General equations for transmitted and reflected waves

In summary, the homework statement says that if you send an incident wave of specified shape down string number 1, it will give rise to a reflected wave, and a transmitted wave. By imposing the appropriate boundary conditions, find the reflected wave and transmitted wave.
  • #1
ELB27
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Homework Statement


Suppose you send an incident wave of specified shape, ##g_I(z-v_1t)##, down string number 1. It gives rise to a reflected wave, ##h_R(z+v_1t)##, and a transmitted wave, ##g_T(z-v_2t)##. By imposing the appropriate boundary conditions [see below], find ##h_R## and ##g_T##.

Homework Equations


The appropriate boundary conditions as given in the book:
##f(0^-,t)=f(0^+,t)## and ##\frac{\partial f}{\partial z}|_{z=0^-} = \frac{\partial f}{\partial z}|_{z=0^+}##.

The Attempt at a Solution


The net displacement of the strings is ##f=g_I(z-v_1t)+h_R(z+v_1t)## for ##(z<0)## and ##f=g_T(z-v_2t)## for ##(z>0)##. Using the first boundary condition: ##g_I(-v_1t)+h_R(v_1t)=g_T(-v_2t)##. For the second condition, I think that the definition of a derivative as a limit is needed. Thus, [tex]\lim_{\Delta z→0}\frac{g_I(-v_1t)-g_I(-v_1t-\Delta z) + h_R(v_1t)-h_R(v_1t-\Delta z)}{\Delta z} = \lim_{\Delta z→0}\frac{g_T(-v_2t-\Delta z)-g_T(-v_2t)}{\Delta z}[/tex]
Rearranging and plugging in the first boundary condition, [tex]\lim_{\Delta z→0}\frac{2g_T(-v_2t)-g_I(-v_1t-\Delta z)-h_R(v_1t-\Delta z)-g_T(-v_2t-\Delta z)}{\Delta z}=0[/tex]
Now I'm stuck. I tried to evaluate the last equation at different times and using the first constraint to attempt to eliminate either ##h_R## or ##g_T##. For instance, evaluating at ##t=\frac{\Delta z}{v_1}## and noting that ##h_R(0) = g_T(0)-g_I(0)## by the first constraint I got: [tex]\lim_{\Delta z→0}\frac{2g_T(\frac{v_2}{v_1}t)-g_T(0)-g_T(-\frac{v_2}{v_1}\Delta z-\Delta z)+g_I(0)-g_I(-2\Delta z)}{\Delta z} = 0[/tex]
I don't see where this leads me and the same happens when I try to evaluate at different times. Any ideas on how to progress from here?

Any suggestions/comments will be greatly appreciated!
 
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  • #2
Hi ELB27,

Hint 1 if [itex]a(u)=b(u+1)[/itex], then [itex]a(u-\Delta z)=b(u-\Delta z + 1)[/itex] (can you see how to use this to eliminate [itex]g_{T}[/itex]?)

Hint 2 [itex]\frac{d}{du}a(w) = \frac{dw}{du}\frac{d}{dw}a(w)[/itex]

Using some clever substitutions and applying the above derivative rule you should be able to arrive at a simple relationship between [itex]g'_{I}(u)[/itex] and [itex]h'_{R}(u)[/itex]

Cheers,
Gabba
 
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  • #3
gabbagabbahey said:
Hi ELB27,

Hint 1 if [itex]a(u)=b(u+1)[/itex], then [itex]a(u-\Delta z)=b(u-\Delta z + 1)[/itex] (can you see how to use this to eliminate [itex]g_{T}[/itex]?)

