So I'm working my way through some basic trig (Khan Academy) - I'm trying to get a better intuition for what graphs of sine and cosine represent.

I've seen the nice unit circle animations that do explain it fairly well.

But would it be fair to say that:

The period of the curve of the sine or cosine function is how fast one full rotation of the circle happens. So as the period decreases I imagine the circle or "wheel" speeding up.

The amplitude of the curve of the sine or cosine function is the radius of this imaginary circle. So as the amplitude of the cure increases so does the size of the circle or "wheel".

So I could imagine speeding up and increases the size of a wheel by decreasing the period and increasing the amplitude of this circle.



4 Answers 4


The $(x,y)$ coordinates of a point that lies on the unit circle are $(\cos\theta,\sin\theta)$.


Yep, your wheel analogy is pretty good. Depending on how detailed you want to look at it, you could explain more about why and how in relation to math, but in general you are correct. Do you need a more detailed explanation? Hope you can figure it out!

  • $\begingroup$ No I think I'm good now, thanks for the feedback. $\endgroup$
    – drc
    Dec 13, 2011 at 20:38

This is a very interesting questions that normally arises in the context of oscialltions and waves.

Consider for instance the ODE: $$m\cdot\dfrac{d^2}{dt^2}x + 2\gamma\dfrac{d}{dt}x+D\cdot x = f\left(t\right)$$ This is the so calles harmonic oscialltor where $m$ denotes the mass, $\gamma$ denotes the friction coefficient (translated from the German "Reibungskoeffizient"), $D$ denotes the spring constant for the spring you're dealing with and $f$ is some external force.

Consider now the case where $f = 0, \gamma = 0$. Then, the equation reduces to: $$\dfrac{d^2}{dt^2}x + \Omega ^{2}\cdot x = 0, \Omega^2 = D/m$$

A solution is then given by: $$x = C\cdot\exp(i\cdot\Omega\cdot t)+C^{\star}\exp(-i\cdot\Omega\cdot t)$$, where the star in the exponent denotes complex conjugation. By choosing now the initial conditions for the movement adequately, you can determine $C$ and hence its complex conjugate.

Now, if the amplitude is increased then this means a different initial condition, i.e., the pendulum is removed a bit further from its state in inertia. If the frequency is increased, then we know, that either the mass $m$ has been decreased or that the spring constant has been increased (e.g. by choosing a different spring).

--Remark: The frequency is given by the formula: $\Omega = \sqrt{D/m}$ with the constants' meaning staying the same as above.--

Now, it is known that the frequency, or to be more precise, angular frequency, is related to the period time as follows:

$$T = 2\pi/\Omega$$

Hence, your intuitive picture was completely right. As frequency speeds up, period time decreases, hence the oscillation speeds up as well.

I am speaking of oscialltions instead of circular movements because it is possible via introducting polar coordinates to consider the latter ones as oscillations as well.

This can easily be verified be regarding the Newtonian gravitational law and applying this for instance to the movement of the earth around the sun. Approximately (indeed, some rough approximation because $\epsilon_{\text{Earth}} \neq 0$), we can regard this as a movement on a circle, regarding the earth and sun both as mass points. Now, the crucial point is, by setting a coordinate system, we can measure angles. By considering e.g. the projection of the radius vector of the earth on one of our coordinates lines, say the "x-axis", we obtain a term like $r_x = r_{\text{middle}} \cdot \cos(\omega_{\text{earth around sun}}\cdot t)$ where use has been made of $\omega = \phi t$ and $\phi$ denotes the angle in the coordinate system transformed to polar coordinates.

Some interesting stuff:

  1. Whenever you deal with a potential minimum in physics, you can locally approximate the underlying equations of motions (in Newtonian dynamics) by some harmonic oscillator.

  2. The gravitational constant $G$ can be measured e.g. by using a torsion pendulum.

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    $\begingroup$ I don't think that the OP would use this answer. The question was simply about the graph of Sine/Cosine, which seems to be a beginner's question. You gave an answer which contains differential equations and advanced physics. $\endgroup$ Dec 13, 2011 at 19:39
  • 1
    $\begingroup$ @BeniBogosel Thanks. You're right. But David Heider, I appreciate the effort. $\endgroup$
    – drc
    Dec 13, 2011 at 20:37

Technically, the $\sin()$ and $\cos()$ functions always have an argument consisting of an angle (i.e. speed of the angle changing does not matter), they always have a period of $2\pi$ and have an amplitude of $1$. We can change the period of the function by scaling the argument. For example $\sin(2\pi x)$ has a period of $1$. We change the amplitude by multiplying by a constant. For example $2\sin(2\pi x)$ produces an amplitude of 2.


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