Why is the velocity of a wave given by $v=\lambda{}f$ different than differentiation?

Question

As obvious from the nature of this question, I am a first-year physics student (with an absent lecturer), and I am having trouble grasping the concept of wave velocity. I've read these answers (1,2) but still don't get it.

Suppose a particular wave is described by the following equation: $$f(x,t)=0.5\sin{(10x+10t)}$$ where $10$ is both the wave number and angular frequency. Thus, the wave velocity can be solved for: $$\lambda=\frac{2\pi}{10}=0.628\text{ m}\\f=\frac{10}{2\pi}=1.59\text{ Hz}\\v=\lambda{}f=0.628\times1.59\approx1\text{ ms}^{-1}$$ The thing that is tripping me up is this velocity does not line up with what you would get if you differentiated the equation: $$\frac{d}{dt}0.5\sin{(10x+10t)}=5\cos{(10x+10t)}$$ This means that the maximum velocity is $\pm5\text{ ms}^{-1}$ which is different from the $1$ that we got from $\lambda{}f$. Why? Clearly these two equations are describing different things but I don't know what.

Answer 1

First things first: units!

$10$ is not a wavenumber or a frequency. $10 \frac{1}{m}$ is a wavenumber and $10$Hz is a wave number, but so can $10 \frac{1}{cm}$ and $10 \frac{1}{\mu s}$.

Alright, so you are confusing two different things to be the same. $v=\lambda f$ is called the phase velocity. If you pick one part of the wave, eg. the maximum, it will travel with velocity $v$.

$\frac{df}{dt}$ on the other hand is something different. The value of f(x,t) describes the amplitude, so when you differentiate it w.r.t time, you calculate the rate at which the amplitude changes. In an example where $f$ describes a wave on a chain (i.e. f(x,t) is the distance of the chain-piece at position x at time t from equilibrium), then $\frac{df}{dt}$ describes the velocity of that chain-piece perpendicular to the wave propagation.

Answer 2

Differentiating the equation is appropriate for particle, where the velocity is the rapidity of changing the coordinate, that is by definition the derivative of the position: $$ v(t)=\dot{x}(t)=\frac{dx(t)}{dt}. $$

Although the same term velocity is used for waves, it means something different: how fast and in which direction the wave front (i.e., the surface of constant phase) moves in time. That is, if we put a point at the surface of the constant phase, its velocity (in particle sense discussed above) woudl be the velocity of the wave.

Equation $$ F(\mathbf{x},t)=F_0\cos(\omega t - \mathbf{k}\cdot\mathbf{x}+\phi_0) $$ gives us the value of the field at point $\mathbf{x},t$, wheres the wave front of a constant phase is defined by $$ \omega t - \mathbf{k}\cdot\mathbf{x}+\phi_0=\phi = const. $$ Assuming now that we are no looking at how position $\mathbf{x}(t)$ shoudl change in time, so that the expression above remains constant, we we can differentiate the equation in respect to time, obtaining $$\omega -\mathbf{k}\dot{\mathbf{x}}(t)=0\rightarrow \mathbf{v}(t)=\dot{\mathbf{x}}(t)=\frac{\omega\mathbf{k}}{k^2} $$

Answer 3

One dimensional harmonic wave is describe with this equation

$$ f(x,t)=A\,\sin(k\,x-\omega\,t-\phi_0) \tag 1$$

we want to obtain the velocity $~v~$ that the zero crossing point move to the right

with equation (1) the position $~x_0~$ of the zero crossing point is :

$$ ~f(x,t)=0,\quad\Rightarrow k\,x-\omega\,t-\phi_0=0,\quad\Rightarrow x_0=\frac{\omega}{k}\,t+\frac{\phi_0}{k}$$

$$v=\frac{dx_0}{dt}=\frac{\omega}{k}$$

with $~\omega=2\,\pi\,f~\quad,k=\frac{2\,\pi}{\lambda}~$ you obtain

$\boxed{~v=\lambda\,f}\quad~,\left[\frac ms\right]=\left[ m\right]\,\left[\frac 1s\right]$

this velocity is call phase velocity

Answer 4

Suppose you are walking in a road, and at the left side there is a decorative wall, with a height that varies in a sinusoidal way. From your frame of reference, the relief is moving. Its velocity (that is minus your velocity), is the wave velocity mentioned in the equation.

For example, if your velocity is $1$ m/s, and the distance of the wall peaks is $0.628$m (the wave length), it will take $0.628$s (the period) to reach the next peak. And in $1$s, $1.59$ peaks (the frequency) passes by you.

If you stop, a static relief is viewed, and the equation of the change of the height of the wall with respect to an average value is $h = 0.5 sin(10x)$.

The only problem with this way of understanding wave velocities is that it can not be used for electromagnetic waves, because there is no frame associated to their velocity.