6
$\begingroup$

I found an intuitive reason for the sine function, but want one that can be used for all kinds of waves. I've found a good explanation summarized in the following figure:

Circle and Sin

It is logical because a circle is symmetric and starts to change in height fast near the x axis, but gets slow near the y axis. But intuitively, why is it just a circle?

$\endgroup$
  • 3
    $\begingroup$ Related: Why can a wave be expressed with a sine function? and links therein. $\endgroup$ – Qmechanic Jun 13 '18 at 20:44
  • 1
    $\begingroup$ Intuitively: if you observe a sinusoidal wave in nature then, indeed, something of fixed radius is rotating. If the sines are the result of Fourier analysis, then someone is simulating net motion by adding a bunch of rotating elements. $\endgroup$ – amI Jun 13 '18 at 23:45
  • 2
    $\begingroup$ related on electronics.SE: What is a sine wave?. Several answers explain why sine waves occur so often, and others point out that decomposing other signals into sine waves (as opposed to square waves or triangle waves...) is natural because any linear filter won't create new frequencies of sine wave. That's what makes sines special, and why Fourier transforms are mathematically nice while decomposing to square waves would be a mess. $\endgroup$ – Peter Cordes Jun 14 '18 at 2:24
  • $\begingroup$ I see two different questions here: a) why can all waves be represented by a circle? b) why are there no other shapes to represent a wave? $\endgroup$ – Jasper Jun 14 '18 at 15:36
  • $\begingroup$ waves(i mean travelling EM wave) aren't represented by circles there $\mathbf {phasors}$ are represented by circles $\endgroup$ – Faraday Pathak Aug 30 '18 at 22:12
14
$\begingroup$

At its core, physics is fundamentally built from differential equations. Solving a particular set of differential equations called the equations of motion allow us to determine the paths of objects acting under various types of forces. As such, if a particular function is the solution to some of the more common differential equations, you might expect to see it everywhere.

Sine and cosine functions are the solution to the ordinary differential equation

$$\frac{d^2 x(t)}{dt^2}=-kx(t)$$

where $k$ is positive. This kind of system is found very commonly in nature, because it's a linear approximation for a ton of other systems (which are generally more complicated, but we can usually ignore the details if we disturb the system gently enough). Basically anything that moves back and forth in a potential well will eventually be describable by sinusoidal motion, for small enough oscillations.

Sine and cosine functions are also the solution to a particular partial differential equation called the wave equation:

$$\frac{\partial^2 f(x,t)}{\partial t^2}=c^2\frac{\partial^2 f(x,t)}{\partial x^2}$$

This describes the motion of traveling waves on a string or another continuous body. This equation is, again, very common in nature, because it's also a simple approximation of lots of more complex behaviors in the limit of small disturbances.

There are also many other differential equations that have sine and cosine functions as the solution, but even with these examples, it shouldn't be surprising that we see sine and cosine everywhere.

On a more technical level, sine and cosine functions are also useful because they form a basis of the space of sufficiently smooth periodic functions. It's a mathematical fact that any repeating signal or pattern can be decomposed into a sum of a bunch of sine and cosine waves, as long as it doesn't include too many "jumps" (this process of decomposition is called the Fourier transform). As such, they're extremely important in the study of any kind of repeated motion, because oftentimes it's much easier to solve such problems after taking the Fourier transform.

On the deepest level, sine and cosine functions get all of the above properties because they are intimately related to the exponential function, through Euler's formula:

$$e^{it}=\cos t + i \sin t$$

The exponential function is the solution to the simplest nontrivial differential equation, $\frac{dx(t)}{dt}=x(t)$ (for which the solution is $x(t)=e^t$). Because of this, the exponential function is the building block for almost all linear differential equations; since we like to work with linear approximations in physics, it's no surprise that you see exponentials, and hence sines and cosines, everywhere you go.

$\endgroup$
  • $\begingroup$ Note: The 1D wave equation is special, because it has a general solution in the form $f(x-ct)$. So I the 3D equation would be a better example. $\endgroup$ – Botond Jun 13 '18 at 19:47
  • $\begingroup$ Nice answer. But I think you should mention explicitly that cos is the derivative of sin, and why that's significant (even though that's implicit in your 1st equation). $\endgroup$ – PM 2Ring Aug 30 '18 at 0:46
10
$\begingroup$

The circle is just one way to see those oscillations (the x component of a circular movement is a sinusoidal function).

But the actual reason is that sines and cosines are the simplest solutions to the wave equation; and, since the wave equation is linear, any sum of solutions is also a solution. So any sum of sines and cosines is also a possible "wave".

Fourier analysys tells you how any shape can be achieved by summing sines and cosines.

So we want to work out with the simplest case, as the complex ones are linear combinations of these.

$\endgroup$
5
$\begingroup$

Thanks to QuirkyTurtle98. I should clarify that this answer does not apply to wave packets. Wave packets are composed of numerous periodic waves, but are not themselves periodic. You would not represent a wave packet as a circle.

