: Transverse

Solution steps:

1. Recall the classification rule: transverse waves have particle displacement perpendicular to wave propagation direction; longitudinal waves have particle displacement parallel to propagation direction.
2. For both waves, propagation direction is along the length of the slinky (from free to fixed end).
3. Wave 1 displacement is perpendicular to slinky length, so it matches the definition of transverse. Eliminate options A and C.
4. Wave 2 displacement is parallel to slinky length, so it matches the definition of longitudinal. Eliminate option D. Final answer: B.

Exam tip: AP exam questions often mix classification axes (mechanical vs electromagnetic and transverse vs longitudinal). Always read the question carefully to confirm which classification it asks for, don’t assume.

3. Core Measurable Properties of Periodic Waves

All periodic (repeating) waves share four core measurable properties that you will use for all subsequent wave analysis. First: Period ($T$), defined as the time required for one full wave cycle to pass a fixed observation point, with units of seconds. Second: Frequency ($f$), defined as the number of full wave cycles that pass a fixed point per unit time, with units of hertz (1 Hz = 1 cycle per second). Period and frequency are inversely related, by definition: $$f=\frac{1}{T}\text{and}T=\frac{1}{f}$$ This relationship makes intuitive sense: if 2 cycles pass per second, each cycle must take 0.5 seconds. Third: Wavelength ($\lambda$), the distance between two identical points on consecutive wave cycles, with units of meters. Only one full cycle counts as one wavelength. Fourth: Amplitude ($A$), the maximum displacement of the medium from its equilibrium (rest) position. A key AP-tested relationship connects amplitude to the energy carried by the wave: energy is proportional to the square of the amplitude, or $E\propto A^2$. This means doubling the amplitude quadruples the energy carried by the wave.

Worked Example

Problem: A floating buoy in the ocean bobs up and down with the passage of periodic water waves. A student records that 5 full up-down cycles of the buoy take 10 seconds total. The student also measures the horizontal distance between the first and third consecutive wave crests passing the buoy as 8.0 m. Calculate the period, frequency, and wavelength of the water wave.

Solution steps:

1. Calculate period $T$: period is time per full cycle. We have 5 cycles in 10 seconds, so: $T=\frac{\text{total time}}{\text{number of cycles}}=\frac{10\text{s}}{5}=2.0\text{s}$
2. Calculate frequency using the inverse relationship $f=1/T$: $f=\frac{1}{2.0\text{s}}=0.50\text{Hz}$
3. Calculate wavelength: the distance between the first and third crest spans two full wavelengths (first to second crest is 1λ, second to third is another 1λ). So $2\lambda =8.0\text{m}$, so $\lambda =\frac{8.0\text{m}}{2}=4.0\text{m}$. Final values: $T=2.0\text{s}$, $f=0.50\text{Hz}$, $\lambda =4.0\text{m}$.

Exam tip: When asked for wavelength from a given distance between crests, always count how many full wavelengths are between the given points, don’t automatically assume the distance given is a single wavelength. AP questions frequently test this common mistake.

4. The Universal Wave Speed Formula

Wave speed is the speed at which the wave disturbance propagates through the medium. The relationship between wave speed, frequency, and wavelength holds for all periodic waves, regardless of type, which is why it is called the universal wave formula: $$v=f\lambda$$ The units check out: frequency has units of 1/s, wavelength has units of m, so multiplying gives m/s, the correct unit for speed. Intuitively, if 2 full cycles pass per second, and each cycle is 3 m long, then the wave travels 6 m per second, so $v=2\times 3=6$ m/s, which matches the formula. A critical AP-tested rule: for mechanical waves in a given medium, wave speed depends only on the properties of the medium, not on frequency or amplitude. For example, sound travels at ~340 m/s in room-temperature air, regardless of how high or low its frequency is. If you increase the frequency of a wave in a fixed medium, speed stays the same, so wavelength decreases proportionally to keep $v=f\lambda$ balanced.

Worked Example

Problem: A loudspeaker produces a sound wave of constant frequency 200 Hz. The speed of sound in air is 340 m/s, and the speed of sound in freshwater is 1500 m/s. What is the wavelength of the sound wave in air, and what is its wavelength in freshwater?

Solution steps:

1. Rearrange the universal wave formula to solve for wavelength: $\lambda =\frac{v}{f}$.
2. Frequency is set by the source (the loudspeaker) and does not change when the wave moves from one medium to another, so $f=200$ Hz in both air and water.
3. Calculate wavelength in air: ${\lambda }_{\text{air}}=\frac{340\text{m/s}}{200\text{Hz}}=1.7\text{m}$.
4. Calculate wavelength in water: ${\lambda }_{\text{water}}=\frac{1500\text{m/s}}{200\text{Hz}}=7.5\text{m}$. Final answer: 1.7 m in air, 7.5 m in water.

