Simple problem on waves (Math query)

[maverick280857]

Homework Statement

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

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

Homework Equations

Continuity at z = 0 implies

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

Also,

\frac{\partial g_{I}}{\partial z} = -\frac{1}{v_{1}}\frac{\partial g_{I}}{\partial t}

\frac{\partial h_{R}}{\partial z} = \frac{1}{v_{1}}\frac{\partial h_{R}}{\partial t}\frac{\partial g_{T}}{\partial z} = -\frac{1}{v_{2}}\frac{\partial g_{T}}{\partial t}

So the differentiability condition

\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^{+}}

is equivalent to

-\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}

The Attempt at a Solution

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

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

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

 

[maverick280857]

Any ideas, anyone?

 

[maverick280857]

To the Moderator: Can you please shift this to the appropriate sub forum?

 

[andrew1982]

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
<br /> \int (\frac{\partial}{\partial t} f(z,t) )dt=f(z,t) + C<br />
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.

 

[maverick280857]

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:

-\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}

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

-\frac{1}{v_{1}}\int\frac{\partial g_{I}(-v_{1}t)}{\partial t}dt

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...

 

[andrew1982]

Hi!

OK, explicitly I think your integral would be (ignoring the first constant):
<br /> \int\frac{\partial g_{I}(-v_{1}t)}{\partial t}dt=<br /> \int -v_1\, g&#039;_I(-v_{1}t)dt<br />
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
<br /> =\int -v_1\, g&#039;_I(x)\frac{-dx}{v_1}<br /> =\int g&#039;_I(x) dx<br /> =g_I(x)+C=g_I(-v_1 t)+C<br />

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.

 

[maverick280857]

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

Cheers
Vivek