Solving Waves on a Metal Rod with kx + Φ

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The discussion revolves around solving wave equations on a metal rod using the function ξ = A*cos(ωt + θ)*sin(kx + Φ). Two approaches are examined: the first involves setting the stress at the ends of the rod to zero, leading to conditions that do not yield a valid solution. The second approach considers the node at the half-length position but struggles with the dependency on Φ. Participants suggest that both methods should be integrated rather than treated separately, emphasizing the need to account for boundary conditions at both ends of the rod. The conversation highlights the importance of correctly interpreting the cosine function's behavior at the boundaries.
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Homework Statement
I will post an image right below.
Relevant Equations
All below.
1595537783294.png

I am trying to solve this question by ξ = A*cos(ωt + θ)*sin(kx + Φ)
Anyway, the two initial terms of the product helps nothing (i think), what matters is sin(kx + Φ)
So, i tried by two ways:

First:

The stress is essentially zero on the ends, that is, something like cte*∂ξ/∂x (strain) would be zero, so
L is the length of the rod.

cos(kx + Φ) need to be 0
Φ = (2n-1)*π
kL + Φ = (2n-1)*π

It is not good.

Second:

The half length position would carry a node, so
ξ = A*cos(ωt + θ)*sin(kL/2 + Φ) = 0
KL/2 + Φ = (n)*π
and with this i can not solve without Φ in the expression.
 
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I assume ξ is displacement. You would find it easier with x measured from the centre of the rod.
 
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haruspex said:
I assume ξ is displacement. You would find it easier with x measured from the centre of the rod.
Definitively is easier deal with x measured from the center of the rod, but what is wrong in the above approach?
 
LCSphysicist said:
Definitively is easier deal with x measured from the center of the rod, but what is wrong in the above approach?
It seems to me you should need to use the knowledge of the state both at the clamped point and one end, so they are not alternative methods. Rather, they should be combined into a single method.
Also
LCSphysicist said:
cos(kx + Φ) need to be 0
Φ = (2n-1)*π
I assume you meant cos(kL + Φ).
I would have thought the free end would have maximum amplitude.
On the other hand, cos((2n-1)*π)=-1, not 0, so you have effectively taken it as max.
 
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