Amplitude, wavelength, frequency, period & wave speed
AQA GCSE Physics 4.6 | Year 11 Higher Tier
🌊 Define amplitude, wavelength, frequency and period for a wave
📐 Identify wave properties from diagrams and oscilloscope traces
⚡ Use the wave speed equation v = fλ to solve problems
🔄 Use the relationship T = 1/f to link period and frequency
🔬 Describe how wave speed, frequency and wavelength are measured
🏆 Apply wave equations to real contexts including sound and light
🌊 What is a Wave?
A wave is a transfer of energy from one place to another through oscillations (vibrations). Importantly, waves transfer energy but not matter — the medium itself does not travel with the wave.
Transverse wave: Oscillations are perpendicular (at right angles) to the direction of wave travel. Examples: light, water waves, electromagnetic waves.
Longitudinal wave: Oscillations are parallel (along the same direction) to the direction of wave travel. Example: sound waves (compressions and rarefactions).
All waves share the same set of measurable properties: amplitude, wavelength, frequency and period. These properties allow us to fully describe any wave mathematically.
Waves transfer energy without transferring matter.
📏 Amplitude and Wavelength
Amplitude (A): The maximum displacement of a point on the wave from its undisturbed (equilibrium) position. Measured in metres (m).
The amplitude is the height from the midline to a crest (peak) or from the midline to a trough. It is not the full height from crest to trough (that would be 2A).
A larger amplitude means more energy is being transferred. For sound, greater amplitude means louder sound. For light, greater amplitude means brighter light.
Wavelength (λ): The distance between two successive identical points on a wave — for example, crest to crest or trough to trough. Measured in metres (m). Symbol: λ (lambda).
One complete wavelength represents one full cycle of the wave pattern. Wavelength can be very large (radio waves: hundreds of metres) or extremely small (gamma rays: 10⁻¹² m).
Symbol
Quantity
SI Unit
A
Amplitude
metre (m)
λ
Wavelength
metre (m)
⏱️ Frequency and Period
Frequency (f): The number of complete waves (cycles) passing a point per second. Measured in hertz (Hz). 1 Hz = 1 wave per second.
Human hearing range is approximately 20 Hz to 20 000 Hz. Ultrasound has frequencies above 20 000 Hz. Radio waves can have frequencies of millions or billions of Hz (MHz or GHz).
Period (T): The time taken for one complete wave cycle to pass a given point. Measured in seconds (s).
Frequency and period are reciprocals of each other — if you know one you can always find the other:
T = 1 ÷ f f = 1 ÷ T
For example, mains electricity in the UK has a frequency of 50 Hz, so its period is T = 1/50 = 0.02 s.
Symbol
Quantity
SI Unit
f
Frequency
hertz (Hz)
T
Period
second (s)
Higher frequency → shorter period. Lower frequency → longer period.
⚡ Wave Speed and the Wave Equation
Wave speed (v): The distance travelled by the wave per second. Measured in metres per second (m/s).
Wave speed depends on the medium through which the wave travels, not on the source of the wave. For example, sound travels at approximately 340 m/s in air, but much faster in water (~1500 m/s) and even faster in steel (~5000 m/s).
Light and all electromagnetic waves travel at 3 × 10⁸ m/s in a vacuum.
The fundamental relationship linking wave speed, frequency and wavelength is:
v = f × λ
Where: v = wave speed (m/s), f = frequency (Hz), λ = wavelength (m).
This equation can be rearranged to find any one of the three quantities if the other two are known:
f = v ÷ λ λ = v ÷ f
Symbol
Quantity
SI Unit
v
Wave speed
m/s
f
Frequency
Hz
λ
Wavelength
m
If wave speed stays constant (same medium), increasing frequency means wavelength decreases — they are inversely proportional.
🔬 Measuring Wave Properties
Wave properties can be measured in several practical ways depending on the type of wave.
Using an oscilloscope: An oscilloscope displays a voltage–time graph of a wave (e.g. from a microphone). The horizontal axis shows time and the vertical axis shows the signal amplitude.
Amplitude — read the maximum displacement from the centre line (in volts for electrical signals).
