The Wave Equation

A single equation ties together the three things you can measure about any travelling wave — how fast it moves, how often it wobbles, and how long each wobble is. It is called the wave equation, and it works for water ripples, sound, light and every other wave you will meet. With it you can find the speed of a sound wave from its frequency, work out the wavelength of a radio station from the speed of light, or measure the speed of waves on a rope in your own classroom. This lesson, part of Topic P4 (Waves and radioactivity) of OCR Gateway Combined Science A, sets out the wave equation $v = f\lambda$ , rearranges it every way, links it to the period, and walks through the required practical of measuring the speed of waves.

By the end of this lesson you should be able to state and use the wave equation $v = f\lambda$ , rearrange it to find frequency or wavelength, use $T = \dfrac{1}{f}$ inside a calculation, describe how to measure the speed of water waves in a ripple tank and the speed of waves on a string, and carry out worked calculations with the correct units.

This lesson combines heavy AO2 application when you rearrange and use $v = f\lambda$ and $T = \dfrac{1}{f}$ in worked calculations and set up the required practicals, with AO3 analysis when you process the measurements to find a wave speed.

The Wave Equation

For any wave, the wave speed equals the frequency multiplied by the wavelength:

$v = f\lambda$

where $v$ is the wave speed in metres per second ( $\text{m/s}$ ), $f$ is the frequency in hertz ( $\text{Hz}$ ), and $\lambda$ is the wavelength in metres ( $\text{m}$ ).

Why does this work? Frequency is the number of waves passing a point each second, and wavelength is the length of each wave. So in one second, $f$ waves go past, and each is $\lambda$ long; the total length of wave that passes in one second is $f \times \lambda$ — and a length passing per second is exactly a speed. For example, if $5$ waves pass each second and each is $2 \text{ m}$ long, then $5 \times 2 = 10 \text{ m}$ of wave passes each second, so the wave travels at $10 \text{ m/s}$ .

A formula triangle helps you rearrange the equation: cover the quantity you want, and the triangle shows the calculation.

Wave speed sits at the top, so $v = f\lambda$ ; frequency and wavelength sit side by side at the bottom, so $f = \dfrac{v}{\lambda}$ and $\lambda = \dfrac{v}{f}$ :

$v = f\lambda \qquad f = \dfrac{v}{\lambda} \qquad \lambda = \dfrac{v}{f}$

Exam Tip: $v = f\lambda$ is on the OCR equation sheet, but you must know what each symbol means and its unit. A common slip is to mix up frequency and wavelength — remember frequency is in hertz (waves per second) and wavelength is a distance in metres.

It is worth pausing on why this one equation is so powerful. There are only three quantities in it, so if a question gives you any two of them, you can always find the third. That is why so many wave questions boil down to reading off the two values you are given, rearranging the equation if necessary, and calculating the third. The wave equation also connects to the idea from the previous lesson that a wave transfers energy at a steady speed through a given material: for a fixed speed, if you make the source vibrate faster (raising the frequency), the waves are packed closer together (a shorter wavelength), but they still travel at the same speed. Speed depends on the material, not on how you shake the source.

Worked Calculations

Worked Example 1 — Finding wave speed

A water wave has a frequency of $4 \text{ Hz}$ and a wavelength of $0.5 \text{ m}$ . Calculate its speed.

Step 1 — write the equation: $v = f\lambda$ .

Step 2 — substitute: $v = 4 \times 0.5$ .

Step 3 — calculate: $v = 2 \text{ m/s}$ .

Answer: the wave travels at $2 \text{ m/s}$ .

Worked Example 2 — Rearranging for frequency

A sound wave travels at $340 \text{ m/s}$ and has a wavelength of $0.85 \text{ m}$ . Calculate its frequency.

Step 1 — rearrange to make frequency the subject: $f = \dfrac{v}{\lambda}$ .

Step 2 — substitute: $f = \dfrac{340}{0.85}$ .