Phase space is a mathematical space with as many dimensions as your system has degrees of freedom. For example, if your system consists of one particle in 3D space, the phase space for that system is 3D. If you have two particles in 3D space, then the phase space is 6D ($2\times3=6$). An N particle system in 3D space is 3N-dimensional. Every possible configuration of your system is a single point in phase space. As the configuration of the system changes with time, the point it occupies in phase space traces out a trajectory.

A wave is periodic. If your system exhibits wave (or any oscillatory) behavior, it is periodic and will continuously return to the same configuration given sufficient time. So the path that your system follows in phase space is closed. That path doesn't have to be a circle, but a circle is the simplest example (any path that is not self-intersecting can be transformed into a circle).

In the specific case of the sine function depicted in your animation, you have one degree of freedom (the y axis). You get your second dimension (the x axis), and therefore a circle, by treating the velocity as an additional degree of freedom. Positive y represents the function's value being positive, and positive x represents movement in the upward direction on the graph, or a positive time derivative (positive because the second graph (depicting the sine function) has it's zero of time ($t=0$) to the right of the x axis, so that time is continuously increasing at the origin).

Simple intuitive answer:

A wave changes as time passes. Imagine that you have many time slices, each one being just one moment. In each one the wave is slightly different. Because the wave repeats itself, every time slice will be identical to some time slice before it. Each point on the edge of a circle represents one time slice. When you move along the edge of the circle, you're moving through all the time slices, and the wave "moves" like in a flip-action animation:

flip action animation

The points make a circle because the wave repeats itself many times. As it repeats you pass through the same time slice (or an identical time slice) over and over. Since each point represents a time slice, you pass through the same points over and over, and end up with a closed circle.

Why a circle, as opposed to e.g. a square (also closed)? Imagine a loop made by tying some string. No matter what shape you twist the loop into, you can always stretch it out into a circle:

enter image description here

So although a wave could be represented as another shape, you can always transform that shape into a circle. It is easiest to just use a circle, though it is correct to note that you could have some waves represented by different shapes.

$\endgroup$
  • $\begingroup$ A wave is not necessarily periodic, eg wave packets are solutions of the wave equation but are not necessarily periodic in time no space. $\endgroup$ – QuirkyTurtle98 Jun 13 '18 at 19:05
  • $\begingroup$ @QuirkyTurtle98 the wave packet itself would not be represented as a circle in any space as it is not periodic. It would seem that wave packets are cases that fall beyond the scope of the question, as the question seems to be asking specifically about waves representable as circles. $\endgroup$ – The Ledge Jun 13 '18 at 19:10
  • $\begingroup$ Yeah, I know. But you say 'A wave is periodic.', I just try to complete your answer :) $\endgroup$ – QuirkyTurtle98 Jun 13 '18 at 19:13
  • $\begingroup$ Thanks, but your explanation it's not sufficiently intuitive and easy for me. All simple situations represent wave as circles, but there are another periodic functions. In case of springs (Hooke's law) it's easy to understand, but what about sound waves?and electromagnetic waves? $\endgroup$ – Paulo Buchsbaum Jun 13 '18 at 19:18
  • 1
    $\begingroup$ @PauloBuchsbaum Springs, sound waves, and electromagnetic waves all can be modeled by the same kinds of differential equations, because they all involve small disturbances in some sort of potential well. Sine and cosine functions are the solutions to those differential equations. $\endgroup$ – probably_someone Jun 13 '18 at 19:21
2
$\begingroup$

After much research, I found a posted answer that does not depend on such arbitrary facts as assuming Hooke's law is true.

Suppose Energy(E) as a continuous function of displacement (y)

$$\color{blue}{E = f(y)}$$

Using Taylor Series for a continuous function we have (where $\color{blue}{'}$ represent a derivative)

$$\color{blue}{E(y) = E(0) + E\,'(0)y + \frac{E\,''(0)y^2}{2\,!} + \frac{E\,'''(0)y^3}{3\,!} + ...}$$

$\color{blue}{E(0)}$ is a constant which depends on reference. It can be considered as $\color{blue}{0}$ and selected as, for instance, the position of the tip of a stretched spring, when the movement changes direction. This is a characteristic of any type of oscillatory movement.

$\color{blue}{E\,'(0)}$ is $\color{blue}{0}$. Reversing the direction of motion in a continuous function means that the first derivative is null in the reference point. Non-oscillatory movements are not included in that reasoning.

$\color{blue}{E\,''(0)}$ as a constant $\color{blue}{k}$, since it is the value of a function at a given point.

$\color{blue}{E\,'''(0)}$ and so on can be dismissed in simpler models.

