derive, using the definitions of speed, frequency and wavelength, the wave equation v = f λ

Progressive Waves

Learning Objective

Derive, using the definitions of speed, frequency and wavelength, the wave equation

$$v = f \lambda$$

Key Definitions

• Wave speed ($v$): the distance a point of constant phase travels per unit time.
• Frequency ($f$): the number of complete cycles that pass a fixed point each second (units: Hz).
• Wavelength ($\lambda$): the spatial distance between two consecutive points that are in phase (e.g., crest to crest).

Derivation of the Wave Equation

Consider a sinusoidal progressive wave travelling along the $x$‑axis. Let a particular crest be observed at position $x_1$ at time $t_1$ and the same crest be observed at position $x_2$ at a later time $t_2$.

1. By definition of speed, $$v = \frac{\text{distance travelled}}{\text{time taken}} = \frac{x_2 - x_1}{t_2 - t_1}.$$
2. The distance $x_2 - x_1$ between two successive occurrences of the same phase is one wavelength, $\lambda$.
3. The time $t_2 - t_1$ for one complete cycle to pass a fixed point is the period $T$ of the wave.

Substituting $\lambda$ for the distance and $T$ for the time gives

$$v = \frac{\lambda}{T}.$$

Frequency $f$ is the reciprocal of the period:

$$f = \frac{1}{T} \quad \Longrightarrow \quad T = \frac{1}{f}.$$

Replacing $T$ in the speed expression:

$$v = \frac{\lambda}{1/f} = f\lambda.$$

Thus the fundamental relationship for a progressive wave is

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

Units and Typical \cdot alues

Quantity	Symbol	SI Unit	Typical A‑Level Example
Wave speed	$v$	metre per second (m·s⁻¹)	Sound in air ≈ 340 m·s⁻¹
Frequency	$f$	hertz (Hz)	Middle C note ≈ 261 Hz
Wavelength	$\lambda$	metre (m)	Visible light ≈ 5×10⁻⁷ m

Worked Example

Find the wavelength of a sound wave travelling at $v = 340\ \text{m·s}^{-1}$ with a frequency of $f = 500\ \text{Hz}$.

1. Start from the wave equation $v = f\lambda$.
2. Re‑arrange for $\lambda$: $\displaystyle \lambda = \frac{v}{f}$.
3. Insert the numbers: $\displaystyle \lambda = \frac{340\ \text{m·s}^{-1}}{500\ \text{Hz}} = 0.68\ \text{m}$.

Therefore the wavelength is $0.68\ \text{m}$.

Conceptual Check

• If the frequency of a wave is doubled while the speed remains constant, what happens to the wavelength?
• Explain why the wave equation holds for both transverse and longitudinal waves.

Summary

The wave equation $v = f\lambda$ links three fundamental properties of a progressive wave. It follows directly from the definitions of speed (distance per time), frequency (cycles per time), and wavelength (distance per cycle). Mastery of this relationship enables analysis of a wide range of wave phenomena encountered in A‑Level physics.