Step 3 — calculate: $f = 400 \text{ Hz}$ .

Answer: the frequency is $400 \text{ Hz}$ .

Worked Example 3 — Rearranging for wavelength

A radio station broadcasts at a frequency of $100 \text{ MHz}$ ( $1 \times 10^8 \text{ Hz}$ ). Radio waves travel at $3 \times 10^8 \text{ m/s}$ . Calculate the wavelength.

Step 1 — rearrange to make wavelength the subject: $\lambda = \dfrac{v}{f}$ .

Step 2 — substitute: $\lambda = \dfrac{3 \times 10^8}{1 \times 10^8}$ .

Step 3 — calculate: $\lambda = 3 \text{ m}$ .

Answer: the wavelength is $3 \text{ m}$ .

Worked Example 4 — Using the period first

A wave on a string completes $20$ oscillations in $4 \text{ s}$ and has a wavelength of $0.6 \text{ m}$ . Calculate its speed.

Step 1 — find the frequency from the number of waves and the time: $f = \dfrac{\text{number of waves}}{\text{time}} = \dfrac{20}{4} = 5 \text{ Hz}$ .

Step 2 — write the wave equation: $v = f\lambda$ .

Step 3 — substitute and calculate: $v = 5 \times 0.6 = 3 \text{ m/s}$ .

Answer: the wave travels at $3 \text{ m/s}$ .

Worked Example 5 — Converting a frequency first

A sound wave has a frequency of $3.4 \text{ kHz}$ and travels at $340 \text{ m/s}$ in air. Calculate its wavelength.

Step 1 — convert the frequency to hertz: $3.4 \text{ kHz} = 3.4 \times 1000 = 3400 \text{ Hz}$ .

Step 2 — rearrange for wavelength: $\lambda = \dfrac{v}{f}$ .

Step 3 — substitute and calculate: $\lambda = \dfrac{340}{3400} = 0.1 \text{ m}$ .

Answer: the wavelength is $0.1 \text{ m}$ ( $10 \text{ cm}$ ). Substituting $3.4$ instead of $3400$ would have given a wavelength $1000$ times too large — which is exactly why the conversion comes first.

Exam Tip: Lay every calculation out in three lines — equation, substitution, answer with unit. Watch the units: frequencies are sometimes given in $\text{kHz}$ ( $\times 1000$ ) or $\text{MHz}$ ( $\times 1\,000\,000$ ), and these must be converted to $\text{Hz}$ before you substitute.

Bringing in the Period

Because the period and frequency are linked by $T = \dfrac{1}{f}$ , a question sometimes gives you the period and expects you to find the frequency first, then use the wave equation. The order is: use $T = \dfrac{1}{f}$ to get $f$ , then use $v = f\lambda$ .

Worked Example 6 — Speed from period and wavelength

A wave has a period of $0.2 \text{ s}$ and a wavelength of $1.5 \text{ m}$ . Calculate its speed.

Step 1 — find the frequency: $f = \dfrac{1}{T} = \dfrac{1}{0.2} = 5 \text{ Hz}$ .

Step 2 — use the wave equation: $v = f\lambda = 5 \times 1.5$ .

Step 3 — calculate: $v = 7.5 \text{ m/s}$ .

Answer: the wave travels at $7.5 \text{ m/s}$ .

Exam Tip: If you are given a period rather than a frequency, do not put the period straight into $v = f\lambda$ . First convert it to a frequency with $f = \dfrac{1}{T}$ , then use the wave equation.

Required Practical: Measuring the Speed of Waves

OCR requires you to measure the speed of waves. There are two standard situations for combined science, and you should be able to describe at least one fully. In each case the plan is the same: find the frequency and the wavelength, then use $v = f\lambda$ . The wave equation is the tool that turns two easy-to-measure quantities — how often the waves come and how long each one is — into the speed, which is much harder to measure directly because the waves move too quickly to time over a short distance.

The Wave Equation

The Wave Equation