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Identifying traveling waves

  1. Aug 27, 2006 #1
    This is for Optics - the third calc-based physics class I've taken. I'm taking calc4 now and this class seems to use differential equations quite frequently..something i am just now learning. This is an example problem that i don't quite understand:

    now, the book attempts to explain the solution in the back of the book, but i don't quite get it. it says both a) and b) are waves since they are twice differentiable functions of (z-vt) and (x+vt) respectively. Therefore, for a) psi = a^2(z - bt/a)^2 and the velocity is b/a in the positive z-direction. For b) psi = a^2(x + bt/a + c/a)^2 and the velocity is b/a in the negative x-direction.

    i don't understand where the bolded equations come from. I understand how they got the velocity and direction from those equations, but no idea how they were derived.
     
  2. jcsd
  3. Aug 27, 2006 #2

    quasar987

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    You said you understand the "theory" of traveling wave, but here's a little refresher anyway...

    f(x,t) describes a travelling wave if the variables x and t always appear as (x±vt) where v is a positive constant that we (rightly) call the velocity of the wave. If the variables a and t appear in f as the "combination" (x-vt), the wave is travelling in the positive x-direction. If they appear in as the combination x+vt, the wave is travelling in the negative x-direction.

    Another way to see it, is this: take any function g(y) of one variable and "create" out of it the function of two variables x and t by f(x,t) = g(x±vt). Then f is travelling wave. Inversely, f(x,t) is a travelling wave if and only if there exists a function g(y) of one variable such that g(y=x±vt)=f(x,t).


    Ok. Now on to your problem...

    a) is psi(z,t) = (az - bt)^2 a travelling wave? This boils down to asking, "Is it of the form g(z±vt)?" or equivalently, "do the variables z and t appear only in the form (z±vt)? We can see that it is by "pulling out" an 'a' out of the square. Step-by-step, this is done in the following way:

    (az - bt)² =[a(z-bt/a)]²=a²(z-bt/a)²

    The variables z and t appear only in the combination (z-bt/a) ==> psi(z,t) = (az - bt)^2 is a travelling wave travelling in the positive z-direction with velocity v=b/a.

    Or we could reach the same conclusion by noting that if we define the function of one variable g(y)=a²y², then g(z-bt/a)=a²(z-bt/a)²=psy(z,t).


    Try working out b) for yourself. You just need to algebraically manipulate psi so it comes out in the form psi(x,t)=g(x±vt).
     
  4. Aug 27, 2006 #3

    quasar987

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    I don'T know about you, but personally, I find the definitions of travelling wave difficult. The first one I have given lacks formalism or "rigidity". It is an attempt at conveying the idea in an intuitive way. On the other hand, the second could be too abstract for the mind to grasp fully.

    In my opinion (and experience), the best way to understand what is and what is not a travelling wave is by seeing a lot of exemples:

    [tex]f(x,t)=Ae^{-b(x-vt)^2}[/tex]

    [tex]f(x,t)=A\sin (b(x+vt))[/tex]

    [tex]f(x,t)=\frac{A\sqrt{x-vt}}{b(x-vt)^{(x-vt)}+1}[/tex]

    Are all travelling waves. However,

    [tex]f(x,t)=Ae^{-b(x^2+vt)}[/tex]

    [tex]f(x,t)=A\sin(x)+ \cos(vt)[/tex]

    are not. One last exemple:

    [tex]f(x,t)=\frac{x-vt}{x+vt}[/tex]

    is NOT a travelling wave. It has to be either x-vt OR x+vt.
     
  5. Aug 27, 2006 #4
    Thanks for all the help!

    After you explained it i had no trouble plowing through the rest of that problem. this was an example similar to a homework problem i was stumped on, and what you said helped me tremendously - but still can't figure out a couple parts of this homework problem.

    Both contain squares inside the function (i.e...psi(x,t) and either x or t or both are squared). What are the rules for this?

    also what happens if the values x and t (for psi(x,t), for example) show up more than once?

    Thanks again :)
     
  6. Aug 27, 2006 #5

    quasar987

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    I'm not sure I know what you mean. If you'd write the exact equations you're talking about, there'd be no ambiguity.
     
  7. Aug 28, 2006 #6
    In your example [​IMG] you said it was not a traveling wave. I was wondering if it was because x was squared.

    you also said that this: [​IMG] was a traveling wave. If I subtracted (b*t*x) from the end of the previous function (still exponentiated to e) would it still be a traveling wave?

    the problems in question are:

    psi(y,t) = e^-(a^2y^2+b^2t^2-2abty)

    and

    psi(z,t) = Asin(az^2-bt^2)
     
  8. Aug 28, 2006 #7
    Not all wave equations will generate two travelling waves simultaneously: f(x,t) = g(x±vt)

    The number of waveforms will depend on the nature of the pde & one waveform at a time is possible.

    desA
     
  9. Aug 28, 2006 #8

    HallsofIvy

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    More correctly because it is NOT just a function of x+vt or x-vt. There is no way to algebraically change x2+ vt into a function of x+vt alone.

    Do you mean [itex]Ae^{-b(x-vt)^2- btx}[/itex]? No, it wouldn't because -b(x-vt)2- btx= -bx2+ 2bvtx- bv2t2- btx cannot be written as a function of x-vt or x+vt only.


    The crucial part is that [itex]a^2y^2+ b^2t^2- 2abty[/itex]. Rewrite it as [itex]a^2y^2- 2abty+ b^2t^2[/itex] and you should be able to recognize that as a "perfect square": it is [itex](ay- bt)^2[/itex]. Now factor out the "a": [itex]a^2(y- (b/a)t)^2[/itex]. Yes, that is a wave function with wave speed b/a.

    az2- bt2 cannot be written in terms of z-vt or z+ vt. It can be written as [itex]a(z- \frac{b}{\sqrt{a}}t)(z+ \frac{b}{\sqrt{a}}t)[/itex] but since those cannot be "separated"- that is psi(z,t) cannot be written as a sum of two functions, one depending only on [itex]z- \frac{b}{\sqrt{a}}t[/itex], the other on [itex]z+ \frac{b}{\sqrt{a}}t[/itex] it is not a wave function.
     
  10. Mar 11, 2009 #9
    Couldn't c) also be considered a wave traveling with phase velocity of 0? I think that's just semantics though..
     
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