E. Waves and Sound · Waves and SoundThe Universal Wave Equation: v = fλ
E2.4 E2.1 A1.6 A1.13About 70 minutes
Learning goals- I can derive the universal wave equation v = fλ from the definitions of speed, period and wavelength.
- I can solve problems for speed, frequency or wavelength, rearranging the equation as needed.
- I can explain that frequency is set by the source while speed is set by the medium.
- I can use a graph to show how wavelength depends on frequency in one medium.
Words to know
universal wave equation
v = fλ: wave speed equals frequency times wavelength. It applies to every kind of wave.
wave speed (v)
How fast the disturbance travels through the medium, in metres per second (m/s).
inverse relationship
When one quantity goes up, the other goes down so that their product stays constant (here f × λ = v).
refraction (of a wave)
The change in speed (and often direction) of a wave as it passes from one medium into another.
One equation for every wave
The same equation works for ripples in a puddle, sound from a snowplough, waves on a guitar string, and the light and radio waves you met in Grade 10. It is called the universal wave equation, and it is the most-used equation in this unit.
Deriving v = fλ
Watch one crest of a periodic wave. In the time it takes the source to make one complete cycle (one period, T), the crest moves forward exactly one wavelength, λ. The next crest has just been made behind it.
Speed is distance ÷ time, so
v = λ ÷ T
Since f = 1/T, dividing by T is the same as multiplying by f:
v = f λ
With units: (cycles per second) × (metres per cycle) = metres per second. ✓
Rearranged: f = v/λ and λ = v/f.
Two snapshots of the same wave, one period apart. In time T the crest marked with a dot has moved exactly one wavelength, so v = λ/T = fλ.Who controls what?
This is the most important idea for using the equation correctly:
- The source sets the frequency. A guitar string vibrating 196 times per second pushes the air 196 times per second. The frequency does not change when the wave enters a new medium.
- The medium sets the speed. Sound in room-temperature air travels at about 343 m/s, whatever made it.
- The wavelength adjusts to fit: λ = v/f.
So, in one medium (constant v), frequency and wavelength are inversely proportional: double the frequency and the wavelength is halved. A high note has a short wavelength and a low note has a long wavelength.
When a wave moves into a new medium, f stays the same, so if v changes, λ changes by the same factor. This change of speed at a boundary is called refraction. Ocean waves slow down as they reach shallow water near a beach, so their wavelength shrinks and the crests bunch up.
Water waves (seen from above) move from deep into shallow water. The frequency stays 0.50 Hz, so when the speed halves, the wavelength halves too.Worked exampleExample 1: Speed on a string
A wave on a stretched spring has a frequency of 12 Hz and a wavelength of 0.50 m. Find its speed.
- Write what you know: f = 12 Hz, λ = 0.50 m, v = ?
- v = fλ = (12 Hz)(0.50 m) = 6.0 m/s.
Worked exampleExample 2: Wavelength of a musical note
The note A above middle C has a frequency of 440 Hz. Find its wavelength in air where the speed of sound is 343 m/s.
- Rearrange: λ = v/f.
- λ = 343 m/s ÷ 440 Hz = 0.7795… m
- Round to three significant figures: λ = 0.780 m (78.0 cm).
About 0.780 m, roughly the length of a child's arm span.
Worked exampleExample 3: Into a new medium
An ocean wave has a wavelength of 12 m in deep water and travels at 6.0 m/s. Near the shore it slows to 3.0 m/s. Find its frequency and its new wavelength.
- Frequency in deep water: f = v/λ = 6.0 ÷ 12 = 0.50 Hz.
- The frequency does not change at a boundary, so f is still 0.50 Hz in the shallows.
- New wavelength: λ = v/f = 3.0 ÷ 0.50 = 6.0 m.
- Check: the speed halved, so the wavelength halved. ✓
f = 0.50 Hz (unchanged); new λ = 6.0 m.
