Changing the period of sine versus arc length

Question

Let's consider $ y = \sin x $. Let $ s \in \mathbb{Q} $ and $ s > 1 $. One may calculate the arc length of sine between $ 0 $ and $ 2\pi s$ using the formula:

$$ L = \int_0^{2\pi s} \sqrt{1 + \cos^2 x}\, dx $$ which after certain manipulations reveals to contain an elliptic integral of second kind, which cannot be solved in elementary terms, but can be computed numerically for given values using certain algorithms (discussed for instance here: What is the length of a sine wave from $0$ to $2\pi$?).

Now, let's assume that one changes the period from $2\pi$ to $\frac{4\pi}{3}$.

Question: how should the amplitude of $\sin x$ be changed (which currently is $|A|=1$), so that after the decrease of the period, the length of the arc from $0$ to $2\pi s$ remains $L$?

Edit #1:

As per suggestion in the comment, we take general form of sine wave equation:

$$ f(x) = y_{max} \sin {\frac{2\pi x }{\lambda}} $$

so the derivative would be:

$$ f'(x) = y_{max} \frac{2\pi}{\lambda} \cos {\frac{2\pi x }{\lambda}} $$

thus the arc length formula is:

$$ L = \int_0^{2\pi s} \sqrt{1 + y_{max}^2\frac{4\pi^2}{\lambda^2} \cos^2 {\frac{2\pi x }{\lambda}} }\, dx $$

Well, that's a start...

Answer 1

Consider the incomplete elliptic integral of the second kind $E(\varphi, k) = \int \limits _0 ^\varphi \sqrt {1 - k^2 \sin^2 t} \ \Bbb d t$. Then, with the first period,

$$L = \int \limits _0 ^{2 \pi s} \sqrt {1 + \cos^2 t} \ \Bbb d t = \int \limits _0 ^{2 \pi s} \sqrt {1 + (1 - \sin^2 t)} \ \Bbb d t = \\ \sqrt 2 \int \limits _0 ^{2 \pi s} \sqrt {1 - \left( \frac 1 {\sqrt 2} \right)^2 \sin^2 t} \ \Bbb d t = \sqrt 2 E \left(2 \pi s, \frac 1 {\sqrt 2}\right) .$$

With the second period and an unknown amplitude $A$, the curve becomes $y = A \sin \frac 3 2 x$, so the same length is now given by

$$L = \int \limits _0 ^{2 \pi s} \sqrt {1 + \frac 9 4 A^2 \cos^2 \frac 3 2 t} \ \Bbb d t = \frac 2 3 \int \limits _0 ^{3 \pi s} \sqrt {1 + \frac 9 4 A^2 \left( 1 - \sin^2 u \right)} \ \Bbb d u = \\ \frac 2 3 \sqrt {1 + \frac 9 4 A^2} \int \limits _0 ^{3 \pi s} \sqrt {1 - \frac {9 A^2} {4 + 9 A^2} \sin^2 u} \ \Bbb d u = \frac 2 3 \sqrt {1 + \frac 9 4 A^2} \ E \left( 3 \pi s, \frac {3A} {\sqrt {4 + 9A^2}} \right) .$$

This means that you want to solve for $A$ in the equation

$$\sqrt 2 E \left(2 \pi s, \frac 1 {\sqrt 2}\right) = \sqrt {\frac 4 9 + A^2} \ E \left( 3 \pi s, \frac {3A} {\sqrt {4 + 9A^2}} \right) .$$

Good luck with that, but don't count on me for it - I guess it's obvious why! :) Even for "nice" values of $s$ this would still be a nightmare: first, because $A$ is inside $E$ and inverting $E$ is not a reasonable project; second, because even knowing the inverse of $E$ with respect to the second parameter, you would still have an $A$ outside of it.

Your problem is like the often asked question about solving $x \Bbb e ^x = 1$ - you may express the root with the aid of some "known" function, but that would only mean to give a name to something that you still will not know! On the other hand, approaching it with numerical algorithms will probably give you useful results.

Answer 2

We take the sine curve more generally as:

$$ y = y_{max}\,\sin \frac{2 \pi x}{\lambda}, \quad \text{and}\quad \mu = 2 \pi y_{max} $$

(We do not take $y_{max}=1, \lambda = 2 \pi $)

Then by the integration procedure you outline we obtain arc length of a single wave as:

which is relation between $\lambda$ and $\mu$.

$$ \pi /2 \cdot\,\text{ Arc length of wave} = \sqrt{ { \lambda^2+ \mu^2 }}\cdot EllipticE \left( \frac{\mu^2}{\mu^2+\lambda^2}\right), $$

and for partial length of sine curve $ 2 \pi x < \lambda $

$$ 2\pi \cdot\,\text{ Arc length of wave} = \sqrt{ { \lambda^2+ \mu^2 }}\cdot EllipticE \left( \frac{2 \pi x}{\lambda},\frac{\mu^2}{\mu^2+\lambda^2}\right). $$

Answer 3

You could use a power series representation of the arc length formula at the end of your question with the variable $t=y_{max}^2\frac{4\pi^2}{\lambda^2}$, then set up the equation $P(t)-\int_0^{2\pi} \sqrt{1 + \cos^2 x}\, dx=0$, where $P(t)$ is the power series, then use a root finding algorithm like Newton's Method to solve for $t$, and then use that value of $t$ to solve for $y_{max}$. I would have a guess for what $t$ would be, and center the power series around that point. But no matter what you do, you will have to solve this numerically.

