Wave length speed and wave types

A wave is the propagation of oscillations through space. Wavelength $\lambda$ is the distance between the two nearest points oscillating in the same phase (for example, between adjacent crests). The period $T$ is the time of one complete oscillation of a particle, and the frequency $f=1/T$ is the number of oscillations per second. These quantities are related to the wave speed by $v=\lambda f=\frac{\lambda}{T}.$ In a uniform medium the speed is set by the properties of the medium (for a string, $v=\sqrt{T_0/\mu}$, where $T_0$ is the tension and $\mu$ the linear density), so when a wave passes into another medium the frequency set by the source stays constant while the wavelength changes. By the direction of particle oscillation we distinguish transverse waves (particles oscillate perpendicular to the propagation direction) from longitudinal ones (oscillation along the propagation direction); by nature, waves are mechanical or electromagnetic.

A source produces a wave with frequency $f=50$ Hz and wavelength $\lambda=0.6$ m. Find the wave speed.

A transverse wave travels along a string with tension $T_0=80$ N and linear density $\mu=0.05$ kg/m at frequency $f=10$ Hz. Find the wave speed and its wavelength.

A sound wave of frequency $170$ Hz travels in air at $340$ m/s and passes into water, where its speed is $1500$ m/s. Find the wavelength in air and in water and explain which quantity does not change at the boundary.

Self-check with simulations (safe, no equipment). Open the PhET "Wave on a String" simulation in "Oscillate" mode and do the following:

1. Set the frequency to $f=1.0$ Hz and measure the wavelength $\lambda$ with the simulation ruler.
2. Raise the frequency to $2.0$ Hz, measure $\lambda$ again and record the pairs of values.
3. For each pair compute $v=\lambda f$ and check that the speed stays approximately the same.
4. Change the tension ("Tension" slider) and see how the speed changes at constant frequency.
5. To see longitudinal particle motion, open a dedicated longitudinal-wave simulation such as PhET "Sound" (compressions and rarefactions along the propagation direction) and compare it with the transverse string view from the "Wave on a String" simulation: in the string, particles oscillate perpendicular to propagation; in the sound wave, along it.

Conclusion: write down what is set by the frequency, what is set by the medium, how this shows up in $\lambda$, and how a transverse wave differs from a longitudinal one.

To show that wavelength is a spatial period and to compare transverse and longitudinal waves visually by the direction of particle oscillation.

Two sketches stacked vertically. Top: a transverse sinusoid with a marked amplitude, one highlighted particle, and arrows showing the distance $\lambda$ between adjacent crests; beside it a vertical double arrow for the particle oscillation. Bottom: a longitudinal wave drawn as alternating compressions and rarefactions with the same $\lambda$ label and a horizontal double arrow for the particle oscillation; the propagation arrows on both waves point to the right.

At a football match a bass drum is struck. The drumhead vibrates at $f = 85\ \text{Hz}$, so it pushes the air 85 times per second. The speed of sound in $20^\circ\text{C}$ air is $v = 340\ \text{m/s}$. The distance between successive compressions (the wavelength) is $\lambda = \frac{v}{f} = \frac{340}{85} = 4.0\ \text{m}.$ The period of one vibration is $T = 1/f = 0.012\ \text{s}$. A spectator $40\ \text{m}$ away hears the beat about $t = 40/340 = 0.12\ \text{s}$ later. The wave is longitudinal: air molecules move back and forth along the direction the sound travels, and the compressions travel 4.0 m in one period.

This scenario uses the relation between wave speed, frequency and wavelength, $v = \lambda f = \lambda/T$, and the distinction between a longitudinal wave (sound in air) and a transverse wave. The period $T$ is the time for one vibration; the wavelength $\lambda$ is the distance the disturbance travels in that time.

A surfer sits on a board in the sea. Successive wave crests pass every $T = 6\ \text{s}$, so the frequency is $f = 1/6 = 0.17\ \text{Hz}$. The distance between two crests is $\lambda = 30\ \text{m}$. The wave speed is $v = \frac{\lambda}{T} = \lambda f = 30 \times 0.17 = 5\ \text{m/s} = 18\ \text{km/h}.$ The surfer bobs up and down through a vertical amplitude of about $1.5\ \text{m}$ but stays in roughly the same place; it is the wave shape, not the water, that moves towards the shore. This is a surface wave: water particles move in circles, so it has both transverse and longitudinal components.

This exercises $v = \lambda/T = \lambda f$ for a surface wave and shows that in a wave the medium oscillates while the wave pattern travels. The amplitude ($1.5\ \text{m}$) does not change the wave speed or wavelength in this simple model; the period $T$ comes from the source (wind and weather), and the wavelength follows from the speed of the wave in water.

A phone connects to a Wi-Fi router at $f = 2.4\ \text{GHz} = 2.4\times 10^9\ \text{Hz}$. These radio waves are electromagnetic and travel through air at almost the speed of light in vacuum, $c = 3.0\times 10^8\ \text{m/s}$. Their wavelength is $\lambda = \frac{c}{f} = \frac{3.0\times 10^8}{2.4\times 10^9} = 0.125\ \text{m} = 12.5\ \text{cm}.$ A 5 GHz Wi-Fi band gives a shorter wavelength: $\lambda = \frac{3.0\times 10^8}{5.0\times 10^9} = 0.060\ \text{m} = 6.0\ \text{cm}.$ The 2.4 GHz signal passes through a doorway (about $0.8\ \text{m}$ wide) easily because $12.5\ \text{cm}$ is smaller than the doorway, but a metal mesh of $1\ \text{cm}$ holes reflects it.

This uses $v = \lambda f$ for an electromagnetic transverse wave, with the wave speed in air taken as $c$. It shows how the same speed with different source frequencies produces different wavelengths, and it introduces the classification of waves (mechanical vs electromagnetic, longitudinal vs transverse). The wavelength compared to obstacle size is what decides whether the wave spreads around the obstacle (diffraction) or is blocked.