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What I am really asking is are there other functions that, like $\sin()$ and $\cos()$ are bounded from above and below, and periodic?

If there are, why are they never used to describe oscillations in Physics?

EDIT:

Actually I have just thought of a cycloid, which indeed is both bounded and periodic. Any particular reason as to why it doesn't pop up in science as much as sines / cosines?

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Related: physics.stackexchange.com/q/108423 –  DumpsterDoofus Apr 18 at 23:42
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Basically, the answer is: yes, there are many other periodic functions, and the reason you typically see harmonics (like $\sin,e^{i\omega t}$) used is because in most simple applications of interest which are easily understandable, either the behavior itself is harmonic, or the behavior is most easily understood in terms of harmonics. There is also a bit of a confirmation bias: the systems which are not easily understood in terms of harmonics are often very difficult, and thus less people know about them, and so less textbooks are written about them. –  DumpsterDoofus Apr 19 at 0:17
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Actually, the simple method is purely calculus: sine (and cosine, which is the same with a lag) is the solution of $\ddot{x}=-x$, which is the canonical, linear equation with an oscillatory solution. This means that this solution will be present in any linear oscillator. Then, the general approach with nonlinear problems is to break them down into linear problems, so you get back to sine. –  Joce Apr 23 at 7:47

8 Answers 8

up vote 26 down vote accepted

Part of it is that since Newtonian mechanics is described in terms of calculus.

When we consider vibrational motions, we're talking about some particle that tends to not be displaced from some equilibrium position. That is, the force on the particle, at displacement $x$, $F(x)$, is equal to some function of displacement $x$, $g(x)$.

There are two ways calculus gets involved here. Firstly, $F=ma$, and $a$, acceleration, is a "rate of change" and therefore a calculus concept. So we have $ma(x)=g(x)$.

Now, dealing with a general function $g$ is too difficult - we won't get anywhere with it. So how can we proceed in the most general way? One fruitful method is to do a Taylor expansion. $g(x)=g(0)+g'(0) x+\frac{1}{2} g''(0) x^2+\frac{1}{3!} g^{(3)}(0)x^3+\cdots$, where these are the $g^{(n)}(x)$ is the nth derivative of g at point x.

If we want $x=0$ to be an equilibrium position, we must have $g(0)=0$ - there isn't any force on the particle at equilibrium. If we want it to be a stable equilibrium that will tend to turn back to its original position, we must have $g'(0)<0$. All other derivatives are fair game. Writing $-k=g'(0)$:

$$m a(x)=-k x+\frac{1}{2} g''(0) x^2+\frac{1}{3!} g^{(3)}(0)x^3+\cdots$$ as is so useful in physics, we now suppose that $x$ is small, so that $x^2$ is very small, and $x^3$ is even smaller. That is, we ignore all powers of $x$ greater than one. We wind up with: $$m a(x)=-k x$$ Hooke's law. The solution to this equation is sinusoidal, always. (that is, it can be written in the form $x=a \cos(\omega t-\varphi)$)

So it is inevitable that, with these definitions of "stable equilibrium", the resulting vibrational pattern at small amplitudes will be sinusoidal. Always. That's what makes $\cos$ and $\sin$ special from a physical point of view.

(of course, we've also tacitly assumed that $g$ is a nice function that is nice and smooth and differentiable, but one generally does that when working on Newtonian style problems)

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IMO, of the four answers presented so far, this is the only correct one. –  garyp Apr 19 at 2:57
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Plus, of course, the fact that any periodic function can be expressed as a decomposition into sine functions (of increasing frequency). It's just, as NeuroFuzzy says, a question of how many terms you want to carry thru the calculation. –  Carl Witthoft Apr 19 at 16:35
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"So it is inevitable that, with these definitions of "stable equilibrium", the resulting vibrational pattern at small amplitudes will be sinusoidal. Always." Well, occasionally you get cases where $k=0$ and the first non-trivial term is the $g^{(3)}$ term (not $g^{(2)}$, because that makes your equilibrium unstable again), but these situations are few and far between. –  dmckee Apr 19 at 20:18
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@dmckee Another situation that occasionally comes up is when the energy function has a minimum, but is not differentiable, say $U(x)=k|x|$, $ma(x)=-k\operatorname{sgn}x$ (like the oscillations due to gravity when you jump in a portal on the ground in the game Portal). –  Mario Carneiro Apr 19 at 21:15
    
It is also common in vectors, Fourier series and are the fundamental aspects of trigonometry. –  user3058846 Apr 20 at 16:02

One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of periodic functions, why worry about the latter, when you can always reexpress any function as a sum of sines and consines, and a solution in this form is completely isomorphic with the general case, assuming your base equation is linear?

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Good point. But the question then becomes; Can you decompose any periodic function using something other than sine/cosine functions? –  Sam Pinkus Apr 20 at 10:59
    
The point is that for a large class of very, very common functions and relations encountered by simply looking around the real world you don't need to express anything as a sum (much less an infinite sum) of sin/cos, but a single sin/cos is sufficient - while another representation might need an [infinite] sum of those terms instead. –  Peteris Apr 20 at 14:42
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@SamPinkus: Absolutely. See for example the wavelet transform. –  Nate Eldredge Apr 20 at 16:16
    
@SamPinkus: See the 'generalized Fourier series,' which allows one to compute the Fourier series of a function in terms of any set of functions, providing they satisfy the orthogonality relations. –  JamalS Apr 26 at 12:58

Because cycles and oscillations and things with periodicity, are all intimately related to the circle. And $sin$ and $cos$ are defined based on the circle.

enter image description here

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This is better than dancing hamsters :-) –  Peter Mortensen Apr 21 at 15:30

A literal response to your title question would simply be "because in the physical world, oscillations behave in ways consistent with $\sin$ and $\cos$." Of course, one then wonders why these functions are so ubiquitous.

