Wave Equation in 1-d Proof/Verify

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The discussion focuses on verifying that the functions Acos(kx-ωt) and Bsin(kx-ωt) are solutions to the one-dimensional wave equation when the wave speed v is defined as v=ω/k. Participants clarify that to verify these solutions, one should apply the wave equation's standard form and check if the functions satisfy it. There is also a question regarding whether the function f(x,t)=(ax+bt+c)^2 represents a propagating wave, with a request for its velocity if it does. The conversation reveals that the verification process is simpler than initially perceived, leading to a better understanding of the wave equation. Overall, the thread emphasizes the importance of applying the wave equation correctly to verify potential solutions.
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Homework Statement


Verify that Acos(kx-ωt) and Bsin(kx-ωt) are solutions of the one dimensional wave eqn. if v=ω/k. Does f(x,t)=(ax+bt+c)^2 represent a propagating wave? If yes what is its velocity?


Homework Equations


I know the partial differ. eqns. for the wave equation are
d^2 U/dz^2 = 1/c^2 d^2E/dt^2 for the function f(x-ct) + g(x+ct)
The lapacians for E is μoεo d2E/dt2


The Attempt at a Solution


I am just confused as to how to show this? And same for part b.). To verify something would I have to just take any value for v and show it for the first part. (I'll be talking to my prof. about this problem today as well) Thanks any help and hints are appreciated.
 
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hitman0097 said:
I know the partial differ. eqns. for the wave equation are
d^2 U/dz^2 = 1/c^2 d^2E/dt^2 for the function f(x-ct) + g(x+ct)
The lapacians for E is μoεo d2E/dt2

I don't really get what you mean here. Usually, the wave equation is just:
\frac{\partial^2 U}{\partial z^2} = \frac{1}{c^2} \frac{\partial^2 U}{\partial t^2}
And for the question, they've given you some possible solutions. Its fairly simple to show that they satisfy the wave equation. Think about it - if you just thought you had worked out a solution, then how would you check that it is correct?
 
Yeah, I think I was just making the problem harder than it was. I understand it better now I think about it. Thanks!
 
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