Investigating the relationship
To show that v = fλ, a class can shake a long spring at different frequencies (counted with a metronome) and measure the wavelength each time. Here is a set of results for one spring at one tension:
| Frequency f (Hz) | Wavelength λ (m) | f × λ (m/s) | 1/f (s) |
|---|
| 2.0 | 2.4 | 4.8 | 0.50 |
| 3.0 | 1.6 | 4.8 | 0.33 |
| 4.0 | 1.2 | 4.8 | 0.25 |
| 6.0 | 0.80 | 4.8 | 0.17 |
| 8.0 | 0.60 | 4.8 | 0.13 |
Sample data: the product f × λ is constant, so the speed in this spring is 4.8 m/s. Graphing λ against f gives a curve (an inverse relationship). Graphing λ against 1/f gives a straight line through the origin, and its slope is the wave speed, 4.8 m/s.Graphing tip
Scientists like straight-line graphs because a straight line is easy to recognise and its slope has meaning. If you suspect y ∝ 1/x, plot y against 1/x. If the points line up through the origin, you have confirmed the relationship. This is called linearizing a graph.
Just for funWhy did the wave refuse to change its frequency when it went into the water?
It said, "I may slow down, and I may shorten up, but I never forget where I came from." (The source sets the frequency!)
Hands-on activityRipples in a baking dish
You need: a clear glass or metal baking dish, water, a pencil, a ruler, a free metronome app (or a friend counting beats), and a phone that can record slow-motion video if possible. Wipe up spills so nobody slips.
- Pour about 1 cm of water into the dish and let it settle. Put the ruler along one side so you can see it in a video.
- Set the metronome to 2 beats per second (120 beats per minute). Tap the water surface gently with the pencil tip on each beat near one end. You are making waves with f = 2 Hz.
- Record a short video from above. Pause it and use the ruler to estimate the distance between ripples (λ).
- Calculate v = fλ.
- Now check the speed a second way: make one tap and time (or count video frames for) how long the ripple takes to cross the dish. Use v = d/t.
- Try 4 beats per second. What happens to the wavelength?
What to look for: in 1 cm of water, ripples move at a few tens of centimetres per second, so at 2 Hz the ripples are roughly 10 cm to 15 cm apart (only two or three fit in the dish). Your two speed values should roughly agree. At 4 Hz the ripples should be about half as far apart: same speed, double the frequency, half the wavelength. Very small ripples are also affected by surface tension, so expect some scatter.
Your turn
1A wave has f = 5.0 Hz and λ = 3.0 m. Find v.
v = fλ = 5.0 × 3.0 = 15 m/s.
2Sound travels at 343 m/s in air. Find the wavelength of a 1000 Hz tone.
λ = v/f = 343 ÷ 1000 = 0.343 m.
3A wave on a rope travels at 8.0 m/s with wavelength 0.40 m. Find its frequency.
f = v/λ = 8.0 ÷ 0.40 = 20 Hz.
4In one medium, the frequency of a wave triples. What happens to its wavelength?
The speed stays the same, so λ becomes one third as long.
5A wave passes from a rope into a heavier rope where it moves more slowly. Which of v, f and λ change?
v decreases and λ decreases; f stays the same (set by the source).
6Find the period of a wave with v = 1500 m/s and λ = 3.0 m.
f = v/λ = 1500 ÷ 3.0 = 500 Hz, so T = 1/500 = 0.0020 s (2.0 ms).
7Human hearing runs from about 20 Hz to 20 000 Hz. Using 343 m/s, find the longest and shortest wavelengths we can hear.
Longest: λ = 343 ÷ 20 ≈ 17 m. Shortest: λ = 343 ÷ 20 000 ≈ 0.017 m (1.7 cm).
More practice: warm-up
1Write the universal wave equation and name each symbol with its unit.
v = fλ: v = speed (m/s), f = frequency (Hz), λ = wavelength (m).
2Rearrange v = fλ to solve for λ.
3What sets the frequency of a wave: the source or the medium?
