Dr Alcoholic
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I'm going out of my mind here but I can't seem to simplify this equation.


 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting E_x(z, t) = E_x(z)e^{-j\omega t} into both sides?

EDIT:  \frac{1}{\mu \mu_0 \varepsilon_0} is a constant.

Would I be right in saying on the left side you would get -\omega^2 e^{-j\omega t} E(z)?
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Old_Simon
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(Original post by Dr Alcoholic)
I'm going out of my mind here but I can't seem to simplify this equation.


 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting E_x(z, t) = E_x(z)e^{-j\omega t} into both sides?
Let me just check with Stephen Hawkins
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Dr Alcoholic
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(Original post by Old_Simon)
Let me just check with Stephen Hawkins
Who?
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Old_Simon
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(Original post by Dr Alcoholic)
Who?
Hawking. typo don't stress.
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Dr Alcoholic
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(Original post by Old_Simon)
Hawking. typo don't stress.
Maybe you can ask him to give me a new brain while you're at it.
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Namch
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That question is a joke.
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Dr Alcoholic
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I don't mean to be rude, but this is stressing me out so could you only leave a comment if you know how to do this please.
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BlueSam3
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#8
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Calculate the second partial derivatives of your given substitution. Substitute it in.
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Reety
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#9
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I had a go, didn't really get it, but here is what i came out with:

\[(z,t) = \frac{1}{\mu \mu_0\varepsilon_0}\frac{(z)e^{-j\omega t}{t^2}}{\ z^2}\]



*I tried my very best, so don't be mad if i didn't help

All the best,
Reety.
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Dr Alcoholic
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(Original post by Reety)
I had a go, didn't really get it, but here is what i came out with:

\[(z,t) = \frac{1}{\mu \mu_0\varepsilon_0}\frac{(z)e^{-j\omega t}{t^2}}{\ z^2}\]



*I tried my very best, so don't be mad if i didn't help

All the best,
Reety.

TBH it's further than I've got.
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Dr Alcoholic
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(Original post by BlueSam3)
Calculate the second partial derivatives of your given substitution. Substitute it in.


Does that imply that  E_x(t) = V_p  E_x(z), where  V_p is the phase speed?
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RhymeAsylumForever
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What is this for? I'm sensing a mechanics module?
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Dr Alcoholic
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(Original post by RhymeAsylumForever)
What is this for? I'm sensing a mechanics module?
No it's to do with transmissions. I'm not a maths student, it's engineering but this is a maths based problem hence I'm asking for help from maths people.
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RhymeAsylumForever
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(Original post by Dr Alcoholic)
No it's to do with transmissions. I'm not a maths student, it's engineering but this is a maths based problem hence I'm asking for help from maths people.
Oh ok, looked a bit like further maths with physics.
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Reety
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I had another crack at it, here is what i got:

\[(z,t) = \frac{(z^{-1})e^{-j\omega t}{t^2}}{\ \mu \mu_0\varepsilon_0}\]

Hope this helped
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BlueSam3
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(Original post by Dr Alcoholic)
Does that imply that  E_x(t) = V_p  E_x(z), where  V_p is the phase speed?
The image is broken, so I can't tell you, sorry.
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Farhan.Hanif93
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#17
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(Original post by Dr Alcoholic)
I'm going out of my mind here but I can't seem to simplify this equation.


 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting E_x(z, t) = E_x(z)e^{-j\omega t} into both sides?

EDIT:  \frac{1}{\mu \mu_0 \varepsilon_0} is a constant.

Would I be right in saying on the left side you would get -\omega^2 e^{-j\omega t} E(z)?
If this hasn't been sorted yet, yes the LHS there is correct.

The RHS is even easier - notice that the "e^{-jwt}" factor in E_x(z,t) = E(z) e^{-jwt} is a constant w.r.t. z. So when you differentiate w.r.t. z twice for the RHS, the e^{-jwt} can be pulled out of the derivative, leaving only the E(z) factor to be operated upon.

This should give you a second order linear differential equation for E(z), which you should be able to solve.
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