Answer 4

TL;DR: What you are asking to do simply does not work. There is no such amplitude.

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Literally, what you are asking to do is to find an amplitude $k$ such that the arc length of $y = k \sin\left(\frac32 x\right)$ from $0$ to $2\pi s$ is always the same as the arc length of $y = \sin(x)$ from $0$ to $2\pi s$ for rational $s > 1,$ that is,

$$ \int_0^{2\pi s} \sqrt{1 + \cos^2(x)}\d x \stackrel?= \int_0^{2\pi s} \sqrt{1 + \tfrac94 k^2\cos^2\left(\tfrac32 x\right)} \d x. $$

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Let's try this first for the particular value $s = 2.\newcommand{d}{\,\mathrm d}$

Let $f(x) = \sin(x)$ and $g(x) = k \sin\left(\frac32 x\right),$ so that $g$ has period $\frac43\pi.$ We have $f'(x) = \cos(x),$ so the arc length of $f$ is

\begin{align} L_{f,2} &= \int_0^{4\pi} \sqrt{1 + (f'(x))^2}\d x \\ &= \int_0^{4\pi} \sqrt{1 + \cos^2(x)}\d x \\ &= 2 \int_0^{2\pi} \sqrt{1 + \cos^2(x)}\d x \\ &\approx 15.2808, \end{align}

since the interval $[0,4\pi]$ contains two copies of the curve over the interval $[0,2\pi]$, identical except for translation. (The integral is an elliptic integral; see What is the length of a sine wave from $0$ to $2\pi$?.)

Since $g'(x) = \frac32k\cos\left(\frac32 x\right),$ the arc length of $g$ is

\begin{align} L_{g,2} &= \int_0^{4\pi} \sqrt{1 + (g'(x))^2}\d x \\ &= \int_0^{4\pi} \sqrt{1 + \tfrac94 k^2\cos^2\left(\tfrac32 x\right)}\d x \\ &= 3 \int_0^{4\pi/3} \sqrt{1 + \tfrac94 k^2\cos^2\left(\tfrac32 x\right)}\d x \end{align}

since the interval $[0,4\pi]$ contains three copies of the curve over the interval $\left[0,\frac43\pi\right]$.

Now do a substitution $u = \frac32 x.$ Then

\begin{align} \int_0^{4\pi/3} \sqrt{1 + \tfrac94 k^2\cos^2\left(\tfrac32 x\right)}\d x &= \int_0^{2\pi} \sqrt{1 + \tfrac94 k^2\cos^2(u)} \cdot \tfrac23 \d u \\ &= \frac23 \int_0^{2\pi} \sqrt{1 + \tfrac94 k^2\cos^2(u)} \d u \end{align}

and therefore the arc length of $g$ is $$ L_{g,2} = 2 \int_0^{2\pi} \sqrt{1 + \tfrac94 k^2\cos^2(u)} \d u. $$

Now to make the arc lengths equal we need to make the integrals $\int_0^{2\pi} \sqrt{1 + \cos^2(x)}\d x$ and $\int_0^{2\pi} \sqrt{1 + \tfrac94 k^2\cos^2(u)} \d u$ equal, and the obvious way to do that is to set $\tfrac94 k^2 = 1,$ that is, $ k = \tfrac23.$ So we set

$$ g(x) = \tfrac23 \cos\left(\tfrac32 x\right). $$

And indeed then the arc length of $g$ is $$ L_{g,2} = 3 \int_0^{4\pi/3} \sqrt{1 + \cos^2\left(\tfrac32 x\right)}\d x \approx 15.2808. $$

Now let's try $s = \frac94.$ Then the arc length of $f$ is $$ L_{f,9/4} = \int_0^{9\pi/2} \sqrt{1 + \cos^2(x)}\d x \approx 17.1909 $$ while the arc length of $g$ is $$ L_{g,9/4} = \int_0^{9\pi/2} \sqrt{1 + \cos^2\left(\tfrac32 x\right)} \d x \approx 17.1222. $$

Oops!

So, why did the function $g$ that worked for $s = 2$ not work for $s = \frac94$? What's happening is that you're trying to set $k$ so that for all rational $s > 1$ you have the equation $$ \int_0^{2\pi s} \sqrt{1 + \cos^2(x)}\d x \stackrel?= \int_0^{2\pi s} \sqrt{1 + \tfrac94 k^2\cos^2\left(\tfrac32 x\right)} \d x. $$

But an antiderivative of a continuous function is continuous, so if you can do this for rational $s > 1$ you can do it for all $s > 1,$ and if you can make the two sides equal then their derivatives also are equal, which can only occur if

$$ \sqrt{1 + \cos^2(2\pi s)} \stackrel?= \sqrt{1 + \tfrac94 k^2\cos^2\left(3\pi s\right)} $$

for all $s > 1,$ which in turn requires that $$ \cos(2\pi s) \stackrel?= \tfrac32 k\cos\left(3\pi s\right), $$ that is, it requires us to find an amplitude $k$ on the right that makes two sinusoids of unequal period equal. And that simply does not happen in general.

The arc lengths of the two curves cannot always accumulate at the same rate: if the are equal at one value of $s,$ then one will be ahead or behind the other at some other value of $s.$