# Sinusoidal Wave Displacement Function

I am learning about waves (intro course) and as I was studying Wave Functions, I got a little confused.

The book claims that the wave function of a sinusoidal wave moving in the $+x$ direction is $y(x,t) = A\cos(kx - wt)$.

However, I see a drawing of the wave and they always seem to be $\cos$ graphs. Are sinusoidal waves always cosine graphs? Or can they be sine? If I ever see a sine wave, then does that mean that this is merely a pulse/wave travelling once and not oscillating in a periodic motion?

Sorry for the basic beginners question.

-
Consider the application of the sum/difference equations for sine and cosine to the above. Or the simple phase shift: $\sin \theta = \cos (\theta - \frac{\pi}{2})$. –  dmckee Sep 11 '12 at 1:15

Since $$\cos\left(x-\frac{\pi}{2}\right)=\sin x,$$ using $\cos$ or $\sin$ does not matter, it depends on the choice of initial conditions.

In addition, in general, there will be a initial phase $\phi$, so sinusoidal wave is written like $$y(x,t)=A \cos(kx-\omega t+\phi).$$

-
I am aware of the identity, but sin(x) and cos(x) on their own, are not equal. Therefore, sin(kx - wt) is not equal to cos(kx - wt). So when I'm given a question say: Write down the displacement function of a sinusoidal wave with A = 2.0, k = 4.0 and w = 1.5, would I write it as y(x,t) = 2.0cos(4x - 1.5t) or y(x,t) = 2.0sin(4x - 1.5t)? Forget about trig identities: I do not understand whether sinusoidal waves are ALWAYS cosine (as the book highlights a cosine function) or may be sine and differs per each function (which I should determine from a graph/given in the question) –  Darksky Sep 11 '12 at 1:22
@Darksky: As I said, the general form of sinusoidal function include an initial phase $\phi$, determined by initial conditions. –  C.R. Sep 11 '12 at 1:24
...which would be written as a cosine function (including the initial phase)? Or can I see the same function above you gave as a sine function, depending on the question? –  Darksky Sep 11 '12 at 1:27
@Darksky, this does not require physics expertise, only a competent rational faculty and desire to know rather than to parrot. –  Alfred Centauri Sep 11 '12 at 1:41
@AlfredCentauri: You write another long answer for him. I don't understand his confusion. –  C.R. Sep 11 '12 at 1:43
show 8 more comments

Here's how I would suggest thinking about it. This is a sinusoidal wave:

Is that described by cosine formula or a sine formula? Well, as it stands, neither. It's just a thing that exists in space, there are no numbers attached.

In order to represent this wave by a formula, you need to find a way to represent points in space and time by numbers. That's what a coordinate system is for. And in order to apply a coordinate system, you need to choose some spacetime point to be labeled $(t, x) = (0, 0)$. For example, you might choose that point like this:

Now that every point is labeled by a number, you can meaningfully talk about what formula describes the wave. To do so, take a look at one specific "slice" of time, and at one specific point, for example the time and location that we chose to be $(t, x) = (0, 0)$.

You'll notice that at $(0,0)$, the wave has an value of zero, and its derivative is at a maximum. That uniquely matches the formula $y = \sin(kx - \omega t)$.

But what if you take the same wave and put a different coordinate system on it - that is, choose a different location to be position zero? You might get something like this:

It's the same wave, we've just chosen different labels for the points in the space the wave is traveling through. Suppose we call these new labels $(t',x')$. So let's look at the snapshot at $t'=0$ (when I write something with a prime like this, it means "$t$ in the primed coordinate system") with this new coordinate system.

Now, at the point we've chosen to call $(0,0)$, the wave is at its maximum. So with this new coordinate system, it doesn't match $y = \sin(kx' - \omega t')$; instead, the formula that matches the wave is $y = \cos(kx' - \omega t')$.

Of course, there are an infinite number of other ways you could choose the coordinate system. For example, suppose you choose a double-primed coordinate system $(t'',x'')$ so that the same time is labeled zero, but instead choose the position coordinates such that the $t'' = 0$ snapshot looks like this:

With this labeling, the wave doesn't match either the sine formula or the cosine formula. So what can you do? Use a shifted coordinate. For example, you can see that the wave crosses zero with a positive derivative at some negative value around $-2$. (The value is actually $-8\pi/11$.) Since you already know how to write the formula for a wave which crosses zero with positive derivative at position zero, you can do that and just use $x'' + 8\pi/11$ as your position coordinate. You can think of this as shifting the coordinate system back so that the zero point falls where the wave crosses zero.

So the formula would be

$$y = \sin\biggl[k\biggl(x'' + \frac{8\pi}{11}\biggr) - \omega t\biggr]$$

Alternatively, since you already know how to write the formula for a wave which hits its maximum at position zero, you could do that instead: shift the coordinate system back only enough so that the zero point falls where the wave hits its maximum.

You would need to shift by $5\pi/22$, which is equivalent to using $x'' + 5\pi/22$ as your position coordinate. So in this case the formula would be

$$y = \cos\biggl[k\biggl(x'' + \frac{5\pi}{22}\biggr) - \omega t\biggr]$$

You may notice that all these different equations for the wave take the general form

$$y = \sin(kx - \omega t + \phi_s)$$

or

$$y = \cos(kx - \omega t + \phi_c)$$

You can use either of these; they both describe the same kind of wave, as long as you pick the value of $\phi$ correctly for your coordinate system. For example, in the latest of the three cases I presented (the double-primed coordinates), you could choose $\phi_s'' = 8\pi/11$ and use the sine formula, or you could choose $\phi_c'' = 5\pi/22$ and use the cosine formula. It's the same thing either way (and you can show this mathematically, too). In the second case, the single-primed coordinates, the cosine formula worked with $\phi_c' = 0$, but you could just as well have used the sine formula with $\phi_s' = \pi/2$. And in the first case, the sine formula worked with $\phi_s = 0$, but you could also have used the cosine formula with $\phi_c = -\pi/2$.

In most resources you will see at your level, the author will choose to use one or the other of these formulas consistently, i.e. always cosine, or always sine. In fact, they might always choose the coordinate system that makes $\phi$ be zero. But it's useful to remember that, in case you ever have to use a different coordinate system (which you usually will, in higher-level physics), you can always use either of the general formulas, the sine or the cosine, and just pick the value of $\phi$ that makes things work out.

-