Reading the content in a sinusoidal wave

Making sense of the equation

The equation for the displacement of an elastic string undergoing sinusoidal oscillation is

$$y(x,t) = A \sin{(kx-ωt)}$$

This is an anchor equation. It can serve as the basis for understanding a lot of interesting phenomena, including beats, interference, and spectral analysis.

What does it all mean? Let's make sense of each part of it.

1. What are we talking about?  The expression $y(x,t)$ means that we are finding the $y$ displacement of the bit of string that is labeled by its position $x$ at a time $t$.  By labeling the displacement as $y$ and the position along the string $x$ we imply that the wave is a transverse wave. More generally we could write:  $f(x,t) = A \sin{(kx-ωt)}$ where $f$ indicates the displacement of the bit of string away from its equilibrium position. The tricky part is that it oscillates in both time and space. 

2. What's $A$?  Since we know that the function "sin" only oscillates between 1 and -1 and is dimensionless, multiplying it by $A$ means that $y$ will oscillate between the values $A$ and $-A$. So we can interpret $A$ as the amplitude of the oscillation. 

3. What's $k$? The constant $k$ was introduced to make units come out right. But it will have implications. Since $k$ is about how fast the argument of sine changes with $x$, let's choose a fixed value of $t$, say for convenience at $t = 0$. Then our function is $A \sin{kx}$. It looks like the figure below.

We've put a little ruler to indicate that the axis is a position measurement.

How does this change as $x$ changes? We know that the sine goes through a full oscillation when its argument (in radians) changes by $2π$ (say, from 0 up to $2π$). If $kx$ changes by $2π$, then $x$ must change by $2π/k$. Therefore, when $x$ changes by $2π/k$ our function goes through one full oscillation. The spatial distance for one full oscillation is called the wavelength, $λ$, and $k$ is called the wave number. Therefore

$$k = 2π/λ$$

and it has dimensionality $[k] =$ 1/L. Choosing $k$ selects how fast the wave will be changing in space.

4. What's $ω$? This was put in to give correct units to what happens when time changes. So, as in part 3,  this time weconsider a fixed $x$ position, say for convenience $x = 0$. When we look at the time variation of a particular bead, we get something proportional to $\sin{ωt}$. It looks like the figure below.

We've put a little clock to indicate that the axis is a time measurement.

We know that the sine goes through a full oscillation when its argument (in radians) changes by $2π$ (say, from 0 up to $2π$). If the argument of sin, $ωt$ changes by $2π$, then $t$ must change by $2π/ω$. Therefore, when $t$ changes by $2π/ω$, the sine goes through one full oscillation. The time for a bit of the string to go through one full oscillation is called the period. $T$ Therefore

$$ω = 2π/T.$$

The inverse of the period is also a convenient variable, the frequency, $f$. It's measured in inverse seconds (cycles per second) which is called Hertz. This gives the equations

$$f = 1/T = ω/2π.$$

It's easy to see that this is right via unit conversion including the units of the angles. The frequency $f$ is in cycles/sec, the angular velocity $ω$ is in radians/sec and 1 cycle = $2π$ radians. So multiplying omega by 1 = (1 cycle)/(2π radians) converts the units from radians/sec to cycles/sec.

Relating the frequency and the wavelength

We've related the frequency and the wavelength to our parameters $ω$ and $k$, but we have a relationship between them: $ω = kv_0$. What does that tell us about the frequency, wavelength, and period?

If we express $ω$ and $k$ in terms of frequency and wavelength in this relation, we get

$$ω = kv_0$$

Expressing $\omega$ and $k$ in terms of $f$ and $\lambda$, we get

$$(2πf) = (2π/λ)v_0$$

so the $2\pi$s cancel and we can get

$$fλ = v_0.$$

So the product of the frequency and the wavelength is the wave speed. This makes more sense if we express it in terms of the period. Since $f = 1/T$, we get

$$λ = v_0T.$$

This relation makes good sense:  If we wiggle a string in our hand to generate a wave, in the time we go through one full oscillation (the period, $T$) a full wiggle will have run out onto the string. In that time the start of the wiggle has already moved along the string for a time $T$.  Since the wiggle is moving with a speed $v_0$, the wavelength, $λ$, must be just $v_0T$.

Joe Redish 3/31/12 , Wolfgang Losert 4/6/13