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 an earthquake wave from its frequency, work out the wavelength of a radio station from the speed of light, or measure the speed of sound in your own classroom. This lesson, part of Topic P5 (Waves in matter) of OCR Gateway Science A, sets out the wave equation $v = f\lambda$ , rearranges it every way, and walks through the required practical of measuring the speed of waves in three different situations.

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, describe how to measure the speed of water waves in a ripple tank, the speed of waves on a string, and the speed of sound in air, and carry out worked calculations with correct units.

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.

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 — A two-step problem

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: $f = \dfrac{\text{number of waves}}{\text{time}} = \dfrac{20}{4} = 5 \text{ Hz}$ .

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 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.

Required Practical: Measuring the Speed of Waves

OCR requires you to measure the speed of waves. There are three standard situations, 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 $v = f\lambda$ — except for the echo method, where you measure speed directly from distance and time.

Method A — Speed of water waves in a ripple tank

A ripple tank is a shallow tray of water with a motor that dips a bar in and out to make straight water waves (or a small dipper to make circular ones). A lamp above shines down so the wave pattern is projected onto a screen or the bench below.

Set the motor running at a steady, low frequency so the waves are clear.
Find the frequency $f$ — read it from the motor's setting, or count how many waves pass a point in (say) $10 \text{ s}$ and divide by $10$ . Using a strobe to "freeze" the waves makes counting easier.
Find the wavelength $\lambda$ — the projected waves appear stationary or slow, so measure the distance across several wavelengths with a ruler and divide by the number of wavelengths (measuring many and dividing reduces the percentage uncertainty).
Calculate the speed using $v = f\lambda$ .

The Wave Equation

The Wave Equation