What do angle have to do with waves? Why do we use trigonometric functions to model waves?

Question

I've just learned SHM and I'm not able to understand why "phi" (initial phase) is an angle. Why are angles used in SHM?

Answer 1

Take a circle, and any function that maps onto this circle by angle, then projecting onto a line produces a wave. This is how angle comes into the analysis of waves. In short, functions on the circle are essentally periodic and can be unbundled into a wave.

Answer 2

SHM and wave motion are not exactly the same thing, although the local motion at each point, in the case of a plane wave, is a SHM.

Then, the core of this question is what is the connection between SHM and angles even in the cases, like the 1D SHM, where there is no obvious angle.

The reason for speaking about angles can be seen from two related complementary points of view.

On the one hand, the independent solutions of the equation of motion for the harmonic oscillator are the functions sinus and cosinus which may be defined with reference to the trigoniometric circle (although it is always possible to define them by using series without direct contact with trigonometry). This way of introducing trigonometric functions induces an angle-related language.

On the other hand, something like $cos( \omega t + \phi)$ can be seen as the projection on an axis of a uniform circular motion.

Answer 3

Circles and trig functions are very closely linked, infact trig functions are essentially measurements of the unit circle.

This may help you to visualise:

There's also a pretty nice Khan Academy section here.

The basic idea is that as you move around the unit circle, you can consider a moving triangle inscribed into that circle, the edges being:

• A vertical line connecting the point that you're at with the horizontal (conventionally $x$-axis)
• A horizontal line from that point on the $x$-axis to the origin
• The hypotenuse connecting your point on the circle to the origin.

The angle in question is the angle between the horizontal line and the hypotenuse. Let's call it $\theta$

If you were to plot out the values of the short sides as you move your point around the circle, starting your point on the positive $x$-axis you'd get a $cos(\theta)$ function for the horizontal and a $sin(\theta)$ function for the vertical.

Effectively the trig functions describe the $x$ and $y$ coordinate of your point.

If you focus your attention on a single axis, say, the $x$-axis it shouldn't be too hard for you to convince yourself that the $x$ coordinate will oscillate.

If instead you were to start your point at some other position on the circle, you'll have a "starting angle" - you can think of this as being some constant offset $\phi$. I.e. you're measuring $\theta$ starting from there instead of starting from 0.

The trig functions are then: $sin(\theta + \phi)$ and $cos(\theta + \phi)$.

An amplitude of anything but 1 is simply a scaled version of the unit circle.

Something a bit more indepth, don't worry if you don't understand this or haven't encountered it yet:

We call the horizontal and vertical lines the "projection" of the point onto the $x$ and $y$ axis respectively. (Think of this as the component of the hypotenuse in the $x$ and $y$ direction.)

Euler's theorem is, in the complex plane $(x, iy)$:

$$Ae^{i(\theta + \phi)} = A(cos(\theta + \phi) + isin(\theta + \phi))$$

So we can represent oscillation aslong the real horizontal axis by the real part of a complex exponential. This is very commonly used to simplify calculations.

Answer 4

The solution to the differential equation of a harmonic oscillator involves rotation.

If you specify your position and velocity as the real and complex components of a complex number $z$, then the solution of the differential equation is $z k^t$ where $k$ is some complex number, which is just rotation and scaling.

If you specify the position and velocity as components of a vector $x$, then you have a matrix $K$ for which the solution is $K^t x$ where $K$ is a scaled, rotation matrix.

You might be interested in this fantastic video in which he solves the differential equation of a harmonic oscillator in order to motivate matrix exponentiation.

Answer 5

It is very helpful to visualize SHM as a projection of circular motion and thus we use angles, trigonometric ratios to represent SHM