Three exact wavelength-frequency pairs show the inverse relation at fixed speed. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Short wavelength needs high frequency

At wavelength 1 metre, the frequency must be 6 hertz to keep speed 6 metres per second.

c=λf=1 m6 Hz=6 m/sc=\lambda f=1\ \text{m}\cdot6\ \text{Hz}=6\ \text{m}/\text{s}
Wavelength-frequency scan rowThe shortest wavelength row carries the largest frequency.speed=6 m/smagneticField=2 TelectricField=12 V/mwavelength=1 mfrequency=6 Hz

Doubling wavelength halves the frequency

Wavelength 2 metres pairs with frequency 3 hertz. The product stays at the same speed.

λf=2 m3 Hz=6 m/s\lambda f=2\ \text{m}\cdot3\ \text{Hz}=6\ \text{m}/\text{s}
Wavelength-frequency scan rowThe middle row keeps the speed product unchanged.speed=6 m/smagneticField=2 TelectricField=12 V/mwavelength=2 mfrequency=3 Hz

Longer wavelength lowers frequency again

The third wavelength is 3 metres and the matching frequency is 2 hertz. The inverse pair is now visible across three rows.

λf=3 m2 Hz=6 m/s\lambda f=3\ \text{m}\cdot2\ \text{Hz}=6\ \text{m}/\text{s}
Wavelength-frequency scan rowThe checked row still lands on the training speed.speed=6 m/smagneticField=2 TelectricField=12 V/mwavelength=3 mfrequency=2 Hz