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When discussing the vibrating string problem with one end (or both) free to move in the vertical direction but constrained in the longitudinal direction (achieved by placing the "free" end in a frictionless sleeve for example), it is generally accepted that the proper boundary condition to impose at that end is the homogeneous Neumann condition, that is $\frac{\partial u} {\partial x} = 0$ where $u$ is the vertical displacement and $x$ is the longitudinal space coordinate.

I have looked in a few books, and couldn't find a rigorous explanation of why this boundary condition should be enforced, the authors usually vaguely stating that this is due to the lack of force at the open end. Has someone encountered a more detailed explanation of this?

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The open boundary condition means, as stated in the question, that at the boundary no force acts on the end of the string in the direction of elongation.

As the tip of the string has infinitesimal mass, we can argue as if we were considering conditions for static equilibrium (if the forces caused by the string would differ from the forces caused by the wall by a finite amount, the end of the string would be accelerated infinitely which is unphysical).

The forces acting on the end of the string can easily be analysed: The force acts in the direction of the string (because an ideal string has, by defintion, no resistance to bending) and is equal in magnitude to the tension of the string.

This means that there is no force parallel to the boundary if the direction of the string is perpendicular to the surface. This condition can obviously be encoded in the requirement that $\partial_x u = 0$ (as $u$ is the $y$ coordinate of the string at position $x$, so the string is perpendicular to the boundary if the slope of the graph is zero).

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  • $\begingroup$ The mass density is uniform, so the mass in the element $l$ at the end of the string has mass $l \sigma$ which gets arbitrarily small when $l \to 0$. $\endgroup$ – Sebastian Riese Apr 17 '18 at 11:57
  • $\begingroup$ The special thing is that the force at the end of the string is not balanced (the other answer demonstrates that), but the entire tension of the string acts there (while only the difference of the tension to the right and to the left acts on any other mass element). $\endgroup$ – Sebastian Riese Apr 17 '18 at 13:54
  • $\begingroup$ As to the experimental evidence: An open organ pipe is governed by the same wave equation and has a discrete spectrum. (A string that is tensioned frictionless is not so easy to implement in the physical world ...). $\endgroup$ – Sebastian Riese Apr 17 '18 at 13:56
  • $\begingroup$ Yes it has an obvious interpretation: It is the tangent of the angle which the string has relative to the static state (and since the angle is small, the tangent is approximately equal to the angle). $\endgroup$ – Sebastian Riese Apr 17 '18 at 16:02
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I think that Sebastian Riese's answer captures the essence of the argument. Just to add to it for those who might be interested, here is a mathematical version of the argument:

Let a string extend from $x = 0$ to $x = L$ and let $u(x)$ be the $y$ coordinate of each point of the string. Consider a portion of the string from $x = 0$ to $x = b$ and apply Newton's second law to it, which states that the sum of external forces on this portion of the string equals the sum of the mass times acceleration of the elements composing the portion of string. Consider the $y$ component of this equation, then the string being considered as a continuous medium, the latter quantity can be expressed as the integral of the string's density times the second order derivative of $u(x)$ with respect to time, $\int_0^b \rho u_{tt}(x) dx$.

Now consider the external forces on the portion of string in the $y$ direction. Internal forces due to the tension of the string do not contribute a net force because of the action-reaction law, so that the net forces only occur at the end points of the string. At $x = 0$, because the end of the string is constrained to remain at $x = 0$ but is otherwise free to move along $y$, the net force along $y$ is null. At $x = b$, a net force will result from the tension in the string (the atoms on the left and right of $x = b$ are pulling on each other, but only the atom on the left are part of the portion of string considered). Tension being a force aligned along the string, the $y$ component of the force at $x = b$ is $T \frac{u_x}{\sqrt{1+u_x^2}} \approx T u_x$ where $T$ is the tension magnitude. Therefore Newton's second law along $y$ reads

$T u_x(b) = \int_0^b \rho u_{tt}(x) dx$

Now assume that $u_x(x)$ is continuous and that $u_{tt}(x)$ is continuous and bounded, taking the limit $b \rightarrow 0$ yields $u_x(0) = 0$, which is the boundary condition.

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