Wave Properties and the Wave Equation
The 2004 Boxing Day Indian Ocean tsunami had a wavelength of 200–300 km, a wave speed of 800 km/h (222 m/s) in deep ocean, and a period of approximately 18 minutes. Applying the wave equation: $f = v/\lambda = 222/250{,}000 = 8.9 \times 10^{-4}$ Hz; period $T = 1/f = 1{,}124$ s $\approx 18.7$ min, consistent with observations that warned coastal areas had 15–30 minutes of lead time.
Practise this lesson
Three printable worksheets that build from foundations to mastery.
Two waves travel through the same rope. One has twice the frequency of the other. Does that automatically mean the higher-frequency wave travels twice as fast?
Warm-up, in the same medium, if frequency doubles, what happens to wavelength?
Know
- The wave equation $v = f\lambda$
- The relationship $T = 1/f$
- How to identify λ on a displacement-distance graph
- How to identify T on a displacement-time graph
Understand
- Why distance graphs and time graphs tell different things
- Why frequency and wavelength trade off when speed is fixed
- What phase difference means
Can Do
- Solve basic wave equation problems
- Convert between period and frequency
- Interpret wave graphs without mixing up λ and T
Core Content
Linking wave speed, frequency, and wavelength in one compact relationship
On 26 December 2004, oceanographic sensors picked up the Indian Ocean tsunami travelling at 800 km/h (222 m/s) across deep water with wave crests spaced roughly 250 km apart. The time between one crest and the next passing a fixed buoy was about 18.7 minutes. You can verify this directly: if $v = 222$ m/s and $\lambda = 250{,}000$ m, then $f = v/\lambda = 8.9 \times 10^{-4}$ Hz, so period $T = 1/f = 1{,}124$ s $\approx 18.7$ min. Three measurable quantities, speed, wavelength, frequency, linked by one equation: $v = f\lambda$.
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.
$v = f\lambda$ · $f = v/\lambda$ · $\lambda = v/f$
$T = 1/f$ · $f = 1/T$
A wave travels along a rope with frequency 8 Hz and wavelength 0.50 m. Find the wave speed and period.
- Speed: $v = f\lambda = 8 \times 0.50 = 4.0 \text{ m/s}$
- Period: $T = 1/f = 1/8 = 0.125 \text{ s}$
What if frequency doubled? Speed stays the same (same rope), so λ halves: $\lambda = 4.0/16 = 0.25 \text{ m}$.
The wave equation $v = f\lambda$ links wave speed (m/s), frequency (Hz) and wavelength (m); in the same medium $v$ is approximately constant so doubling frequency halves wavelength. Period $T = 1/f$ (s).
Pause, copy the highlighted definition and formula into your book before moving on.
A wave has frequency 10 Hz and wavelength 0.30 m. What is its speed?
Distance graphs show wavelength; time graphs show period, they look identical but mean different things
We just saw that $v = f\lambda$ links wave speed, frequency and wavelength. That raises a question: given a wave graph, how do you tell whether the horizontal axis gives wavelength or period? This card answers it → always check the axis label: distance → read $\lambda$; time → read $T$.
A displacement-distance graph is a snapshot across space. A displacement-time graph is a history of one point over time.
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. Amplitude is read the same way on both graphs.
Top: read wavelength across space. Bottom: read period across time. Always check the horizontal axis label first.
On a displacement-distance graph, the crest-to-crest spacing is the wavelength $\lambda$ (m); on a displacement-time graph, the crest-to-crest spacing is the period $T$ (s). Amplitude is the maximum displacement from equilibrium on either graph.
Add the highlighted graph-reading rule to your notes before the check below.
On a displacement-time graph, the horizontal spacing between crests gives the wavelength.
Amplitude is measured the same way on both displacement-distance and displacement-time graphs.
Phase difference and why the medium is the boss of wave speed
We just saw how to read $\lambda$ and $T$ from graphs. That raises a question: what happens to wavelength when a wave crosses into a different medium? This card answers it → frequency stays constant (set by source) while speed and wavelength change together via $\lambda = v/f$.
