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1-D wave equation

  • Thread starter SWiTCHRiDE
  • Start date
1. Homework Statement
Express the solution
P(t, x1) = cos(!t − kx1)
as the superposition of two complex exponentials. Show that each complex
exponential is also a solution of the 1-D wave equation.



2. Homework Equations
just that THETA=P
!=w

whoops, made a type



3. The Attempt at a Solution
im not really sure how to tackle this, i do know that a solution to the wave equation added to another solution is also a solution.
 
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Answers and Replies

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0
well, for the first part, what is the complex exponential equivalent of cos?

Hint: Remeber that Euler's Formula igives [tex]e^{i\theta}=cos\theta + isin\theta[/tex] and [tex]e^{-{i\theta}}=cos\theta - isin\theta[/tex]. How can you derive a formula for [tex]cos\theta[/tex] from this (ie how do you get cos in terms of _only_ complex exponentials.

You know that from the first part you should be able to express your wave function as a sum of two complex exponentials. Take each one and sub it into the wave equation and see if the LS equates to the RS for each one. It may also be useful to remember that [tex]\frac{\omega}{k}=v[/tex] where v is the velocity of the wave.
 
218
0
The solutions to the wave equation are http://en.wikipedia.org/wiki/Harmonic_function" [Broken]. Their sums, differences and scalar multiples are also solutions to the wave equation.

You should use "[URL [Broken] formula[/URL] to write the cosine as a sum of two complex exponentials.

EDIT: crossposted with Warr
 
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Gib Z
Homework Helper
3,344
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Ahh dont you guys love it when you see a formula for the first time, and it looks obscure and you don't really understand it, eg Euler formula. And then you take a second to think, and from Eulers Formula you can get the good old cos^2 x + sin^2 x=1? I love it :p
 
guys, thanks for the help but im so lost in all this stuff. Is there somewhere where i can learn about harmonic functions to get up to speed. i did crappy in the math course before the geophys course im taking so im pretty lost as it is. i do not want the answer (obviously) but an easier explanation leading me down the right path please....
 
218
0
guys, thanks for the help but im so lost in all this stuff. Is there somewhere where i can learn about harmonic functions to get up to speed. i did crappy in the math course before the geophys course im taking so im pretty lost as it is. i do not want the answer (obviously) but an easier explanation leading me down the right path please....
You can't get easier than this:

Euler's Formula igives [tex]e^{i\theta}=cos\theta + isin\theta[/tex] and [tex]e^{-{i\theta}}=cos\theta - isin\theta[/tex]. How can you derive a formula for [tex]cos\theta[/tex] from this (ie how do you get cos in terms of _only_ complex exponentials.
Try adding the two equations.
 
ok, i get (in words i dont know how to type it fancy...)

cos (theta) = [ e^j(theta) + e^-j(theta) ] /2

then just sub back in for (theta)

because adding euler together was adding 2 complex equations right?
 
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218
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The problem asks to express the solution [tex]P(t,x_{1})=cos({\omega}t-kx_{1})[/tex] as the superposition of two complex exponentials and to show that each complex exponential is also a solution of the 1-D wave equation. Now using Euler's formula you found that

[tex]cos({\omega}t-kx_{1}) = \frac{e^{j({\omega}t-kx_{1})}+e^{-j({\omega}t-kx_{1})}}{2}[/tex]

right?

http://en.wikipedia.org/wiki/Superposition_principle" [Broken] , if you didn't know, means algebraic sum.

So what would be the two complex exponentials which summed give you [tex]P(t,x_{1})=cos({\omega}t-kx_{1})[/tex] ?

I'm sure you know the answer. Next you need to verify that the two complex exponentials are also solutions to the wave equation, which is very easy since you know the expression of the 1D wave equation.
 
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