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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)?

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    $\begingroup$ Do you have a reference to one of those derivations? $\endgroup$
    – nicoguaro
    Dec 23, 2021 at 18:52
  • $\begingroup$ @nicoguaro see p.2 of this link web.mit.edu/1.138j/www/material/chap-1.pdf $\endgroup$
    – user246795
    Dec 23, 2021 at 22:16
  • $\begingroup$ (a) $\mathcal{O}[(\partial \psi / \partial x)^2]$ is the integrand, not the evaluated integral, surely. (b) Why don't you try evaluating the integral for (say) a $\psi$ of a small amplitude that varies sinusoidally with distance? $\endgroup$ Jan 1, 2022 at 13:13
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    $\begingroup$ Yes. The estimate $O((\partial \psi/\partial x)^2$ is correct. Just use the binomial theorem on the integrand: physics.stackexchange.com/questions/685858/… $\endgroup$
    – mike stone
    Jan 1, 2022 at 13:29
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    $\begingroup$ You can assume that $|\partial \psi/\partial x|^2$ is uniformly bounded if you like. But you do not need to get it out of the integral because the change in the tension $T$ is given by $\delta T =Y \delta L/L$, (Y= Young's modulus) so it is the integrand that matters not the integral. $\endgroup$
    – mike stone
    Jan 1, 2022 at 13:46

2 Answers 2

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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.

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  • $\begingroup$ So is there no way to show that $$ \int_0^x\sqrt{1 + (\partial \psi / \partial x')^2} \, dx' - x = \int_0^x \frac 12 (\partial \psi / \partial x')^2 + ... \, dx' = \mathcal{O}[(\partial \psi / \partial x)^2], $$ in general, without assuming a sinusoidal $\psi$? I guess if you show it for a sinusoidal solution then Fourier methods can be applied for more general functions? $\endgroup$
    – user246795
    Jan 2, 2022 at 13:35
  • $\begingroup$ (a) I'd like to agree about the Fourier methods, but the connection between arc length and $\psi(x)$ is seriously non-linear and you can't just add arc lengths that each Fourier component would have by itself. (b) I found that, for the same amplitude and wavelength, a triangular waveform (easy maths without Fourier!) gives an extra length about 80% of that for the sinusoid – not that much different. (c) But with a square wave you're really in trouble: far higher arc length for a given $A$ and $\lambda$, as you can calculate in your head in a couple of seconds. Short wavelength Fourier comp's! $\endgroup$ Jan 2, 2022 at 15:22
  • $\begingroup$ @user246795 A possible way forward, though extremely crude, would be to look at the Fourier component that would, by itself, give the maximum arc length out of all the components. Writing $\lambda=\lambda_1/n$ and 𝐴=𝐴(𝑛) in which 𝑛 is the 𝑛'th harmonic, we have $$\frac{\Delta l}l=\frac{\pi^2[A(n)]^2 n^2}{\lambda_1^2}.$$ We therefore look at the harmonic for which 𝑛𝐴(𝑛) is a maximum. For a square wave, 𝑛𝐴(𝑛) is the same for all harmonics, so no wonder we're in trouble! but for a triangle wave, 𝑛𝐴(𝑛) falls off as 1/𝑛, so the fundamental dominates. $\endgroup$ Jan 2, 2022 at 19:20
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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 $L$.

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  • $\begingroup$ Is there equality in the second expression, or is that an upper bound? $\endgroup$
    – user246795
    Jan 4, 2022 at 9:02
  • $\begingroup$ You are right. It should have been written as an upper bound. I'll fix it. Thanks. $\endgroup$
    – C Tong
    Jan 4, 2022 at 11:57

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