Once we can name amplitude, wavelength, period and frequency precisely, we can connect them with one compact relationship: $v = f\lambda$. The challenge is not just plugging in numbers. It is reading the wave correctly first.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Two waves travel through the same rope. One has twice the frequency of the other. Does that automatically mean it travels twice as fast? Predict your answer before using any formula.
Type your prediction below. You will revisit it at the end.
Write your prediction in your book. You will revisit it at the end.
Try to justify your prediction in words, not with a guess.
Wrong: A machine can produce more work output than work input.
Right: Energy is conserved; machines can only transform energy, never create it (efficiency ≤ 100%).
📚 Core Content
Wave speed depends on how quickly the source oscillates and how far apart repeating points on the wave are.
The wave equation links the three most important measurable properties of a wave. If a source vibrates more times each second, frequency increases. If the wave still travels through the same medium, the wavelength must usually decrease so that the product $f\lambda$ stays equal to the same speed. This inverse relationship between frequency and wavelength is one of the most tested ideas in HSC wave questions.
This is why the medium matters. A rope, air column, or water surface largely determines wave speed. The source changes frequency. The wave adjusts its wavelength in response. Think of a guitar string: tightening the string (changing the medium) increases wave speed and therefore raises the pitch for the same wavelength. Plucking harder increases amplitude but does not change speed, frequency, or wavelength significantly.
| Medium | Speed fixed by medium? | If frequency increases... | Then wavelength... |
|---|---|---|---|
| Same rope under same tension | Approximately yes | Increases | Decreases |
| Sound in same air conditions | Approximately yes | Increases | Decreases |
| Different medium | No — speed may change | May stay same at boundary | Often changes with speed |
Frequency counts oscillations each second. Period measures the time for one oscillation. They are reciprocals.
If a wave has frequency 5 Hz, it completes 5 oscillations every second. That means each oscillation takes $1/5$ of a second, so the period is 0.2 s. If frequency rises, period falls. This relationship is mathematically simple but conceptually deep: frequency tells you how crowded the oscillations are in time, while period tells you the duration of one complete cycle from start to finish.
It is essential to remember that period and frequency describe the source, not the medium. When a wave crosses from air into water, its speed and wavelength change, but its frequency (and therefore period) stay the same because they are determined by whatever is creating the wave. A tuning fork vibrating at 440 Hz produces sound at 440 Hz regardless of whether that sound travels through air, water, or steel.
"How many cycles happen per second?"
"How long does one cycle take?"
A displacement-distance graph is a snapshot across space. A displacement-time graph is a history of one point over time.
These two graphs look almost identical — a smooth sinusoidal curve — but they mean completely different things. On a displacement-distance graph, the horizontal axis represents position, so the distance between two adjacent crests is the wavelength. On a displacement-time graph, the horizontal axis represents time, so the distance between two adjacent crests is the period. Misreading the axis is the single biggest source of lost marks in wave graph questions.
Amplitude is read the same way on both graphs: it is the maximum vertical displacement from the equilibrium line. But never assume the horizontal spacing is wavelength unless you have confirmed the axis label. A good habit is to write "axis = distance → λ" or "axis = time → T" on your working before doing any calculation.
Top read wavelength across space. Bottom read period across time. The shape may look similar, but the horizontal axis changes the meaning.
Two points are in phase if they are at the same stage of oscillation. Phase difference tells us how far "out of step" they are.
On a wave graph, points separated by one full wavelength are in phase. They have the same displacement and are moving the same way. Points separated by half a wavelength are in antiphase: they have opposite displacement and opposite motion direction. The same idea applies over time: one full period means back in phase. A phase difference of $\pi$ radians (or 180°) means antiphase; $\pi/2$ radians means quarter-cycle difference.
Phase is a subtle but powerful concept. When two sound waves arrive at your ear in phase, they constructively interfere and sound louder. When they arrive in antiphase, they can cancel each other out. Engineers use phase differences to design noise-cancelling headphones and to tune concert hall acoustics. At this level, you need to be able to identify in-phase and antiphase points on a diagram and explain what that means for their motion.
| Separation | Phase Relationship | Meaning |
|---|---|---|
| $\lambda$ or $T$ | In phase | Same stage of oscillation |
| $\lambda/2$ or $T/2$ | Antiphase | Half a cycle apart |
| $\lambda/4$ or $T/4$ | Quarter-cycle difference | One point reaches an extreme one quarter cycle later |
Wave speed is not something the wave decides for itself. It is determined by the properties of the medium through which the wave travels.
For a wave on a string, speed depends on tension and linear density: tighter strings and lighter strings give faster waves. For sound in air, speed depends on temperature and the molecular composition of the gas: warmer air means faster sound because molecules collide more frequently. For light in glass, speed is reduced compared to vacuum because the electromagnetic fields interact with the atoms in the material. In every case, the medium is the boss.
