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I know it sounds easy for most of you all but why is b) not a travelling wave because from what I know b) can be differentiated twice so shouldn't it be a travelling wave?

For d) Can I find the phase velocity using the equation $$\psi(x,t) = f(kx-\omega t)$$ So for d) it is

$$A\cos^2 (-2\pi x-(-2\pi t)) $$

and from the formula $v=\omega/k$ where w is angular frequency and $k$ is the wave number, $v= -2\pi/-2\pi =$ 1 m/s? Is my method correct?

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  • $\begingroup$ Does the equation for $\psi(x,t)$ satisfy the equation $\dfrac{\partial ^2\psi}{\partial x^2} = \dfrac{1}{c^2} \dfrac{\partial ^2\psi}{\partial t^2}\,\,$ or the equivalent with $z$ as the position variable.? $\endgroup$ – Farcher Mar 25 '18 at 8:57
  • $\begingroup$ hi @Farcher can I say b) satisfy because it can be differentiated twice but it does not satisfy travelling wave(which is what the question is asking for) because v=w/k does not work for b)?? $\endgroup$ – john Mar 25 '18 at 9:13
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    $\begingroup$ Being able to differential twice has not a lot to do with it. You have to make sure that if you differentiate $\psi$ twice with respect to $z$ and then you differentiate $\psi$ twice with respect to $t$ the wave equation is satisfied for all values of $z$ and $t$ which it is not whereas the other equation for $\psi$ does satisfy the wave equation.. $\endgroup$ – Farcher Mar 25 '18 at 9:27
  • $\begingroup$ In (d) the units seem wrong. How can I subtract a length from a time? Isn't that apples and oranges? In order to be able to do that subtraction, both $x$ and $t$ must be dimensionless. That means that the units of space and time have been set up in advance to be multiples of some characteristic length and time. This adds a technical twist to the question. But the resolution is this: since $x$ and $t$ are dimensionless, the velocity of that wave is $v=1$, no units. That means $1$ characteristic length per characteristic time. $\endgroup$ – garyp Mar 25 '18 at 18:57
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(b) is not a travelling wave because it is not of the form $$f(x-vt)+g(x+vt)$$ It does not satisfy the wave equation.

(d) is correct.

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  • $\begingroup$ hi @velut luna, thanks for commenting but can I say it is a wave but it is not a travelling wave because it does not satisfy v=w/k ?? $\endgroup$ – john Mar 25 '18 at 8:35
  • $\begingroup$ What is your definition of wave? $\endgroup$ – velut luna Mar 25 '18 at 8:37
  • $\begingroup$ hi @velut luna from what I know something is a wave when its function can be differentiated twice like psy(x,t) = 2(x^2)( t^2) is a wave but is not a travelling wave? $\endgroup$ – john Mar 25 '18 at 8:40
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    $\begingroup$ I can't find a rigorous definition of wave on wikipedia. So may be you can say so. For me, wave is what satisfies the wave equation. $\endgroup$ – velut luna Mar 25 '18 at 10:06
  • $\begingroup$ @velutluna Then any damped wave is not a "wave"? $\endgroup$ – freecharly Mar 25 '18 at 15:09
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First, I'll try a definition of a wave. In one dimension (e.g., with coordinate z), any solution of (a system of) a linear partial differential equation of the form $$\psi(z, t)=\psi_0 \exp (ikz-i\omega t) \tag 1$$ where $\psi_0$ is the (complex) amplitude, $k$ is the wave vector, and $\omega$ is the angular frequency, can be considered to be a "wave" with phase velocity $$v_{ph}=\frac {\omega }{k}$$ and group velocity $$v_{gr}=\frac{\partial \omega}{\partial k}$$ Any linear superposition of such sinusoid waves is again a wave. The functional relation between the frequency $\omega$ and the wave vector $k$, $$D(\omega, k)=0$$ is called the dispersion relation, where $\omega $ and $k$ can be complex. It follows from the partial differential equations defining the wave propagation. Solutions (1) of the proper wave equation have the simple linear dispersion relation $$\omega =v_{ph} k$$ with a constant phase velocity, which is equal to the group velocity $$v_{ph}=v_{gr}$$ More complex systems, like a medium with damping, have more complicated (non-linear) dispersion relations so that both phase and group velocities can become frequency (or wave length) dependent. Usually, solutions as eq. (1) are considered propagting wave in frequency and wave vector ranges where both $\omega$ and $k$ have real parts, so that real phase and group velocities ($v_{gr} \neq 0$) can be defined, which usually signifies the propagation of a signal/energy.

The function (b) of the question seems not to be expressible as a linear combination of $t$ an $z$. It thus cannot be regarded to represent a propagating wave. Function (d) should be a propagating wave.

Anybody who thinks that this is a still wanting description of a traveling wave is asked to improve it or provide a better definition.

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  • $\begingroup$ I wish that those luminaries, who downvote this, have also the guts to give an explanation for that, or provide a better answer. $\endgroup$ – freecharly Mar 25 '18 at 20:36

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