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Simple problem on waves (Math query)

  1. Apr 6, 2007 #1
    1. The problem statement, all variables and given/known data

    (Griffiths) Suppose you send an incident wave of specified shape [itex]g_{I}(z-v_{1}t)[/itex], down string number 1. It gives rise to a reflected wave, [itex]h_{R}(z+v_{1}t)[/itex], and a transmitted wave, [itex]g_{T}(z-v_{2}t)[/itex]. By imposing the boundary conditions 9.26 and 9.27, find [itex]h_{R}[/itex] and [itex]g_{T}[/itex].

    (Boundary conditions 9.26 and 9.27 are equations expressing continuity and differentiability at z = 0).


    2. Relevant equations

    Continuity at [itex]z = 0[/itex] implies

    [tex]g_{I}(-v_{1}t) + h_{R}(v_{1}t) = g_{T}(-v_{2}t)[/tex]

    Also,

    [tex]\frac{\partial g_{I}}{\partial z} = -\frac{1}{v_{1}}\frac{\partial g_{I}}{\partial t}[/tex]

    [tex]\frac{\partial h_{R}}{\partial z} = \frac{1}{v_{1}}\frac{\partial h_{R}}{\partial t}[/tex]


    [tex]\frac{\partial g_{T}}{\partial z} = -\frac{1}{v_{2}}\frac{\partial g_{T}}{\partial t}[/tex]

    So the differentiability condition

    [tex]\left(\frac{\partial g_{I}}{\partial z}\right)_{z=0^{-}} + \left(\frac{\partial h_{R}}{\partial z}\right)_{z=0^{-}} = \left(\frac{\partial g_{T}}{\partial z}\right)_{z=0^{+}}[/tex]

    is equivalent to

    [tex]-\frac{1}{v_{1}}\frac{\partial g_{I}}{\partial t} + \frac{1}{v_{1}}\frac{\partial h_{R}}{\partial t} = -\frac{1}{v_{2}}\frac{\partial g_{T}}{\partial t}[/tex]

    3. The attempt at a solution

    I am just stuck with the integration of the last equation. I know I should get something like

    [tex]g_{I}(-v_{1}t) - h_{R}(v_{1}t) = \frac{v_{1}}{v_{2}}g_{T}(-v_{2}t) + C[/tex]

    C = constant

    But I can't exactly get to this by integrating the last equation...I get velocity squares or something (which I know from analysis of sine waves, is wrong)...I think I am wrongly applying the chain rule. Can someone please help me out here? (:zzz:)

    Thanks and cheers
    vivek
     
  2. jcsd
  3. Apr 8, 2007 #2
    Any ideas, anyone? :confused:
     
  4. Apr 9, 2007 #3
    To the Moderator: Can you please shift this to the appropriate sub forum?
     
  5. Apr 10, 2007 #4
    Maybe I'm wrong, but I don't think you need to use the chain rule. Since you have no hidden t dependence in your functions I guess you can say
    [tex]
    \int (\frac{\partial}{\partial t} f(z,t) )dt=f(z,t) + C
    [/tex]
    Differentiation and then integration with respect to t should give the same function back. Then you get directly the answer you wanted...

    Using this together with the continuity at z=0 should allow you to express transmitted and reflected waves in terms of the incident wave.
     
  6. Apr 10, 2007 #5
    Hi Andrew

    Thanks for your reply. Sorry, I meant to say "change of variable" instead of "chain rule". This is where I am getting stuck:

    [tex]-\frac{1}{v_{1}}\frac{\partial g_{I}}{\partial t} + \frac{1}{v_{1}}\frac{\partial h_{R}}{\partial t} = -\frac{1}{v_{2}}\frac{\partial g_{T}}{\partial t}[/tex]

    Now, integrate both sides with respect to time. The first term is really

    [tex]-\frac{1}{v_{1}}\int\frac{\partial g_{I}(-v_{1}t)}{\partial t}dt[/tex]

    I just want to do this integral explicitly, to convince myself that the equation I ought to be getting is obvious... I am making some silly mistake each time with it...
     
  7. Apr 10, 2007 #6
    Hi!

    OK, explicitly I think your integral would be (ignoring the first constant):
    [tex]
    \int\frac{\partial g_{I}(-v_{1}t)}{\partial t}dt=
    \int -v_1\, g'_I(-v_{1}t)dt
    [/tex]
    using the chain rule (yes if you want to write out the detalis we need it). Now putting x=-v1 t, and with dt=-dx/v1
    [tex]
    =\int -v_1\, g'_I(x)\frac{-dx}{v_1}
    =\int g'_I(x) dx
    =g_I(x)+C=g_I(-v_1 t)+C
    [/tex]

    But this I think we can see directly also since by definition the primitive function of the derivative gives the same function back... Hope I got the details correct.
     
  8. Apr 10, 2007 #7
    Hi Andrew

    Thanks...I understand my mistake now. I was factoring out the 1/v1 term without a multiplicative v1 term so I would get something like 1/v1 squared :rolleyes:

    Cheers
    Vivek
     
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