How does this hypotenuse/derivative/polynomial approximation work?

Question

I am trying to understand an article regarding the forces of a vibrating string in terms of both longitudinal and transverse waves.

Here is the pertinent excerpt:

Equation 1 is obvious. It is just Pythagorean theory. Equation 2 is also obvious as it just converts this to derivatives. Equation 3 also makes sense - tension in a string is equal to the base tension plus the product of length deviation (extra stretch), Young's Modulus, and cross sectional area.

The part I don't get is equation 4. If you put equation 2 in equation 3 you definitely don't get equation 4.

I know equation 4 is some type of approximation, but I don't know how it works. I found another reference that seems to discuss this same approximation and they say of it: "It has been shown that by expanding both y and ξ as a series of polynomials and truncating at third order, the added force per unit length on the element in the longitudinal direction caused by the transverse displacement is given by [this equation]."

But I'm not a mathematician and I don't understand how this is represented by a series of polynomials or how they describe this approximation.

Does this make sense? Can anyone explain it in simple terms?

Thanks for any help. Very appreciated.

Answer 1

For the sake of simplicity, let $$a = \frac{\partial \xi}{\partial x}, \quad b = \frac{\partial y}{\partial x}.$$ Then we can write $(3)$ as $$\frac{ds}{dx} = \sqrt{(a+1)^2 + b^2}.$$ The approximation in $(4)$ is then equivalent to the claim $$\sqrt{(a+1)^2 + b^2} - 1 \approx a + \frac{1}{2}b^2.$$ Is this reasonable? Well, consider $$\left(a + \frac{1}{2} b^2 + 1\right)^2 = (a+1)^2 + (a+1)b + \frac{1}{4} b^4.$$ Although we do get the $(a+1)^2$ term, the rest doesn't seem to fit.

Is there another approach? What about a series expansion about $b = 0$? We get $$\sqrt{(a+1)^2 + b^2} = (a+1) + \frac{b^2}{2(a+1)} - \frac{b^4}{8(a+1)^3} + O(b^6).$$ So if $a \approx 0$, a low-order series expansion seems to be the motivation for the approximation. But without the underlying physical context, I cannot say whether it is reasonable.

Answer 2

If you have $a << 1$ , then

$$(1 + a)^n \approx 1 + na$$

This is because the terms after the linear term are so small that they can be ignored in comparison

$$\frac{ds}{dx} = \left[\left(1 + \frac{d \xi}{dx}\right)^2 + \left(\frac{\partial y}{\partial x}\right)^2\right]^{1/2}$$

$$\implies \frac{ds}{dx} = \left[1 + \left(\frac{d \xi}{dx} \right)^2 + 2 \frac{d \xi}{dx} + \left(\frac{\partial y}{\partial x}\right)^2\right]^{1/2}$$

Now I guess the assumptions taken are

$$\frac{d \xi}{dx}^2 \text{is ignored compared to other terms}$$

$$2 \frac{d\xi}{dx} + \frac{\partial y}{\partial x}^2 <<1$$

Hence you will get the terms you see. Whether these are reasonable I wouldn't know