This is why a question that says "the wave enters a new medium" is really telling you that the speed has changed. If the source stays the same, frequency stays constant, which means wavelength must change to keep $v = f\lambda$ true. Blue light entering water from air slows down, so its wavelength decreases. The colour stays blue because frequency is unchanged, but the waves are now more compressed.
✏️ Worked Examples
Scenario: A wave travels along a rope with frequency 8 Hz and wavelength 0.50 m. Find the wave speed and the period.
The source frequency doubled while the rope and tension stayed unchanged? The speed would stay about the same, so the wavelength would halve.
Scenario: A displacement-time graph shows 3 full oscillations in 0.60 s. The maximum displacement is 4 cm. Find the period, frequency and amplitude.
If the graph were displacement-distance instead, the same horizontal spacing would represent wavelength instead of period. Always read the axis first.
Visual Break
🏃 Activities
Find the missing quantity in each case:
A student says the horizontal spacing between two crests on a displacement-time graph is the wavelength. Explain why this is incorrect, and state what that spacing actually represents.
On a transverse wave, points A and B are separated by $\lambda/2$, and points A and C are separated by $\lambda$. State the phase relationship between A and B, and between A and C.
A sound wave of frequency 500 Hz travels through air at $340\ \text{m/s}$. It then enters water where the wave speed is $1500\ \text{m/s}$.
Earlier you were asked: If two waves on the same rope have different frequencies, does that automatically mean the higher-frequency wave travels faster?
The full answer: not necessarily. In the same medium, wave speed is mainly set by the medium. If the source frequency increases, the wavelength usually decreases so that $v = f\lambda$ still fits the same speed. Frequency and wavelength can change together without changing speed.
Now revisit your prediction. What assumption did you make about speed, frequency or wavelength?
Annotate your original prediction in your book with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
✅ Check Your Understanding
1. A wave has frequency 10 Hz and wavelength 0.30 m. What is its speed?
2. Which quantity is read directly from the horizontal axis of a displacement-time graph?
3. A wave has period 0.25 s. What is its frequency?
4. Two points on a wave are separated by one full wavelength. Their phase relationship is:
5. In the same medium, the source frequency doubles. What happens to the wavelength?
6. Which statement is the best correction to "amplitude is the distance from crest to trough"?
7. Explain the difference between wavelength and period by referring to the correct type of graph for each. 3 MARKS
8. A wave travels at 12 m/s and has wavelength 1.5 m. Calculate its frequency and period. 3 MARKS
9. Two points on a wave have the same displacement at one instant. Does that prove they are in phase? Explain using phase difference and motion direction. 4 MARKS
1. $v = f\lambda = 12 \times 0.25 = 3.0\ \text{m/s}$
2. $\lambda = v/f = 330/660 = 0.50\ \text{m}$
3. $f = 1/T = 1/0.05 = 20\ \text{Hz}$
4. $T = 1/f = 1/2.5 = 0.40\ \text{s}$
1. $\lambda_{\text{air}} = v/f = 340/500 = 0.68\ \text{m}$
2. The frequency stays at 500 Hz in water because frequency is determined by the source and does not change when the wave enters a different medium.
3. $\lambda_{\text{water}} = v/f = 1500/500 = 3.0\ \text{m}$
4. The student is incorrect. Sound travels faster in water ($1500\ \text{m/s}$) than in air ($340\ \text{m/s}$). The wave speeds up in water, not slows down. This is why marine animals can communicate over long distances underwater.
1. C — $v = f\lambda = 10 \times 0.30 = 3.0\ \text{m/s}$.
2. B — a displacement-time graph gives period along the time axis.
3. D — $f = 1/T = 1/0.25 = 4\ \text{Hz}$.
4. A — one wavelength means back in phase.
5. C — in the same medium, doubling $f$ halves $\lambda$ if speed stays fixed.
6. B — crest-to-trough is twice the amplitude.
Q7 (3 marks): Wavelength is a spatial quantity. It is read from a displacement-distance graph as the horizontal spacing between repeating points such as crest to crest. Period is a time quantity. It is read from a displacement-time graph as the time for one full oscillation. They may look visually similar on graphs, but they describe different things because the horizontal axis is different.
Q8 (3 marks): $f = v/\lambda = 12/1.5 = 8\ \text{Hz}$. Then $T = 1/f = 1/8 = 0.125\ \text{s}$.
Q9 (4 marks): No. Equal displacement at one instant does not prove two points are in phase. To be in phase, the points must be at the same stage of oscillation, which means same displacement and same direction of motion. For example, two points on opposite sides of the equilibrium line can both pass through zero displacement, but one may be moving upward while the other moves downward. In that case they are not in phase even though the instantaneous displacement matches.
Sprint through questions on wave properties and the wave equation. Pool: lessons 1–2.
Tick when you have finished the activities and checked the answers.