Transmitted and reflected waves in a wire with 2 knots

[Verdict]

Homework Statement

Consider a string that admits waves with velocity v. Near the origin, i.e. in the
interval -a < z < a, the string has a different weight, and as a result a different
propagation velocity u. The amplitude of the reflected and transmitted wave are
proportional to the amplitude of the incident wave. Therefore it is customary
to definene the coeffcient of reflection, R, and of transmission, T, as the ratio of
the absolute squared amplitudes:
R = |A_R²/A_I²|
And T = |A_T²/A_I²|

Calculate R and T, and the corresponding phase shifts δR and δT.

Homework Equations

I don't know exactly what is used here, but I suppose the wave equation is essential here.

The Attempt at a Solution

I can't really think of how to do this. You'd think that the wave 'enters' at the first knot, of which a part is reflected and then a part is transmitted, and the same happens at the second knot. So a part of the wave gets 'stuck' between the knots, a part is transmitted through the entire thing, and a part is reflected off of the first knot.

Conceptually it is similar to the finite square well from quantum mechanics (or the opposite of a well, that's not specified), but I don't know how to translate this to actual wires.

 

[TSny]

Everything you say in "the attempt at a solution" is spot-on. Follow the method of the finite square well. The algebra is not fun and one should only have to suffer through it once in a lifetime .

 

[Verdict]

Hm alright, just let me check to make sure:
I can set up wave equations at all the locations.
Two waves before -a: One incident, one reflected. They both have the same wavenumber.
Between -a and a: Two waves, one transmitted from -a, one reflected from a. They both have the same wave number, but different from the ones outside of the region. They can be related using the different velocities. Then, after a, there is only 1 wave, the transmitted wave. This again has the same wave number as before.

Now, imposing continuity and continuity of the first derivative at -a and a, I get ' quite a few' equations, which I should solve for, and will give me my answer?

 

[TSny]

Yes, that's the way to do it. You can use complex exponential functions for your wave functions.
[EDIT: Also, you can let the incident wave coming in from the left have unit amplitude since you are only interested in ratios of amplitudes.]

 

[Verdict]

That's a nice touch, i'll try and see if R+T adds up to one, and if not I'll be back :) Thanks for the help!

 

[Verdict]

Alright, so solving the system of 4 equations with 5 unknowns became VERY complicated rather quickly, doing it on paper.

Instead, I let mathematica do it for me, and it gave me the following expression for the wave that gets reflected off of the first knot, wave B:
B = \frac{\text{A} \left(-1+e^{4 i a \text{k2}}\right) \left(u^2-v^2\right) e^{-\frac{2 i a \text{k2} u}{v}}}{u^2 e^{4 i a \text{k2}}-2 u v<br /> e^{4 i a \text{k2}}+v^2 e^{4 i a \text{k2}}-u^2-2 u v-v^2}

Where u is the velocity between -a and a, v is the velocity outside of -a and a, k1 is the wave number outside of that range, and k2 is the wave number between -a and a. Oh, and A is the amplitude of the incident wave.

And for the wave that gets transmitted through the last knot, wave E:
E = -\frac{4 \text{A} u v e^{2 i A \text{k2}-\frac{2 i A \text{k2} u}{v}}}{u^2 e^{4 i A \text{k2}}-2 u v e^{4 i A \text{k2}}+v^2 e^{4 i A<br /> \text{k2}}-u^2-2 u v-v^2}


Now, letting mathematica calculate R and T, they do indeed add up to one.
R = \frac{2 \left(u^2-v^2\right)^2 \sin ^2(2 a \text{k2})}{-\left(u^2-v^2\right)^2 \cos (4 a \text{k2})+u^4+6 u^2 v^2+v^4}
T = \frac{8 u^2 v^2}{-\left(u^2-v^2\right)^2 \cos (4 a \text{k2})+u^4+6 u^2 v^2+v^4}
However, in the question, I am not only supposed to give an expression for R and T, but also their corresponding phase shifts. This, I do not know how to do.

Could someone help me 'extract' these from the equations I've written here?

 

[ehild]

As T and R are absolute values, you can not find phase shifts from them. Determine the phase of the amplitude ratios At/Ai and Ar/Ai.

ehild

 

[Verdict]

As T and R are absolute values, you can not find phase shifts from them. Determine the phase of the amplitude ratios At/Ai and Ar/Ai.

ehild

Well, the first gives me At/Ai = \frac{\left(u^2-v^2\right) \sin (2 a \text{k2}) \left(\cos \left(\frac{2 a \text{k2} u}{v}\right)-i \sin \left(\frac{2 a \text{k2}<br /> u}{v}\right)\right)}{\left(u^2+v^2\right) \sin (2 a \text{k2})+2 i u v \cos (2 a \text{k2})}

And the second gives me Ar/Ai = \frac{2 i u v \left(\cos \left(\frac{2 a \text{k2} (v-u)}{v}\right)+i \sin \left(\frac{2 a \text{k2} (v-u)}{v}\right)\right)}{(\cos (a<br /> \text{k2})+i \sin (a \text{k2}))^2 \left(\left(u^2+v^2\right) \sin (2 a \text{k2})+2 i u v \cos (2 a \text{k2})\right)}

Not all that pretty either I guess..