Exam tip: Remember that frequency is determined by the source of the wave, while speed is determined by the medium. When a wave crosses from one medium to another, $f$ stays the same, $v$ changes, so $\lambda$ changes. This is the most common test of the wave speed formula on the AP exam.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calling sound a transverse wave because it is represented as a pressure graph with peaks and troughs on the y-axis. Why: Students confuse the graphical representation of longitudinal wave pressure with the actual direction of particle displacement. Correct move: Always check the direction of particle displacement relative to propagation; for sound, particles oscillate parallel to motion, so it is longitudinal.
• Wrong move: Assuming that increasing the frequency of a wave in a fixed medium increases its wave speed. Why: Students confuse source properties (frequency) with medium properties (speed) and incorrectly extrapolate from $v=f\lambda$ that higher $f$ means higher $v$. Correct move: Always remember wave speed depends only on medium properties for mechanical waves; if medium is unchanged, $v$ is constant, and higher $f$ only reduces $\lambda$.
• Wrong move: Calculating wavelength as the distance between a crest and the next trough instead of two identical points. Why: Students forget wavelength is the length of one full cycle, and confuse half a cycle with a full cycle. Correct move: Always measure wavelength between two consecutive points at the same displacement and same direction of motion (crest to crest, trough to trough, compression to compression).
• Wrong move: Thinking that waves transfer matter from the source to the detector, not just energy. Why: Students see water moving toward shore and assume the water itself travels from the open ocean to the coast. Correct move: Memorize the core definition: waves transfer energy without net transfer of matter, and confirm this on any definition-based question.
• Wrong move: Calculating energy proportional to $A$ (amplitude) instead of $A^2$. Why: Students mix up the relationship between amplitude and energy with linear relationships for other wave properties. Correct move: Always apply $E\propto A^2$ for wave energy: double $A$, $E$ quadruples; triple $A$, $E$ is 9 times larger.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A student generates a periodic transverse wave on a stretched string. The student then increases the amplitude of the wave by a factor of 2, while keeping frequency, string tension, and string mass per unit length unchanged. What happens to the wave speed and the energy carried by the wave? A. Wave speed doubles, energy doubles B. Wave speed stays the same, energy doubles C. Wave speed stays the same, energy quadruples D. Wave speed quadruples, energy quadruples

Worked Solution: First, recall that wave speed on a string depends only on the medium properties (tension and mass per unit length, which are unchanged here). Changing amplitude and frequency does not change the wave speed, so we can eliminate options A and D. Next, the energy of a wave is proportional to the square of the amplitude: $E\propto A^2$. If amplitude increases by a factor of 2, energy increases by $2^2=4$, so energy quadruples. This matches option C. Correct answer: C.

Question 2 (Free Response)

A longitudinal sound wave travels through air at a speed of 343 m/s. The frequency of the wave is 440 Hz, the standard A4 pitch used to tune musical instruments. (a) Calculate the period and wavelength of the sound wave in air. (b) The sound wave travels from air into a solid block of aluminum, where the speed of sound is 6420 m/s. Does the frequency of the sound increase, decrease, or stay the same? Justify your answer. (c) Calculate the wavelength of the sound wave in aluminum.

Worked Solution: (a) Period is the inverse of frequency: $T=1/f=1/(440\text{Hz})\approx 2.3\times 10^{-3}\text{s}$. Use the wave speed formula $v=f\lambda$ rearranged to solve for wavelength: $\lambda =v/f=343\text{m/s}/440\text{Hz}\approx 0.78\text{m}$. (b) The frequency stays the same. Frequency of a wave is determined by the source of the wave (the vibrating object creating the sound, in this case the instrument), not by the medium the wave travels through. When the wave moves into a new medium, only speed and wavelength change. (c) Frequency remains 440 Hz. Calculate wavelength with the wave speed formula: $\lambda =v_{\text{Al}}/f=6420\text{m/s}/440\text{Hz}\approx 15\text{m}$.

Question 3 (Application / Real-World Style)

Ultrasound is used to measure the depth of water below a research boat. The ultrasound transmitter sends a high-frequency sound wave straight down through the water, and the echo reflected off the lake bottom returns to the detector on the boat 0.020 seconds after it was sent. The speed of sound in freshwater is 1500 m/s, and the frequency of the ultrasound is 2.0 × 10⁶ Hz. What is the depth of the water below the boat, and what is the wavelength of the ultrasound in water?

Worked Solution: The sound wave travels down to the lake bottom and back up to the boat, so the total distance traveled by the sound is twice the depth of the water ($d_{\text{total}}=2h$, where $h$ is depth). Use constant speed kinematics $v=d/t$ to find total distance: $d_{\text{total}}=v\times t=1500\text{m/s}\times 0.020\text{s}=30\text{m}$. Solve for depth: $h=30\text{m}/2=15\text{m}$. Next, find wavelength using the universal wave formula $\lambda =v/f=1500\text{m/s}/(2.0\times 10^6\text{Hz})=7.5\times 10^{-4}\text{m}=0.75\text{mm}$. In context, the very short wavelength of ultrasound allows it to detect small objects and produce accurate depth measurements for underwater surveying.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all subsequent wave topics in Unit 7. Next, you will apply the wave speed formula and core wave properties from this chapter to the analysis of wave interference, standing waves on strings, and standing sound waves in open and closed tubes. Without mastering wave classifications, wavelength-frequency relationships, and the universal wave speed formula here, you will not be able to solve for standing wave wavelengths or harmonics, which make up the majority of FRQ points on wave topics. This topic also builds on simple harmonic motion from earlier in the unit, as both describe periodic motion that transfers energy, and it provides background for conceptual questions about wave behavior across the exam. Follow-on topics to study next: Wave Interference and Superposition Standing Waves Simple Harmonic Motion Basics