Wave speed and measuring waves
The wave equation
In one second, f waves leave the source, each one λ long.
So the front of the wave has moved f × λ metres in that second.
wave speed = frequency × wavelength
v = f λ
v in m/s, f in Hz, λ in metres. Recall this one for every board. Rearranged: f = v ÷ λ and λ = v ÷ f.
- A dolphin call: 6 kHz in sea water at 1500 m/s
- 6 kHz = 6000 Hz
- λ = v ÷ f = 1500 ÷ 6000 = 0.25 m
- 0.25 m = 25 cm
Prefixes and standard form
Change every quantity into base units (Hz, m, s) before it goes into the equation.
- k (kilo) = 10³, M (mega) = 10⁶, G (giga) = 10⁹
- c (centi) = 10⁻², m (milli) = 10⁻³, μ (micro) = 10⁻⁶, n (nano) = 10⁻⁹
- Light: λ = 600 nm = 600 × 10⁻⁹ m = 6.00 × 10⁻⁷ m
- f = v ÷ λ = 3.00 × 10⁸ ÷ 6.00 × 10⁻⁷
- f = 5.00 × 10¹⁴ Hz
On a calculator, type standard form with the ×10ˣ (EXP) key, not "× 10 ^", or brackets go wrong.
Required practical: ripple tank and string
Ripple tank. A lamp above the tank throws the ripples' shadows onto paper below.
- Wavelength: lay a ruler beside the shadows (or photograph them) and measure across 10 waves, then divide by 10. One wave is too small to measure well.
- Frequency: count the waves passing a mark in 10 s and divide by 10 (or read the setting on the signal generator).
- Speed: v = f λ.
Waves on a string. A vibration generator shakes a string that runs over a pulley to hanging masses. Adjust the frequency until the string shows a steady pattern of loops.
- Each loop is half a wavelength: measure across all the loops, divide by the number of loops, and double it.
- Read f from the signal generator, then v = f λ. The hanging mass (the tension) and the string are control variables.
At a fixed tension, a higher frequency gives a shorter wavelength, but f × λ, the speed, stays the same.
Edexcel core practical: sound in a solid
Edexcel's core practical measures waves in a fluid (the ripple tank above) and in a solid.
AQA and OCR do not ask for the solid part.
Hang a metal rod so it can vibrate freely and tap one end with a hammer. It rings with a clear note. A microphone and oscilloscope, or a frequency app on a phone, gives the frequency of the note. When the rod vibrates like this, its length is half a wavelength, so the wavelength is twice the length of the rod.
- a steel rod 1.00 m long; the app reads 2550 Hz
- λ = 2 × 1.00 = 2.00 m
- v = f λ = 2550 × 2.00
- v = 5100 m/s
Why not just time the sound along the rod? At 5100 m/s it crosses a 1 m rod in about 0.0002 s, a thousand times shorter than a reaction time. Choosing equipment that suits the size of what you measure is part of the practical: a frequency meter for the rod, a ruler and a count of waves for the ripple tank.
Sound travels about 15 times faster in steel than in air. Repeat the taps and take a mean, leaving out any reading far from the rest.
Waves on a string
Set the frequency to 1 Hz and the wavelength to 2 m and read the wave speed: 2 m/s. Now set the frequency to 2 Hz and the wavelength to 1 m. The wave looks different, but the speed readout is unchanged. Try 0.5 Hz with 4 m. What do the three settings have in common, and what is the period at 0.5 Hz?
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