Justification for constant tension in transverse wave on a string It is common in the derivation of the transverse wave equation on an ideal string to assume the tension along the rope is uniform in the limit that $$|\partial \psi / \partial x|\ll 1.$$ However, what are the justifications for why this is?
I have seen derivations reason that the instantaneous extension of the string is of second-order smallness:
$$
\int_0^x\sqrt{1 + (\partial \psi / \partial x')^2} \, dx' - x = \mathcal{O}[(\partial \psi / \partial x)^2],
$$
but what intermediate steps have they taken in concluding this?
Also, are there any alternative methods to argue why tension can be assumed to be constant (to first order approximation)?
 A: Assume for simplicity
$$\psi= A\sin (2\pi x/\lambda).$$
The extra length of the string over half a wavelength is
$$(\Delta l)_{\lambda/2}=\int^{\lambda/2}_0 \left[\sqrt{1+(d\psi/dx)^2}-1\right]dx\approx \int^{\lambda/2}_0 \frac12\left(\frac{d\psi}{dx}\right)^2 dx$$
This uses the binomial expansion of the square root to first order in $(d\psi/dx)^2$.
Substituting our simple $\psi(x)$ and evaluating the integral gives
$$(\Delta l)_{\lambda/2} =\frac{\pi^2A^2}{2\lambda}$$
So the fractional increase in length of the string is
$$\frac{\Delta l}{l}=\frac{\pi^2 A^2/2\lambda}{\lambda/2}=\frac{\pi^2 A^2}{\lambda^2}$$
So, for example, if $\lambda$ = 100 mm  and  $A$=0.2 mm, then $\Delta l/l \approx 4 \times 10^{-5}.$
You can show that subsequent terms in the binomial expansion are much smaller still! [The next term contributes a fractional change in length of $-\frac 34 \frac{\pi^4 A^4}{\lambda^4}$]
Suppose that the string (of length $l$) had initially been put under tension by stretching it a distance $\Delta l_0$ before clamping its ends. Then (assuming Hooke's law) the fractional change in tension due to $\psi(x)$ will be
$$\frac{\Delta T} T=\frac{\Delta l}{\Delta l_0}=\frac{\pi^2A^2}{\lambda^2}
\frac l {\Delta l_0}=\frac{\pi^2A^2 /\lambda^2}{\text{original strain}}$$
A typical original strain in a violin string is $3 \times 10^{-3}$, so with our value for $\Delta l/l$ due to $\psi(x)$, this gives a fractional change in tension of about 1%.
So what should we conclude? Certainly not that the string tension is always practically unaffected by any wave (stationary or progressive) that it carries. The extent to which the tension is affected can be calculated approximately from the last equation. The figures that I've used in order to give numerical estimates are to some extent arbitrary; I see no way to avoid this. I have a feeling, though that $A$ = 0.2 mm may be rather a large amplitude.
A: You can simply assume $|\frac{\partial\psi}{\partial x'}|<\epsilon$. Then
$$\int_0^x \sqrt{1+\left(\frac{\partial\psi}{\partial x'}\right)^2}dx'-x\le \int_0^x (1+\frac12\epsilon^2+o(\epsilon^4))dx'-x=O(\epsilon^2),$$
since the integral is always over a local neighborhood, i.e., $|x|<L$ for some constant $L$.
