Physics 4.6 - Wave equations

Frequency, wavelength and speed

Wave speed describes how quickly a wavefront or disturbance changes position. It is not the
instantaneous speed of one oscillating particle. The same two relationships apply to water,
sound, electromagnetic and seismic waves.

The relationship follows from the meanings of the quantities. During each complete cycle,
the wave pattern advances by one wavelength. If $f$ complete cycles occur each second, the
wave advances $f$ wavelengths each second, so $v = f \times \lambda$.

Worked example: calculate speed with a unit conversion

A wave on a rope has frequency $12\,\mathrm{Hz}$ and wavelength $25\,\mathrm{cm}$. Calculate its speed.

1. Identify and convert the quantities.

$f = 12\,\mathrm{Hz}$

$\lambda=25\,\mathrm{cm}=0.25\,\mathrm{m}$ (divide the centimetre reading by 100)

2. Select the relationship and substitute.

$v = f \times \lambda = 12\,\mathrm{Hz} \times 0.25\,\mathrm{m}$

3. Calculate and give the unit.

$v = 3.0\,\mathrm{m\,s^{-1}}$

Sense-check: twelve wavelength intervals pass per second, and each is one quarter of a
metre. $12 \times 1 \div 4 = 3$, so the order of magnitude and unit are sensible.

Worked example: rearrange to find wavelength

A sound wave travels at $340\,\mathrm{m\,s^{-1}}$ and has frequency $680\,\mathrm{Hz}$. Calculate its wavelength.

Start from the taught relationship, then rearrange before substituting:

$v = f \times \lambda$

$\lambda = \frac{v}{f}$

$\lambda = 340\,\mathrm{m\,s^{-1}} \div 680\,\mathrm{Hz} = 0.50\,\mathrm{m}$

Sense-check: $680 \times 0.50 = 340$, so substituting the result back into the original
equation reproduces the stated speed.

For a wave travelling at constant speed in one medium, increasing frequency means wavelength
must decrease so that $f \times \lambda$ remains equal to the same speed. This conclusion requires
the constant-speed condition. Amplitude does not appear in the equation, so a larger amplitude
does not by itself mean a faster wave.