Depending on your level of physics background, you may be familiar with the harmonic oscillator - that is, a system for which there exists a restoring force proportional to displacement. For example, motion of a spring is simple harmonic (since by Hooke's Law, the restoring force is proportional to the amount a string is stretched) and motion of a pendulum for small angle amplitudes is simple harmonic. As a matter of fact, any object in stable equilibrium will move harmonically for small perturbations.

Quantitatively speaking, we mean to say that for simple harmonic motion, $F = - k x$ for some displacement $x$. Moreover, $F = ma = m \frac{d^2x}{dt^2}$, so combining these two equations, we find that

$$\frac{d^2x}{dt^2} = -\frac{k}{m} x$$

This is a differential equation that must be solved to find $x(t)$. It turns out that the solution to this equation is an expression of the form $A \sin( \omega t - \phi)$ for constants $A$, $\omega$, and $\phi$ - to verify this yourself, plug in a function like $x(t) = 2 sin\left(\sqrt{\frac{k}{m}}t - \frac{\pi}{2}\right)$.

Since simple harmonic motion is the most common form of oscillation, and simple harmonic motion is described using $\sin$ and $\cos$, most oscillations in physics follow these trigonometric functions.

The cycloid doesn't appear as much as $\sin$ and $\cos$ simply because there's no reason for it to. There aren't many physical phenomena which follow cycloid paths, since the cycloid is such a complex shape compared to the fairly simple $\sin(\theta) = \operatorname{Im}({e^{i \theta}})$ and $\cos(\theta) = \operatorname{Re}({e^{i \theta}})$.

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Yes, there are alternatives. But a big part of the reliance on sines and cosines is historical. The analysis of oscillating mechanical systems naturally focused on sines, since that's the way things vibrate. With that framework in place, it turned out the electrical and magnetic systems also respond. Plus (or perhaps you might say, alternatively) circular motion decomposes easily into sines and cosines. And the list goes on.

As for alternatives, well-behaved, bounded, periodic waveforms can be decomposed into sines and cosines via the Fourier Transform, and the FT can be powerfully extended into the Laplace Transform. Cycloids are interesting, but they are badly behaved - they require infinite dy/dx. And that, in turn, makes them ill-suited to apply to problems which DON'T have infinite dy/dx - which is just about everything.

That's not to say there are NO alternatives for some applications - see wavelet theory.

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This question reminds me of some remarks in pages 14-16 of Peter Woit's quantum mechanics and representation theory notes

Basically, imagine you are looking at all periodic functions from the real to the complex numbers. This is equivalent to looking at all functions from the unit circle to the complex numbers. Let's add the property that we want our function $f$ to be such that $f(\theta_1 + \theta_2) = f(\theta_1)f(\theta_2)$ (it's admittedly arbitrary to do this at this point, but at least it's an elegant property : ) ).

Then, it is a fact that our function has to be of the form $\theta \rightarrow e^{ik\theta} = \cos(k\theta) + i\sin(k\theta)$, for an integer number $k$. So this introduces why one would care about trigonometric options in particular and not at others ways of describing oscillations.

If one looks at what is going on in the proof, it basically reduces to the fact that the trigonometric functions have nice properties in terms of differentiation, and relation to each other, e.g. $$\sin(\theta_1 + \theta_2) = \sin(\theta_1)\cos(\theta_2) + \sin(\theta_2)\cos(\theta_1)$$

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+1 I don't believe your $f(\theta_1+\theta_2)=f(\theta_1) f(\theta_2)$ is too far out of left field: you can think of seeking a flow as a way to make your physics time shift invariant, so this may be a better answer than you seem to give yourself credit for! –  WetSavannaAnimal aka Rod Vance Apr 26 at 9:08

Another reason is that we perceive time as always advancing, and we see many examples of rotation with consistent revs, so it is natural for us to express oscillations in terms of time to angle plus radius and from that x,y position.

Sin and Cosine by definition give us the x,y coordinates given an angle and radius of 1, so it is convenient to use them.

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Simply put, our universe tends to operate in a manner where the presence of something alters the rate of some related variable. You could say the universe consistently operates off interest for every action, where every instantaneous change affects the future of the system. It turns out that mathematically this is a real world example of exponential growth. You could say that in essence most properties of things we observe in the universe occur exponentially. The canonical exponential we think about is e^x.

Now I bet you are wondering why I am bringing up exponentials in a question about sines and cosines. The reason is that sines and cosines are actually exponential functions when the exponent of our interest involves imaginary numbers. Exactly like how the rate of growth of something in classical exponential growth increases as the independent variable increases, when we deal with sines and cosines the exact amount of material present is creating a system that responds by increasing at certain intervals and decreasing at others.

In essence, you can think of it like this a special form of exponential growth where all your previous value outside the current wavelength you are examining do not matter, and its actually just the amount of your variable in excess of some multiple of your material that affects bulk properties.

To think about it pictorially, draw a graph with a real and imaginary axis. Draw a point on the numberline, then multiply that number by a real number and then an imaginary number and repeat this. You will notice the real number scales the value while the imaginary number rotates the arrow. Mapping this behavior to exponentials creates exponentials where the growth 'scales' (the classical e^x we think of), or the growth 'rotates' (sines and cosines).

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