So

$$\color{blue}{E(y) = \frac{k\,y^2}{2}}$$

This is the formula of the potential energy of a spring, when someone pull a spring and hold it. Now we can derive it in relation to x (Energy = Force * Distance):

$$\color{blue}{E\,'(y) = F = -ky}$$

As the force acts for something to return to a stable position, the constant $k$ shoud be preceded by a - signal.

The above expression can be expressed as a function of the acceleration, 2nd. derivative of displacement function, when expressed as a function of time t (m is mass).

$$\color{blue}{ m\,y\,''(t) = -k\,y(t)}$$

The most general solution of the above differential equation where the second derivative is the function itself, with the changed signal corresponds to:

$$\color{blue}{y(t) = A\,sin(\omega t + \Phi)}$$ $$\mathsf{or}$$ $$\color{blue}{y(t) = A\,cos(\omega t + \Phi)}$$ $$\mathsf{or}$$ $$\color{blue}{y(t) = A\,sin(\omega t) +B\,cos(\omega t)}$$

Where amplitude is maximum value ($\color{blue}{A}$ in the first two, and $\color{blue}{\sqrt{A^2+B^2}}$ in the third ) , $\color{blue}{\Phi}$ is phase and it is shown easily that $\color{blue}{\omega = 2\pi/T}$ ($\color{blue}{T}$ is the period, time for a full lap).

There are 2 different constants because are 2 freedom degrees in double derivation.

For instance, if one derives the first solution twice, one gets

$$\color{blue}{y''(t) = -A\,\omega^2 cos(\omega t)}$$

That can be rewritten as

$$\color{blue}{y''(t) = -\omega^2 y(t)}$$

So

$$\color{blue}{\omega = \sqrt { k/m}}$$

Remembering that it is indifferent to use sine or cosine because

$$\color{blue}{sin(y+\pi /2) = cos(y)}$$

Let's forget the other 2 solutions and let's focus on the first solution ($\color{blue}{\sin}$)

To visualize better, if we imagine this function as expressing the vertical oscillation of an longitudinal oscillatory motion. Let's consider $\color{blue}{\Phi = 0}$ so $\color{blue}{y(t) = A\,sin(\omega t)}$.

It's possible interpret the $\color{blue}{y}$ value as $\color{blue}{\sin}$ value in a uniform circular motion in a circle of radius $\color{blue}{A}$, see in a profile view, as the eye in the below figure:

Circle & Wave

Any above expression represents simple harmonic motion (SHM), that can be displayed as circular movement with constant angular speed. Over time, the amplitude draws a sine graphics.

Most oscillating system will behave like a vibrating spring, so long as the oscillations are small enough. For this reason, the vibrating spring, or simple harmonic oscillator (SHO) as it is called is very important in Physics.

$\endgroup$
  • 1
    $\begingroup$ We have MathJax running on the site so that mathematics can be typeset neatly in-line using a LatEX-mathmode alike language, and this is *strongly *preferred over images of equations. $\endgroup$ – dmckee Jun 14 '18 at 17:17
  • 2
    $\begingroup$ That said, your assertion that your answer "does not depend on such arbitrary facts as assuming Hooke's law is true." isn't actually true. After defining the Taylor series you make use of the form of the potential energy that comes from Hooke's law as the starting place for talking about physics. Same underlying assumption. $\endgroup$ – dmckee Jun 14 '18 at 17:19
  • $\begingroup$ I've deduced a potencial energy formula using Taylor expansion derived from a generic energy function of displacement, dismissing some terms. I know that it' not a real proof, but it help us convince that is true. It's best than throw potential energy formula without any explanation. $\endgroup$ – Paulo Buchsbaum Jun 14 '18 at 17:37
  • $\begingroup$ Se also in web.physics.ucsb.edu/~fratus/phys103/LN/SHM.pdf, classes notes written by Keith Fratus (Stanford) $\endgroup$ – Paulo Buchsbaum Jun 14 '18 at 17:52
  • $\begingroup$ I've made MathJax convertion. It was much better and easier. Formulas was placed in the middle, helping to hightlight them. $\endgroup$ – Paulo Buchsbaum Jun 14 '18 at 18:32
0
$\begingroup$

Here's a differential-equations-free perspective. Linear nondispersive waves are superpositions of moving profiles which are undeformed as time goes on. To be precise, in one dimension they are functions $$f(x - v t)$$ where $v$ is the wave speed. Any such function may be considered a collection of ``moving point masses" $\delta(x-vt+\phi)$ where $\phi$ is a constant phase shift.

If the profile $f$ is periodic, meaning that $x \mapsto x + \lambda$ is a symmetry of $f(x-vt)$, then our point masses are collected into families $$\sum_{n \in \mathbb{Z}} \delta(x - vt + n\lambda + \phi)$$ which all have the same weight in $f$. Using the method of images, this is equivalently a single particle moving at a constant speed around a circle.

So it doesn't really have anything to do with the motion, just the relationship between periodic functions and the circle.

$\endgroup$

protected by Qmechanic Jun 13 '18 at 20:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.