4f = 2.0 Hz and λ = 1.5 m. Find v.
5v = 340 m/s and f = 170 Hz. Find λ.
More practice: on your own
1An FM radio station broadcasts at 103.5 MHz. Radio waves travel at 3.00 × 10⁸ m/s. Find the wavelength.
λ = v/f = (3.00 × 10⁸) ÷ (1.035 × 10⁸) = 2.90 m.
2A dolphin's click has a frequency of 100 kHz. In sea water, sound travels at about 1530 m/s. Find the wavelength.
λ = 1530 ÷ (1.00 × 10⁵) = 0.0153 m ≈ 1.5 cm.
3A wave on a string has a period of 0.040 s and a wavelength of 1.2 m. Find its speed.
f = 1/0.040 = 25 Hz; v = 25 × 1.2 = 30 m/s. (Or v = λ/T = 1.2 ÷ 0.040 = 30 m/s.)
4A sound wave of 512 Hz passes from air (343 m/s) into water (1480 m/s). Find its wavelength in each medium.
Air: λ = 343 ÷ 512 = 0.670 m. Water: λ = 1480 ÷ 512 = 2.89 m. The frequency is the same; the wavelength is longer in water because the speed is greater.
5A student graphs λ against 1/f for waves on a spring and gets a straight line with slope 2.5 m/s. What does the slope mean, and what wavelength would a 5.0 Hz wave have?
The slope is the wave speed, 2.5 m/s (because λ = v × (1/f)). λ = 2.5 ÷ 5.0 = 0.50 m.
Challenge
1A boat bobs up and down 12 times per minute as waves pass. The crests are 18 m apart. How long does a crest take to travel the length of a 45 m dock?
f = 12 ÷ 60 s = 0.20 Hz. v = fλ = 0.20 × 18 = 3.6 m/s. t = d/v = 45 ÷ 3.6 = 12.5 s (about 13 s).
2A wave travels from medium A (v = 4.0 m/s) into medium B (v = 6.0 m/s). In A the wavelength is 0.80 m. Find f and the wavelength in B. Then explain why the ratio of the wavelengths equals the ratio of the speeds.
f = 4.0 ÷ 0.80 = 5.0 Hz (same in both). λ_B = 6.0 ÷ 5.0 = 1.2 m. Since λ = v/f and f is the same, λ_B/λ_A = (v_B/f)/(v_A/f) = v_B/v_A = 6.0/4.0 = 1.5 = 1.2/0.80. ✓
3A tuning fork makes a sound with wavelength 1.30 m in air at a temperature where sound travels at 338 m/s. On a warmer day the speed of sound is 346 m/s. What is the wavelength now? Does the pitch change?
f = 338 ÷ 1.30 = 260 Hz (the fork sets this). New λ = 346 ÷ 260 = 1.33 m. The pitch (frequency) does not change; only the wavelength does.
Check your understandingQuiz
1. Which equation is the universal wave equation?
2. A wave has f = 4.0 Hz and λ = 2.5 m. Its speed is…
3. In one medium, the frequency of a wave is doubled. The wavelength…
4. When a wave passes into a new medium, which quantity stays the same?
5. Sound travels at 343 m/s. The wavelength of a 686 Hz tone is…
6. A graph of wavelength against 1/frequency for waves in one medium is a straight line. Its slope is…
For parents and teachersBig idea: expectation E2.4 asks students to investigate the relationship among wavelength, frequency and speed and to solve problems with it. The equation is simple; the hard part is knowing that the source sets f and the medium sets v.
How to help: Ask, "If the wave goes into a new material, what stays the same?" until the answer "frequency" is automatic. Check that every answer has a unit.
Common difficulties:
- Rearranging incorrectly (λ = f/v is a common slip). Units are a good check: m/s ÷ Hz = m.
- Unit prefixes (kHz, MHz). Convert to hertz first.
- Thinking a higher frequency travels faster. In one medium all frequencies travel at the same speed (that is why the whole orchestra reaches your ears together).