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Simple problem on waves (Math query)

  • #1
1,789
4

Homework Statement



(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).


Homework 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]

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
 

Answers and Replies

  • #2
1,789
4
Any ideas, anyone? :confused:
 
  • #3
1,789
4
To the Moderator: Can you please shift this to the appropriate sub forum?
 
  • #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.
 
  • #5
1,789
4
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...
 
  • #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.
 
  • #7
1,789
4
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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