Period — measure the horizontal length of one complete cycle and multiply by the timebase setting (in seconds per division).
Frequency — calculated using f = 1/T once the period is known.
Measuring wavelength in a ripple tank: For water waves, a stroboscope can freeze the pattern, and a ruler is used to measure the distance across several wavelengths, then divide to find λ.
Measuring sound speed: A starting pistol is fired at a known distance; the time delay between the flash and the sound heard is measured using a stopwatch. v = distance ÷ time.
Using two microphones: A signal generator drives a loudspeaker. Two microphones connected to an oscilloscope detect the wave at different positions. Moving one microphone until the signals are in phase again gives the wavelength directly.
Always use multiple wavelengths when measuring with a ruler to reduce percentage uncertainty.
Example 1: A sound wave has a frequency of 440 Hz and travels through air at 340 m/s. Calculate the wavelength of the sound wave.
1 Write down the wave equation and identify known quantities. v = f × λ v = 340 m/s, f = 440 Hz, λ = ?
2 Rearrange for wavelength. λ = v ÷ f
3 Substitute values. λ = 340 ÷ 440
4 Calculate and include units. λ = 0.773 m (3 s.f.)
λ = 0.773 m — this is the wavelength of the musical note A above middle C.
Example 2: An oscilloscope trace shows that one complete wave cycle occupies 4 divisions on the horizontal axis. The timebase is set to 5 ms/div. Calculate (a) the period and (b) the frequency of the wave.
1 Calculate the period T from the oscilloscope trace. Number of divisions for one cycle = 4 Timebase = 5 ms/div = 5 × 10⁻³ s/div
2 T = number of divisions × timebase setting T = 4 × 5 × 10⁻³ = 0.020 s = 20 ms
3 Calculate frequency using f = 1 ÷ T f = 1 ÷ 0.020
4 f = 50 Hz
(a) Period T = 0.020 s (b) Frequency f = 50 Hz — the same as UK mains electrical frequency.
Example 3: A radio station broadcasts on a frequency of 97.3 MHz. The speed of radio waves is 3.0 × 10⁸ m/s. Calculate the wavelength of the radio waves.
1 Convert frequency to Hz. f = 97.3 MHz = 97.3 × 10⁶ Hz = 9.73 × 10⁷ Hz
2 Write the wave equation and rearrange for λ. v = f × λ → λ = v ÷ f
λ ≈ 3.08 m — FM radio waves have wavelengths of a few metres, which is why FM aerials are about 1.5 m long.
Example 4: Water waves in a ripple tank are measured. A student measures the distance across 10 complete wavelengths and finds it to be 24 cm. The frequency of the waves is 5 Hz. Calculate (a) the wavelength, (b) the wave speed and (c) the period of the waves.
1 Find wavelength from the measurement. Distance for 10 wavelengths = 24 cm = 0.24 m λ = 0.24 ÷ 10 = 0.024 m
2 Calculate wave speed using v = f × λ. v = 5 × 0.024 = 0.12 m/s
3 Calculate the period using T = 1 ÷ f. T = 1 ÷ 5 = 0.20 s
(a) λ = 0.024 m (b) v = 0.12 m/s (c) T = 0.20 s
Question 1. Which of the following correctly defines wavelength?
Question 2. A wave has a period of 0.004 s. What is its frequency?
Question 3. A microwave has a frequency of 2.45 GHz and travels at 3.0 × 10⁸ m/s. Calculate its wavelength in metres. Give your answer to 3 significant figures.
Question 4. On an oscilloscope, one wave cycle spans 5 divisions and the timebase is 2 ms per division. What is the frequency of the wave?
Question 5. Sound travels at 1500 m/s in water. A sound wave has a wavelength of 0.75 m in water. Calculate the frequency of the sound wave in Hz.
Challenge 1. A student investigates sound waves using a signal generator, a loudspeaker and two microphones connected to a dual-beam oscilloscope. Microphone A is fixed. Microphone B is moved away from the loudspeaker until the two traces are in phase again. The distance moved by microphone B is 0.68 m, and the signal generator is set to 500 Hz.
(a) What does it mean for the two traces to be "in phase"?
