# Meaningless sinusoidal wave?

1. Dec 23, 2005

### Mr. Tambourine Man

Forgive me for my ignorance on these matters. I know almost nothing about math or physics. My only knowledge on such things is what I have been able to grain from Popular Science/Math books. I have never even graduated from college. But please allow me to post a question. Although I want to get more in-depth as to why I am asking this question, unfortunately I am pressed for time. Can anyone give me a quick answer?

Is it meaningful to speak of a sinusoidal wave with infinite frequency and infitessimal amplitude? Would such an entity effectively be the same as a straight line or would it not?

Say one were to draw a "ray" straight down the middle of a wave such that its its crests and troughs have an equal perpendicular distance from said ray. I think it would be correct to say that this wave's "deviance" from this straight line is proportional to the total area of crests and troughs (i.e. the area whose boundaries are the ray and the wave itself). But if one decreases the amplitude ad infinitum, would it not be correct to say that the wave's deviance from the straight line is zero, thus equaling a straight line?

Thank you.

David.

2. Dec 23, 2005

### dimensionless

You could speak of a sin wave with an infinite frequency and infitessimal amplitude. A periodic wave with an infitessimal amplitude would basicaly be a straight line. In other words, as the amplitude aproached zero, the sin wave would aproach a straight line.

I don't think that the frequency has anything to do with thus. I might go so far as to say that a wave with low frequency is straighter than one with a high frequency, at least over a finite interval. At most points, however, it would not be parallel to ray drawn down the middle.

3. Dec 24, 2005

### Mr. Tambourine Man

The reason why I ask is that Joseph Campbell posed a paradox which intrigued me in his book "Oriental Mythology". It goes something like this:

Take a straight line of , say, 2 units long. Now draw a perfect sine wave over that line such that one phase (one crest, one trough) is covered over that line, the wave beginning with one end of the straight line and ending at the other end (in other words, the straight line is a "ray" parallel to the wave's direction of motion and cutting the wave in half). Now the crest and the trough are both perfect semicircles in such a juxtaposition with the straight line (this would all be so much easier if we could draw things on these message boards). The circumference of each such half-circle is satisfied by the relation c = d * (pi)/2, where "d" is the diameter, that being half the straight line (which is "1" in this case). Thus the circumference of one semicircle is (pi)/2. Since there are two semicircles, the sum of the two circumferences (which is the same as the distance covered by the wave) is pi.

Now double the frequency and halve the amplitude of this wave such that four, instead of two, semicircles appear. Now the circumference of each semicircle is c = .5 * (pi)/2, or (pi)/4. But there are four of them, so the total distance that the wave covers is still pi. This doubling of frequency and halving of amplitude can be continued indefinitely till the wave's frequency is infinite and its amplitude is infitessimal....and the distance the wave covers will still be pi units. But if a wave is effectively a straight line when it has zero amplitude (presumably whatever the frequency does should be irrelevant), then the distance it covers should be 2 units in this case. Therefore, strangely pi = 2, which is absurd.

What went wrong?

4. Dec 24, 2005

### HallsofIvy

Staff Emeritus
First, while Joseph Cambell is an excellent source on mythology, he is not perfect in mathematics! If you construct a true sine wave along a line, each loop is not a semi-circle. Also, I take it you meant to say that the length of a semi-period is proportional to the circumference of the loop, not proportional to the area.

Basically, what is happening here is that the limit of a sequence does NOT necessarily have the same properties as the terms of the sequence themselves. Here's a very simple example: every term in the sequence {1/n} is positive but the limit of the sequence, 0, is not.

A more classical version of your example is this: Draw a line with length 1 then construct a perpendicular also of length 1 (so you have two sides of a square). The total distance from one end to the other is 2. Now draw a perpendicular from the midpoint of the horizontal line to the diagonal, then horizontal then vertical again (in other words a stair step). Again, the total length is 2.
Keep repeating that "halving" getting a "stair" of more and more steps. At each division the total distance is 2 but in the limit it approaches the diagonal which has length $\sqrt{2}$.

5. Dec 24, 2005

### Mr. Tambourine Man

OK, thanx. btw, how did you add that "square root" symbol?

6. Dec 24, 2005

### krab

Just click on it and you'll see.

In your question, it matters whether the amplitude first goes to zero and then the frequency goes to infinity, or vice versa, or simultaneously: each will give a different limit.

