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This Image is from Electrodynamics by Griffiths. Here also a monochromatic electromagnetic wave is considered.

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  • $\begingroup$ See Section 9.1.2(iii) of Griffiths: "Although the sinusoidal wave function is a very special wave form, the fact is that any wave can be expressed as a linear combination of sinusoidal ones. The formula... can be obtained from the theory of Fourier transforms, but the details are not relevant to my purpose here. The point is that any wave can be written as a linear combination of sinusoidal waves, and therefore if you know how sinusoidal waves behave, you know in principle how any wave behaves. So from now on, we shall confine our attention to sinusoidal waves." $\endgroup$ Commented Mar 28, 2022 at 15:36

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Note: This is answer is not specific to electromagnetic waves.

Short answer: yes, this an over simplification, but a useful one.

In general, the 1-dimensional wave equation (where $c$ is the speed of propagation) $$\frac{\partial^2u}{\partial t^2} = c^2\frac{\partial^2u}{\partial x^2}$$ has solutions of the form $$u(x,t) = F(x-ct) + G(x+ct)$$ where $F(x)$ and $G(x)$ are arbitrary functions.

The usefulness of representing a plane wave as a sinusoid results from being able to represent pretty much all such functions as a linear superposition of sinusoids as these represent the plane wave frequency eigenmodes. This means we can write the wave equation solution as $$u(x,t) = \int^\infty_{-\infty}s_+(\omega)e^{-i(kx+\omega t)}d\omega + \int^\infty_{-\infty}s_-(\omega)e^{-i(kx-\omega t)}d\omega$$ $$= \int^\infty_{-\infty}s_+(\omega)e^{-i(x+ct)}d\omega + \int^\infty_{-\infty}s_-(\omega)e^{-i(x-ct)}d\omega$$

where $F(x-ct) = \int^\infty_{-\infty}s_+(\omega)e^{-i(x+ct)}d\omega$ and $G(x+ct) = \int^\infty_{-\infty}s_-(\omega)e^{-i(x-ct)}d\omega$

$s_+(\omega)$ and $s_-(\omega)$ represent the frequency components of the forward and reverse propagating waves.

A lot more information is available in the Wikipedia article Wave equation.

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A plane wave means a field where the field vector points in the same plane everywhere in the field. For instance, the field

$$ \vec{E}(z,t) = \sin(kz-\omega t)\vec{e_x}$$

is a field that oscillates in time, and oscillates along the z-axis, and always points along the (positive or negative) x-axis, as in your diagram. Only a single sine is necessary to model this.

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  • $\begingroup$ Your example is not a generic plane wave, but rather a monochromatic plane wave. $\endgroup$ Commented Mar 25, 2022 at 11:05
  • $\begingroup$ @SebastianRiese I agree. I don't see the question as asking for a generic plane wave. The answer to the question "There should be many, right?" obviously depends on specifics, but I give an example where "there should not be many", e.g. a single sine suffices. $\endgroup$ Commented Mar 25, 2022 at 12:25
  • $\begingroup$ Of course your answer is not wrong, but in my opinion incomplete – and it could easily point out more clearly that the sines are a (very useful) special case, but just a choice for example's sake. $\endgroup$ Commented Mar 25, 2022 at 13:49
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The reason is mostly practicality & convenience

  1. Sines, cosines and complex exponentials are solutions to most wave equations
  2. For linear systems, you can break down any problem into solving for monochromatic first and than assembling the result using superposition

The hard way

  1. Solve wave equation for arbitrary input signal

The easier way

  1. Break down arbitrary signal into complex exponentials (or equivalent)
  2. Solve the wave equation for the complex exponentials
  3. Sum the results for the final solution.
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