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Context

Considering a finite string of length $L$. We suppose gravitation is negligible. The string is only moving transversaly, in $y$ direction, with small amplitude. The tension of a string is $T$.

We're looking for $y(x,t)$. Right end is fixed $y(x=L,t)=0$. Left end is driven with a $y$ force $f(t)= Fe^{i\omega t} $. We are looking for solution of the form $y(x,t) = Ae^{i(\omega t -kx)} + Be^{i(\omega t +kx)}$

Question

In the book (see source below), the boundary condition on left end is defined as $Fe^{i\omega t} + T \frac{\partial y}{ \partial x}(x=0) = 0$

I don't get this boundary condition: it assumes that at the left end the driving force is equal to the force exerted from the string (coming from the tension). In my opinion, this is a contradiction with the solution we are looking for : if left end is at equilibrium (forces are compensating), it shouldn't move (or at least at constant speed). However, at $x=0$, the solution we are looking for is oscillating with term $e^{i\omega t}$. How do you understand that ? Am I wrong ? is this formulation wrong ?

Source

Example taken from Fundamentals of acoustics, 4th edition, p43.

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3 Answers 3

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The equation states that the transverse applied force at one point in the string is equal to the transverse force at that point expressed in terms of the string tension. It is not stating that there is no net force on a portion of string. If it did say the latter, then it really would contradict the string being able to accelerate.

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  • $\begingroup$ There is no other force applied, only the driving force, and the tension of the string. So if those two are compensating, I assume that there is no net force on the left end. $\endgroup$
    – Setoh
    Commented Sep 10, 2021 at 8:58
  • $\begingroup$ That's right. But the left hand end is only one infinitesimally small slice of string and, having no mass, needs no net force to accelerate it. This is the same point that I was making in my answer. $\endgroup$ Commented Sep 10, 2021 at 16:02
  • $\begingroup$ got it, didn't read it like this on my 1st pass :) $\endgroup$
    – Setoh
    Commented Sep 10, 2021 at 20:06
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I think your mistake is that you consider that the force exerted by the string on the outside is a force acting on the string.

At the left end, the only force acting on the string is the external force $F$. And by Newton's third law, this force $F$, from the outside on the string, is the opposite of the force exerted by the rope on the outside: hence the boundary condition.

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After thinking for some time on this, I conclude that my reasoning is wrong : I cannot treat the end of the string as a particle with a mass with a finite force exerted on it.

Since it's an infinitesimal element with mass $\rho dx$, any non vanishing finite force should cause an infinite acceleration. So we must ensure that finite force are vanishing.

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  • $\begingroup$ That's exactly the point I was making! $\endgroup$ Commented Sep 10, 2021 at 16:10

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