Two points are in phase if they are at the same stage of oscillation. Points separated by one full wavelength (or period) are in phase. Points separated by half a wavelength are in antiphase.
Wave speed is determined by the properties of the medium: tension and linear density for strings, temperature and composition for sound in gases, refractive index for light. When a wave crosses into a new medium, speed changes. If the source stays the same, frequency stays constant, which means wavelength must change: $\lambda = v/f$.
When a wave crosses a boundary, frequency remains constant (set by the source) and speed changes, so wavelength adjusts via $\lambda = v/f$. Two points separated by $\lambda$ (or $T$) are in phase; separated by $\lambda/2$ they are in antiphase.
Pause, write the highlighted boundary rule into your book.
A wave has period 0.25 s. Its frequency is _____ Hz.
Find the missing quantity:
- $f = 12\ \text{Hz}$, $\lambda = 0.25\ \text{m}$ → find $v$
- $v = 330\ \text{m/s}$, $f = 660\ \text{Hz}$ → find $\lambda$
- $T = 0.05\ \text{s}$ → find $f$
- $f = 2.5\ \text{Hz}$ → find $T$
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.
A sound wave of frequency 500 Hz travels through air at 340 m/s. It then enters water where wave speed is 1500 m/s.
- Calculate the wavelength in air.
- State what happens to the frequency in water, and explain why.
- Calculate the wavelength in water.
- A student claims the wave slows down in water. Explain whether this is correct.
Two points on a wave are separated by one full wavelength. Their phase relationship is:
Wave Equation
- $v = f\lambda$ · $f = v/\lambda$ · $\lambda = v/f$
- In same medium: $v$ constant; $f\uparrow$ → $\lambda\downarrow$
Period & Frequency
- $T = 1/f$ · $f = 1/T$
- Frequency set by source; stays constant across boundaries
Graph Reading
- Displacement-distance → read λ
- Displacement-time → read T
Phase
- In phase: separated by λ or T
- Antiphase: separated by λ/2 or T/2
A wave travels at 12 m/s and has wavelength 1.5 m. What is its frequency?
Five questions drawn from the lesson bank.
UnderstandBand 3(3 marks) 1. Explain the difference between wavelength and period by referring to the correct type of graph for each.
ApplyBand 4(3 marks) 2. A wave travels at 12 m/s and has wavelength 1.5 m. Calculate its frequency and period.
AnalyseBand 6(4 marks) 3. 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.
Show all answers
Activity 1, Quick Conversions
1. $v = f\lambda = 12 \times 0.25 = 3.0$ m/s 2. $\lambda = v/f = 330/660 = 0.50$ m 3. $f = 1/T = 1/0.05 = 20$ Hz 4. $T = 1/f = 1/2.5 = 0.40$ s
Activity 4, Boundary
1. $\lambda_\text{air} = 340/500 = 0.68$ m 2. Frequency stays 500 Hz (set by source) 3. $\lambda_\text{water} = 1500/500 = 3.0$ m 4. The student is incorrect, sound speeds up in water (1500 m/s > 340 m/s).
Short Answer, Model Answers
Q1: Wavelength is a spatial quantity read from a displacement-distance graph as the horizontal spacing between in-phase points. Period is a time quantity read from a displacement-time graph as the time for one full oscillation.
Q2: $f = v/\lambda = 12/1.5 = 8$ Hz; $T = 1/f = 1/8 = 0.125$ s.
Q3: No. Equal displacement at one instant does not prove in-phase. Two points with the same displacement can be moving in opposite directions. In-phase requires same displacement AND same direction of motion.
The 2004 Boxing Day tsunami makes the answer to the Think First question concrete. A higher-frequency wave on the same rope does not automatically travel faster, wave speed in a fixed medium is determined by the medium, not the source. The tsunami in deep ocean travelled at 222 m/s regardless of frequency: with wavelength 250 km, $f = 222/250{,}000 = 8.9 \times 10^{-4}$ Hz and $T \approx 18.7$ min. Halving the wavelength would double the frequency but leave the speed unchanged. That 18.7-minute period was what gave coastal regions their evacuation window.