FrogPad
Nov26-06, 11:06 PM
I don't understand something on this problem:
Q: A uniform plane wave with \vec E = \hat x E_x propagates in a lossless simple medium ( \epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0 ) in the +z-direction. Assume that E_x is sinusoidal with a frequency of 100 (MHz) and has a maximum value of +10^{-4} (V/m) at t=0 and z = \frac{1}{8} (m)
Write the instantaneous expression for \vec E for any t and z.
A:
Using \cos \omega t as a reference, we find the instantaneous expression for \vec E to be:
\vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi)
Since E_x equals +10^{-4} when the argument of the cosine function equals zero, that is when:
2 \pi 10^8 - kz + \psi = 0
we have at t = 0 , z = \frac{1}{8}
\psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6}
My question follows from this:
Why do we introduce the phase unknown phase \psi and set E_x according to this. Why couldn't we do this instead,
A:
Using \cos \omega t as a reference, we find the instantaneous expression for \vec E to be:
\vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz)
Set the according t=0, z=1/8 so that,
\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4} and meet the initial condition of a max of 10^-4 with E_0 instead of setting E_0 originally and changing the phase.
Q: A uniform plane wave with \vec E = \hat x E_x propagates in a lossless simple medium ( \epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0 ) in the +z-direction. Assume that E_x is sinusoidal with a frequency of 100 (MHz) and has a maximum value of +10^{-4} (V/m) at t=0 and z = \frac{1}{8} (m)
Write the instantaneous expression for \vec E for any t and z.
A:
Using \cos \omega t as a reference, we find the instantaneous expression for \vec E to be:
\vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi)
Since E_x equals +10^{-4} when the argument of the cosine function equals zero, that is when:
2 \pi 10^8 - kz + \psi = 0
we have at t = 0 , z = \frac{1}{8}
\psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6}
My question follows from this:
Why do we introduce the phase unknown phase \psi and set E_x according to this. Why couldn't we do this instead,
A:
Using \cos \omega t as a reference, we find the instantaneous expression for \vec E to be:
\vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz)
Set the according t=0, z=1/8 so that,
\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4} and meet the initial condition of a max of 10^-4 with E_0 instead of setting E_0 originally and changing the phase.