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| ### Statement |
| ### Statement |
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| $12.1.3.$ [Insert the problem statement] |
| $12.1.3.$ The figure shows the electric field of a plane sinusoidal wave at the initial moment of time $t=0$. The direction of wave propagation is indicated by an arrow. How does the electric field strength depend on the coordinate $z$ and time $t$? |
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| ### Solution |
| ### Solution |
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| For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: |
| For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: |
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| $$E(z,t) = E_0 \sin(At + Bz + \varphi_0)$$ |
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| where $A$, $B$ and $\varphi_0$ are constants. |
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| From the condition of the problem (based on the wave profile at $t=0$), we know that the initial spatial distribution is: |
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| $$E(z,0) = E_0 \sin\left(\frac{2\pi}{\lambda} z\right)$$ |
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| where $\lambda$ is the wavelength. |
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| \begin{equation} |
| By substituting $t=0$ into the general equation, we equate the two expressions: |
| E(z,t)=E_0\sin(At+Bz+\varphi_0) |
| $$E_0 \sin(Bz + \varphi_0) = E_0 \sin\left(\frac{2\pi}{\lambda} z\right)$$ |
| \end{equation} |
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| where $A$,$B$ and $\varphi_0$ are constants. |
| By comparing the phases, we obtain the spatial constant and the initial phase: |
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| $$B = \frac{2\pi}{\lambda}, \quad \varphi_0 = 0$$ |
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| Since the wave travels in the positive $z$-direction with a phase velocity $c$, the phase of the wave $\varphi = At + Bz + \varphi_0$ must remain constant for a fixed point on the wave profile. Taking the time derivative of the phase yields: |
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| $$\frac{d\varphi}{dt} = A + B \frac{dz}{dt} = 0$$ |
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| Given that the wave propagation speed is $\frac{dz}{dt} = c$, we get: |
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| $$A + Bc = 0 \implies A = -Bc = -\frac{2\pi c}{\lambda}$$ |
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| From the condition of the problem we know that: |
| Finally, plugging the derived constants $A$, $B$, and $\varphi_0$ back into the general equation gives us: |
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| $$E(z,t) = E_0 \sin\left(-\frac{2\pi c}{\lambda} t + \frac{2\pi}{\lambda} z\right) = E_0 \sin\left[\frac{2\pi}{\lambda}(z - ct)\right]$$ |
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| \begin{equation} |
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| E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) |
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| \end{equation} |
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| Using eq(1) we get: |
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| \begin{equation} |
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| E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) |
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| \end{equation} |
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| By comparing the phases we obtain that: |
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| \begin{equation} |
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| B=\frac{2\pi}{\lambda};\varphi_0=0 |
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| \end{equation} |
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| As the wave travels in the positive direction: |
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| \begin{equation} |
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| \frac{dz}{dt}=c |
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| \end{equation} |
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| Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): |
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| \begin{equation} |
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| A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 |
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| \end{equation} |
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| So we get the constant $A$: |
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| \begin{equation} |
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| A=-\frac{2\pi c}{\lambda} |
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| \end{equation} |
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| Plugging all the constants into the eq(1) gives us: |
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| \begin{equation} |
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| E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) |
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| \end{equation} |
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| #### Answer |
| #### Answer |
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| $$E(z,t) = E_0 \sin\left[\frac{2\pi}{\lambda}(z - ct)\right]$$ |
| $E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))$ |
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