Transversal Waves: Solving Vibrating String Equations | Easy Guide

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The discussion focuses on solving the vibrating string equation using various mathematical representations of waves. Participants emphasize the importance of verifying if these representations satisfy the 1D D'Alembert's equation by substituting them into the wave equation. A specific example, u(x,t)=5Sin(x)Cos(vt), is confirmed as a solution, but confusion arises regarding its expression in D'Alembert form, which results in an imaginary component. The use of trigonometric identities is suggested to clarify the representation and ensure it remains real, as string waves should not have an imaginary part. The conversation highlights the need for careful manipulation of wave equations and identities in wave mechanics.
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Dear friends,

I need some help with transversal waves, to be precise: the vibrating string. I’ve been given many mathematical representations of what can be a wave (e.g: 10(x^2-v^2*t^2)
or this one, 5 Sinx cosv*t)

I have to argue which of them can be a solution of the vibrating string wave equation. And if it’s a solution I’ve been asked to write it in the D’Alembert form (that’s f(x-vt)+g(x+vt)).

Just tell me if what I think is correct: I insert the possible solution in the wave equation to see if both sides of the equation match (is that all or am I missing something?).

Thanks
 
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Yeah,it needs to satisfy 1D d'Alembert's equation.

Daniel.
 
Thanks Daniel.

Now I have a more specific question on this vibrating string problem. Given u(x,t)=5Senx*Cos(vt), I've proved that it's a solution. However, while trying to write it in the D'Alembert form I get this: 5i cos(x+vt)cos(x-vt)

Does it make sense that it is imaginary? Aren't string waves supposed to be real. I don't know if I'm mixing up things.
 
It can't be complex (with a nonzero imaginary part,that is).U should use a trigonometric identity

\sin x\cos y\equiv \frac{1}{2}\left[\sin\left(x+y\right)+\sin\left(x-y\right)\right]

Daniel.
 
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