# I don't get why the $f(x, t) = A*sin(kx-wt+p)$ sine wave equation works [closed]

I really don't get how this function for calculating the Y of a sine wave based on a particular point in the X-axis and time works.

$$f(x, t) = A\sin(kx-wt+p)$$

$$x=$$ position on X-axis

$$t=$$ point in time

$$a=$$ amplitude

$$k=$$ wave number

w=angular frequency

p=phase

I'm writing down all of the letter meanings mostly as a just in case if the letters you use are different to mine. I've noticed a lot of different letters being thrown around.

this is quite simple, so as you might expect I'm new to physics. incredibly new.

I just don't get how the equation works. I get why y = Asin(wt+p) works - the angular speed times the time creates the angle, plus a "headstart angle" as I like to call the phase, then you have the sin of that angle times the amplitude, to amplify it to it'll have the right 'hight'. but I just don't understand why f(x, t) = Asin(kx-wt+p) works. I can't wrap my head around it. is there any explanation you can think of?

• Try plotting the function, as a function of z, for a few values of t. Use some simple values such as k=1, w=1, p=0 to begin with. First plot for t=0, then for t=0.02, then t a bit bigger, say 0.1, and hopefully you will begin to get a feel for this function. Next try a different pair of values for k and w. Always think about what you are doing of course. Jul 16, 2020 at 22:56
• I posted a (much) more detailed version of @AndrewSteane's comment, then deleted it when I saw the comment, which I think will be considerably more instructive for the OP. Jul 16, 2020 at 23:06

## 5 Answers

The first thing to understand is purely mathematical. Suppose we have a function $$g(t)$$. Then $$g(t+a)$$, for some constant $$a$$ (taken to be greater than zero without loss of generality), is the function shifted $$a$$ to the left. To see this, think of it this way: the new argument is $$t+a$$, so if originally I needed to be at $$t_0$$ to have the value $$g_0 = g(t_0)$$, now I must instead be at the point $$t_0-a$$ so that $$g(t-a+a) = g_0$$.

Now for your question, let us throw away the phase $$p$$, you can see it is just a constant shift. So we are concerned with $$y(x,t) = A\cos(kx - \omega t)$$. To make it more physical, suppose you are looking at a fish tank and the height of a water wave is $$y(x,t)$$, where $$t$$ is the time and $$x$$ is the position along the tank.

Now, let us focus on a single point along the tank. Then $$y(x,t)$$ for $$x$$ fixed is just a function of $$t$$, and indeed when you look at just one point you see the height of the water oscillating in time.

Now suppose we freeze time instead. Then we get a snapshot of the wave, and it will instead vary with position, since time is fixed. $$y(x,t)$$ is then a function of $$x$$ when $$t$$ is fixed.

Notice we have the term $$-\omega t$$ and $$t$$ is increasing. So $$-\omega t$$ is responsible for shifting to the right, and as we increase $$t$$, we are moving further and further right. This gives us the direction the wave propagates along $$x$$.

A waveform (any shape, not necessarily a periodic sine wave) that propagates in space without changing shape can always be written in the form $$f(x, t) = g(x - vt),$$ where $$g(x)$$ is a function with only one parameter and $$v$$ is the propagation velocity. You can try this out for yourself for any function (try sketching $$g(x)$$).

In your case, $$g(x) = A\sin(kx+p)$$ and $$f(x,t)=g(x-ct)$$, where $$c$$ is the speed of light. In your case, $$c$$ is not written explicitly, but hidden in the relation $$\omega=ck$$.

If you find $$\sin(\omega t)$$ more intuitive than $$\sin(kx)$$, you can do the same reasoning starting from here: $$F(x, t) = G\left(t - \frac xv\right).$$

It would help if you could pinpoint exactly what you don't understand about the equation, but as an overview:

1. It doesn't really matter what kind of complicated (real) function you have inside the sine function, it will always output some value between $$-1$$ and $$1$$.

2. If you left the function as $$f(x,t)=\sin(kx-wt + p)$$ and let time pass, at each fixed value of $$x$$ the output (height) will oscillate between $$-1$$ and $$1$$. If you want the function to oscillate between, say, $$-5$$ and $$5$$ you need to set $$A=5$$ in which case when the sine function is equal to $$1$$ the overall function is equal to $$5$$.

3. $$w$$ is the angular frequency, if you take two wave equations, one with $$w=1$$ and second with $$w=10$$, when you allow $$t$$ to run between say $$0$$ and $$1$$, the second wave equation runs through $$10$$ times as many cycles. This effectively determines how quickly the wave oscillates in time, hence the name.

4. As for the phase factor, this will move the entire function along by some amount. For instance if we compare $$\sin(x)$$ and $$\sin(x+\pi)$$, we have just moved the second function over by an amount $$\pi$$. The phase factors allows us to effectively choose where in the wave cycle we start at $$t=0$$.

Starting with your general equation y=Asin(kx-wt+p), y=Asin(-wt+p) gives the values of y at any time at x=0. y = Asin(kx+p) gives the values of y at any x for t=0.

In other words, if you took a photo of the wave at any particular time, there would be a variation in y with x given by the general equation with that fixed value of t. If you focused on a particular x and tracked the variation of y there, it would be given by the general equation with that fixed value of x.

Imagine you are experimenting with a wave on a rope. Here is a nice visual:

You need to describe everything about the wave that could be important to someone across the world so that they can recreate the EXACT wave in their lab.

First of all it should be a sine wave shape:

$$y(x,t)=\sin(????)$$

And as you said, the wave might not have a height of one, so you multiply it by a number "A" that will give you whatever maximum height and depth you desire.

$$y(x,t)=A\sin(????)$$

If your wave's peaks and valleys range from $$[-3,3]$$, "A" should be $$3$$.

Now, you need to write down the frequency of the wave in time $$f$$. This works exactly as it does in the first equation you learned.

It is convenient to multiply it by $$2\pi$$ since you plan to put it inside a sine function:

$$y(x,t)=A\sin(2\pi f t)$$

Now you measure your frequency in SPACE, by taking a picture of the wave and seeing how many wavelengths fit in a meter. You get a number $$\nu$$. It is important to understand that this is the cycles per unit distance, NOT unit time like before: cycles/meter not cycles per second. Now, you multiply it by $$2\pi$$ to stuff it into a sine function:

$$y(x,t)=A\sin(2\pi f t - 2\pi \nu x)$$

We call the $$2\pi f$$ the "angular frequency in time" and shorten it to $$\omega$$ for convenience. Notice, $$\omega$$ is now in radians/second not cycles per second.

We call the $$2\pi \nu$$ the "angular frequency in space" and shorten it to $$k$$ for convenience. Notice, $$k$$ is now in radians/meter not cycles per meter. We often call $$k$$ the wave number.

$$y(x,t)=A\sin(2\pi f t - 2\pi \nu x) \Rightarrow y(x,t)=A\sin(\omega t - kx)$$

Finally you need to communicate the "headstart angle" $$p$$ like you said. $$p$$ is the angle the wave starts at in it's cycle when your experiment begins:

$$y(x,t)=A\sin(\omega t - kx+ p)$$

And that's where the wave equation comes from! It contains the minimum amount of information needed to describe any regular wave.