Another example is 0/0. Take the ratio r/s. Now if r=0, this ratio will be zero even if we let s get smaller and smaller. OTOH, if s approaches zero, this ratio will approach infinity (be unbounded), no matter how small r is to start with. On the third hand, if r=s, it doesn't matter that you let them both become as small as you like, the ratio is always 1. So you have 3 answers to the same question.

7. Dec 24, 2005

### Tide

Halls,

I don't believe that Campbell referred to it as a "sine wave." I think use of that term was the OP's creation.

8. Dec 25, 2005

### uart

You can calculate the length along the sine curve and show that's it's most definitely different to a circle.

$$dl = \sqrt{dx^2 + dy^2} = \sqrt{ 1 + (y^{'})^{2}} dx$$

If you let $$y = 1/2 \sin(\pi x)$$ then you get something that's roughly similar to the two halves of a circle with diameter 1.

Then $$y^{'} = \pi/2 \cos(\pi x)$$

and $$l = \int_0^2 \sqrt{ 1 + (\pi/2 \cos(\pi x))^2} dx$$

Numerically evaluating the above integral gives the length along this particular sine curve (y) as $$l \simeq 2.93$$ which is obviously different than Pi.

The authors assertion that the length of the curve remains unchanged when you half the amplitude and double the frequency (and so on) is however correct.

Let $$y1 = 1/4 \sin(2 \pi x)$$

Now $$y1^{'} = \pi/2 \cos(2 \pi x)$$

and $$l1 = \int_0^2 \sqrt{ 1 + (\pi/2 \cos(2 \pi x))^2} dx$$

Substituting $$u=2x$$ (and $$dx = du / 2$$) gives,

$$l1 =1/2 \int_0^4 \sqrt{ 1 + (\pi/2 \cos(\pi u))^2} du$$

Since the integral part of the expression for $$l1$$ is exactly the same as that in $$l$$, except taken over two full periods instead of one, then it's value is exactly twice as large. This cancels the 1/2 out the front of the integral to give $$l1$$ exactly equal to $$l$$.

Last edited: Dec 25, 2005
9. Jan 24, 2008

### wisling

Bump - Found interesting :P

It seems that there is a contradiction in your statement. A wave does not essentially become a straight line unless the (inverse frequency)/amplitude goes to 0. If it does go to 0, the greatest slope of the wave also becomes 0. In the case where the amplitude decreases more quickly than th frequency, the greatest magnitude of slope of the line goes to 0 and the height above and below the normal becomes negligible. In the case where the frequency decreases faster than the amplitude, the distance over which the greatest slope decreases to 0 is negligible. In this case however, the frequency is doubled and the amplitude halved. This essentially keeps the ratio constant and neither of these cases apply so it's length can easily be calculated by multiplying the circumference of the semicircle by however many semicircles there are, which remains pi.

The example used by Joseph Campbell simply uses a wave whose length is easier to calculate. Whether it is a sinusoidal wave or a semi-circular wave is irrelevant.

10. Jan 24, 2008

### ObsessiveMathsFreak

I believe what you're looking for is something like the following.

If you consider a circular helix given by;
$$\mathbf{x}(s) = \left( \frac{\sqrt{1-a^2}}{\omega}\cos\left(\omega s\right), \frac{\sqrt{1-a^2}}{\omega}\sin\left(\omega s\right) , a s \right)$$

Then, regardless of the value of $$\omega$$, i.e. the frequency, the helix will always have length 1 after it has risen $$a$$ meters. $$a$$ is if you like, the rate of ascent.

With increasing $$\omega$$, the helix loops around more and more, but these loops have smaller and smaller radius, so over the same height, the curves will have the same length.

The curvature and torsions of the curve are (correct me if I'm wrong)

$$\kappa = \omega \sqrt{1-a^2}$$

$$\tau = a \omega$$

Which means as $$\omega \rightarrow 0$$, the curve becomes a straight line. However, as $$\omega \rightarrow \infty$$ both of these values become undefined (though their ratio is still constant, mean the curve is still a helix). Also, you can see that simply putting the loops radius equal to zero, with an "infinite" $$\omega$$, would give a discontinuity, as s would no longer be the arc length parameter.

For this curve at least, moving the frequency towards infinity will not result in a straight line or ray, though if we attempt to draw it we will find that it might eventually look like one from far enough away.

11. Jan 24, 2008

### disregardthat

Cant you just consider the function y=a sin(x) and let a --> 0 => y=0

12. Jan 24, 2008

### Gib Z

Yes, many straight lines can be approximated arbitrarily well by sinusoidal functions, within a certain interval at least. Just ask Joseph Fourier.