Wave parameters and behaviours: the meaning of wavelength, frequency, period, amplitude and speed, the wave equations, the difference between transverse and longitudinal waves, and wave behaviours such as reflection and diffraction.

What this key area is asking

The SQA wants you to define the wave quantities (wavelength, frequency, period, amplitude and speed), use the wave relationships, tell transverse from longitudinal waves, and describe wave behaviours such as reflection and diffraction.

Describing a wave

A bigger amplitude means a louder sound or a brighter light (more energy); a higher frequency means a higher-pitched sound or, for light, a colour nearer the blue end. The speed is how fast the wave energy travels.

The wave relationships

The first relationship links speed, frequency and wavelength; the second links frequency and period (they are reciprocals, so a frequency of $50 \text{ Hz}$ means a period of $\frac{1}{50} = 0.02 \text{ s}$). You can also find frequency as the number of waves divided by the time, and speed as distance divided by time, $v = \frac{d}{t}$.

Transverse and longitudinal waves

Wave behaviours: reflection and diffraction

Waves can be reflected (bounced off a barrier, as with an echo or a mirror) and diffracted (spread out as they pass through a gap or round an edge). Diffraction is greater when the gap is about the same size as the wavelength, which is why long-wavelength radio waves bend around hills and buildings far better than short-wavelength signals. The SQA expects you to recognise diffraction as the spreading of waves and to know that longer wavelengths diffract more.

Reflection is used in echo-sounding (sonar), where a sound or ultrasound pulse is sent out, bounces off a surface such as the sea bed, and returns. Timing the echo and using $v = \frac{d}{t}$ gives the distance; remember the pulse travels there and back, so the distance to the surface is half the total path. This is a common SQA calculation, used for finding the depth of water or the distance to an object.

Counting waves to find frequency

A practical way to find the frequency is to count the number of complete waves passing a point in a measured time and divide: frequency equals the number of waves divided by the time in seconds. For example, if $30$ waves pass a buoy in $6.0 \text{ s}$, the frequency is $\frac{30}{6.0} = 5.0 \text{ Hz}$. This links directly to $v = f\lambda$, because once you have the frequency and you know the wavelength you can find the speed.

Try this

Q1. State the relationship between frequency and period. [1 mark]

• Cue. $f = \frac{1}{T}$ (they are reciprocals).

Q2. A wave has a frequency of $25 \text{ Hz}$ and a wavelength of $4.0 \text{ m}$. Calculate the speed. [2 marks]

• Cue. $v = f\lambda = 25 \times 4.0 = 100 \text{ m s}^{-1}$.

Q3. State whether sound is a transverse or a longitudinal wave. [1 mark]

• Cue. Longitudinal.