# Standing waves

#### Prerequisites

In our study of Beats we considered the superposition of two traveling waves of slightly different frequencies going in the same direction. We found the interesting phenomenon that the alternating adding up and canceling of the two waves resulted in broad wave packets that oscillated at a much lower frequency (basically the difference of the two frequencies). If we consider the case of two waves of the same frequency but moving in opposite directions, we find another (and perhaps even more important) phenomenon: standing waves.

In our analysis of beats we relied on the trigonometric identity

$$\sin{a} + \sin{b} = 2\sin{\frac{a+b}{2}} \cos{\frac{a-b}{2}}$$

Let's consider what this tells us if we have two identical waves but going in opposite directions. From our reading on sinusoidal waves, we know that the wave

$y_+(x,t) = A \sin{(kx - \omega t)} = A \sin{k(x - v_0 t)}\quad$ with $\quad v_0 = \frac{\omega}{k}$

propagates in the positive $x$ direction, and

$y_-(x,t) = A \sin{(kx + \omega t)} = A \sin{k(x + v_0 t)}\quad$ with $\quad v_0 = \frac{\omega}{k}$

is the wave that propagates in the negative x direction.

The notation might look a bit strange at first, but it makes sense if you think about what it's telling you. The wave labelled "$y_+$" has the negative signs inside its arguments — but that means that as time grows, $x$ has to increase (get more positive) to keep up with it. So if you want to stay at the same point on the wave, you have to move towards more positive values of $x$. The wave labelled "$y_-$" has the positive signs inside its arguments  but that means that as time grows, $x$ has to decrease to keep up with it. So if you want to stay at the same point on the wave, you have to move towards more negative values of $x$.

If we add $y_+$ and $y_-$ together and use our trig identity with $a=kx-ωt$ and $b=kx+ωt$, we get an interesting result: a product instead of a sum! And the space ($x$) and time ($t$) parts are separated in the product.

$$y_+ + y_- = 2A\sin{kx}\cos{\omega t}$$

If we consider how it depends in space, it always looks like $A\sin{kx}$. It always has the same shape. But its amplitude is $2A\cos{ωt}$. It varies with time! It doesn't look as if anything is moving left or right at all, even though we built this out of two moving waves.

To see what this looks like, run the PhET simulation of a standing wave in a 10 bead model. To get the particular case shown at the left, slide the slider in the green box to the maximum number of beads and move the amplitude up under "normal mode 1" as shown. Then run the simulation by pressing the "Start" button.

You'll see the shape oscillating — pinned down at the ends, but never changing its shape.

The trick is that because the shape always looks like $\sin(kx)$, if you choose $k$ so that $\sin(kL) = 0$, then the oscillating string will ALWAYS be 0 both at 0 and at $L$. You will be able to pin the string down there and it won't matter that you built the original solution out of traveling waves that are assumed to be running on a string of unlimited length.

However, choosing $\sin(kL) = 0$ gives many different possibilities. Since $\sin(nπ) = 0$ for any integer value of $n$, we can choose a whole range of different values of $k$ that will produce an oscillation that keeps the endpoints fixed. The n-th one is:

$k_nL = nπ\quad n = 1, 2, 3, ...\quad\quad$ or $\quad \quad k_n = nπ/L$

(Why doesn't $n=0$ work?) These different values are called the normal modes of the system and the frequencies associated with them, $ω_n = k_nv_0$ are called the natural frequencies of the system.

We can figure out what this means by looking for the wavelength corresponding to each particular $k$. Since Reading the content in a sinusoidal wave shows us that

$$λ = 2π/k$$

we see that for the n-th mode,

$$λ_n = 2π/k_n = 2πL/nπ = 2L/n.$$

This means for the first mode (n = 1) the wavelength is twice $L$ — so one half of a wavelength fits perfectly between 0 and $L$. For the second mode ($n =2$), the wavelength = $L$, so a whole wavelength fits in. For the third mode ($n = 3$), the wavelength = $2L/3$ so one-and-a half wavelengths fit in, and so on. These are shown in the image at the right.

Explore these oscillations using the PhET sim to see how it works.

Joe Redish 4/1//16

Article 689