# Textbook Wave Example Clarification

My textbook gives the following wave example:

$$\psi(y, t) = (0.040)\sin\left[ 2 \pi \left( \dfrac{y}{6.0 \times 10^{-7}} + \dfrac{t}{2.0 \times 10^{-15}} \right) \right]$$

It then claims that (1) the wave travels in the negative $$y$$ direction and (2) the speed $$v = \nu \lambda = (5.0 \times 10^{14} Hz)(6.0 \times 10^{-7} m) = 3.0 \times 10^8 m/s$$.

I know that the equation of a progressive wave travelling in the positive $$x$$ direction is $$\psi(x, t) = A\sin[k(x - vt)]$$. So I rearrange to get $$\psi(y, t) = (0.040)\sin\left[ \dfrac{2 \pi}{6.0 \times 10^{-7}} \left( y + \dfrac{t (6.0 \times 10^{-7})}{2.0 \times 10^{-15}} \right) \right]$$. This gives us the progressive wave equation of the form $$\psi(y, t) = A\sin[k(y + vt)]$$. With regards to (1), why is this travelling in the negative $$y$$ direction? It seems to me that $$y + vt$$ is a movement in the positive $$y$$ direction, no? With regards to (2), since we have $$y + \dfrac{t (6.0 \times 10^{-7})}{2.0 \times 10^{-15}}$$, it seems to me that we have speed $$v = \dfrac{6.0 \times 10^{-7}m}{2.0 \times 10^{-15}s} = 3.0 \times 10^8 m/s$$, since the numerator is wavelength ($$m$$) and the denominator is temporal frequency ($$s$$), right?

I would greatly appreciate it if people could please take the time to clarify this.

• Your equation has a positive coefficient for t. Your pattern equation has a negative coefficient which is for a positive v, Your equation must be for a negative v, which is why the wave travels in the negative direction. – Bill Watts Jul 4 at 5:05
• @BillWatts $y + vt$ is in the positive $y$-direction, is it not? And $x - vt$ is in the positive $x$-direction. – The Pointer Jul 4 at 5:35
• $y-vt$ is in the positive y direction just like $x-vt$ is in the positive x direction. – Bill Watts Jul 4 at 5:44
• @BillWatts Yes, I seem to have confused myself with the notation. Thanks for the clarification. – The Pointer Jul 4 at 5:45

Consider what we mean when we say the wave velocity is positive:

In my diagram the wave is moving to the right (i.e. in the positive $$x$$ direction). This means if we consider any point on the wave like the one I have marked with an arrow this point moves right.

The wave equation is:

$$\psi(t,x) = A \sin(\omega t - kx)$$

which we can write as:

$$\psi = A \sin(\theta)$$

where $$\theta = \omega t - kx$$ and we call $$\theta$$ the phase. The point I've marked is where $$\theta=0$$ so $$\psi = \sin(0) = 0$$. So when we say the wave moves right we mean the point where $$\theta=0$$ moves right. And $$\theta$$ is given by:

$$\theta = \omega t - kx$$

So if $$\theta$$ is zero we get:

$$\omega t - kx = 0$$

or:

$$x = \frac{\omega}{k} t$$

And this is just the equation $$x = vt$$ for an object moving with velocity $$v$$ with the velocity equal to:

$$v = \frac{\omega}{k}$$

This is why the equation with $$-kx$$ describes a wave moving to positive $$x$$ or conversely if you put $$+kx$$ you get a wave with negative velocity i.e. moving to negative $$x$$.

The velocity defined this way is the phase velocity i.e. it is the velocity that a point of constant phase moves. There is another way to define the velocity called the group velocity, though for waves where the velocity is constant the two definitions are the same.

• But how does this relate to my case? – The Pointer Jul 4 at 5:00
• @ThePointer it explains why a $-kx$ in the equation means a positive velocity. Isn't that what you were asking? – John Rennie Jul 4 at 5:26
• I was asking why it is that the wave $\psi(y, t) = A\sin[k(y + vt)]$ (the one given in the textbook example) travels in the negative $y$ direction. Because it seems to me that $y + vt$ would be in the positive $y$-direction, no? And $x - vt$ is in the positive $x$-direction. – The Pointer Jul 4 at 5:32
• @ThePointer yes, the velocity is $-v$ where in this case $v$ is a positive number (the magnitude of the velocity) so it's a negative velocity. I agree it's often confusing whether quantities like $v$ are intended to be vectors or just the magnitude of a vector. Experienced physicists tend to be careless about this since it's obvious to us. – John Rennie Jul 4 at 5:36
• Ok, thanks for the clarification. – The Pointer Jul 4 at 5:40