(b) Calculate the wave speed of sound in this experiment.
(c) The student repeats the experiment at 1000 Hz. Predict the new distance microphone B would need to move. Explain your answer.
(a) In phase means the two waves have the same displacement at the same time — the crests and troughs align perfectly on the oscilloscope. The distance moved equals one wavelength (λ = 0.68 m).
(b) v = f × λ = 500 × 0.68 = 340 m/s
(c) At 1000 Hz with the same wave speed: λ = v ÷ f = 340 ÷ 1000 = 0.34 m. The distance would be 0.34 m. Doubling the frequency halves the wavelength, so microphone B needs to move half as far.
Challenge 2. The speed of light in a vacuum is 3.00 × 10⁸ m/s. Violet light has a wavelength of 400 nm and red light has a wavelength of 700 nm.
(a) Convert 400 nm and 700 nm into metres.
(b) Calculate the frequency of violet light.
(c) Calculate the frequency of red light.
(d) Which colour of visible light has the higher frequency? Explain what this tells us about the energy carried by each colour.
(a) 400 nm = 400 × 10⁻⁹ m = 4.00 × 10⁻⁷ m; 700 nm = 7.00 × 10⁻⁷ m
(b) f (violet) = v ÷ λ = (3.00 × 10⁸) ÷ (4.00 × 10⁻⁷) = 7.50 × 10¹⁴ Hz
(c) f (red) = v ÷ λ = (3.00 × 10⁸) ÷ (7.00 × 10⁻⁷) = 4.29 × 10¹⁴ Hz
(d) Violet light has the higher frequency. Higher frequency means higher energy (E = hf). Violet light carries more energy per photon than red light. This is why UV radiation (even higher frequency) can cause more biological damage than visible light.
Challenge 3. A student uses a ruler and stopwatch to measure the speed of water waves in a ripple tank. She counts 8 complete waves fitting into a distance of 16 cm. She also counts 20 complete waves passing a point in 5 seconds.
(a) Calculate the wavelength of the water waves.
(b) Calculate the frequency of the water waves.
(c) Calculate the wave speed.
(d) The student estimates her ruler measurements have an uncertainty of ±2 mm per wavelength measured. Suggest one way she could reduce the percentage uncertainty in her wavelength measurement without buying new equipment.
(a) λ = 16 cm ÷ 8 = 2.0 cm = 0.020 m
(b) f = 20 waves ÷ 5 s = 4.0 Hz
(c) v = f × λ = 4.0 × 0.020 = 0.080 m/s
(d) She could measure across a greater number of wavelengths (e.g. 20 instead of 8). The absolute uncertainty in the total length stays roughly the same (±2 mm), but when divided by more wavelengths, the uncertainty per wavelength is much smaller. For example, measuring 20 wavelengths over 40 cm with ±2 mm uncertainty gives ±0.1 mm per wavelength instead of ±0.25 mm.
Challenge 4 (Extended). A guitar string vibrates at 330 Hz when plucked. The sound wave produced travels through air at 340 m/s and then through a concrete wall at 3400 m/s.
(a) Calculate the wavelength of the sound in air.
(b) Calculate the wavelength of the sound in concrete.
(c) What happens to the frequency of the sound when it passes from air into concrete? Explain using the wave equation.
(d) Explain why you can sometimes hear low-frequency sounds (bass) through walls more easily than high-frequency sounds (treble), even though both travel at the same speed in the wall.
(a) λ (air) = v ÷ f = 340 ÷ 330 = 1.03 m
(b) λ (concrete) = v ÷ f = 3400 ÷ 330 = 10.3 m
(c) The frequency does not change. Frequency is determined by the source (the vibrating string). When the wave crosses the boundary, v = fλ must still hold. Since v increases and f stays constant, λ must increase — the wavelength increases by a factor of 10 (same ratio as the speed change).
(d) Low-frequency (bass) sounds have long wavelengths. Long-wavelength waves are better at diffracting around obstacles and through gaps, and are less easily absorbed or reflected by dense materials like walls. High-frequency sounds have short wavelengths, which are more likely to be absorbed by or reflected from the wall material before they can pass through.