$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$
$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$
Let the volume of the region be V. The angular frequency of the vibrations of the electric and magnetic fields are same and are equal to $\omega$ Therefore, their frequency, $\text{f}=\frac{\omega}{2\pi},$ is same. The electrical energy in the region,$\text{U}_\text{E}=\Big(\frac{1}{2}\in_0\text{E}^2\Big)\times\text{V}$
It can be written as,$\text{U}_\text{E}=\Big(\frac{1}{2}\in_0\big(\text{E}^2_0\sin^2(\text{kz}-\omega\text{t})\big)\Big)\times\text{V}$
$\text{U}_\text{E}=\Bigg(\frac{1}{2}\in_0\text{E}_0^2\times\frac{\big(1-\cos2(\text{kz}-\omega\text{t})\big)}{2}\Bigg)\times\text{V}$
$\text{U}_\text{E}=\Big(\frac{1}{4}\in_0\text{E}_0^2\times(1-\cos2(\text{kz}-\omega\text{t}))\Big)\times\text{V}$
The magnetic energy in the region,$\text{U}_\text{B}=\Big(\frac{\text{B}^2}{2\mu_0}\Big)\times\text{V}$
$\text{U}_\text{B}=\bigg(\frac{\text{B}^2_0\sin^2(\text{kz}-\omega\text{t})}{2\mu_0}\bigg)\times\text{V}$
$\text{U}_\text{B}=\Bigg(\frac{\text{B}^2_0\big(1-\cos(2\text{kz}-2\omega\text{t})\big)}{4\mu_0}\Bigg)\times\text{V}$
The angular frequency of the electric and magnetic is same and is equal to $2\omega$ Therefore, their frequency,$\text{f}'=\frac{2\omega}{2\pi}=2\text{f}$
Will be same. Thus, the electric and magnetic fields have same frequencies and the electrical and magnetic energies will have same frequencies.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