Hint 2 [itex]\frac{d}{du}a(w) = \frac{dw}{du}\frac{d}{dw}a(w)[/itex]

Using some clever substitutions and applying the above derivative rule you should be able to arrive at a simple relationship between [itex]g'_{I}(u)[/itex] and [itex]h'_{R}(u)[/itex]

Cheers,
Gabba
Thank you very much for the hints.
I'm afraid that I still don't see the solution you were suggesting. Also, what is the justification for hint no.1? Why is it true?
However, your two hints gave me another idea. Using your hint no.2 (the chain rule) ##\frac{\partial g_T}{\partial z} = \frac{d g_T}{d (z-v_2t)}\frac{\partial (z-v_2t)}{\partial z} = \frac{d g_T}{d (z-v_2t)}## and ## \frac{\partial g_T}{\partial t} = \frac{d g_T}{d (z-v_2t)}\frac{\partial (z-v_2t)}{\partial t} = -v_2\frac{d g_T}{d (z-v_2t)}##. Thus, ##\frac{\partial g_T}{\partial z} = -\frac{1}{v_2}\frac{\partial g_T}{\partial t}##. Acquiring a similar expression for ##g_I## and ##h_R##, and noting that ##\frac{\partial g(z-vt)}{\partial t}|_{z=0} = \frac{\partial g(-vt)}{\partial t}## since the derivative is with respect to ##t##, the 2nd condition can be rewritten as [tex]-\frac{1}{v_2}\frac{\partial g_T(-v_2t)}{\partial t}=-\frac{1}{v_1}\frac{\partial g_I(-v_1t)}{\partial t}+\frac{1}{v_1}\frac{\partial h_R(v_1t)}{\partial t}[/tex]Integrating with respect to time, [tex]-\frac{1}{v_2}g_T(-v_2t)=-\frac{1}{v_1}g_I(-v_1t)+\frac{1}{v_1}h_R(v_1t) + c[/tex] where ##c## is a constant of integration. Now, by the first condition, ##g_T(-v_2t) = g_I(-v_1t)+h_R(v_1t)##. Using these two equations, we can isolate both ##h_R## and ##g_T##. To wit: ##h_R(v_1t)=\frac{v_2-v_1}{v_2+v_1}g_I(-v_1t) + c## and ##g_T(-v_2t) = \frac{2v_2}{v_2+v_1}g_I(-v_1t)+d##. Now your first hint comes into play. If I understood it correctly, I can just add ##+z## into the brackets of each function and get the final solution.
I'm not sure about this last step as there is also a constant in front (although it doesn't depend on ##t##). Also, I would like to solve this problem with the original method.
Thanks in advance!
 
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  • #4
ELB27 said:
Thank you very much for the hints.
I'm afraid that I still don't see the solution you were suggesting.
There's more than one way to skin a cat

ELB27 said:
Also, what is the justification for hint no.1? Why is it true?
Well, [itex]a[/itex] and [itex]b[/itex] are functions (although I realize now I could have made that clearer in the Hint), so it is just a substitution [itex]u \rightarrow u +\Delta z[/itex] (specifically, it can be thought of as a "shift" of the [itex]u[/itex] axis)

If it is not intuitively obvious still, consider an example like the functions [itex]\cos[/itex] and [itex]\sin[/itex] which are related by [itex]\cos\left(x \right) = \sin \left( \frac{\pi}{2} - x\right)[/itex], and compare the plots of [itex]\cos\left(x + a\right)[/itex] and [itex]\sin \left( \frac{\pi}{2} - (x +a)\right)[/itex] for some random values of [itex]a[/itex]

ELB27 said:
To wit: ##h_R(v_1t)=\frac{v_2-v_1}{v_2+v_1}g_I(-v_1t) + c## and ##g_T(-v_2t) = \frac{2v_2}{v_2+v_1}g_I(-v_1t)+d##.

Not sure why you have two different constants here.

ELB27 said:
Now your first hint comes into play. If I understood it correctly, I can just add ##+z## into the brackets of each function and get the final solution.
I'm not sure about this last step as there is also a constant in front!

Well, you can't "just add ##+z## into the brackets of each function and get the final solution", but you can make a substitution (carefully) to express the relationships between the functions in a different form if desired. Are you expected to provide your final answer in a specific format?
If not, I would personally write it in the form [itex]h_{R}(u) = ...[/itex] and [itex]g_T(u) = ...[/itex]
 
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  • #5
gabbagabbahey said:
Well, [itex]a[/itex] and [itex]b[/itex] are functions (although I realize now I could have made that clearer in the Hint), so it is just a substitution [itex]u \rightarrow u +\Delta z[/itex] (specifically, it can be thought of as a "shift" of the [itex]u[/itex] axis)
I understand now, thanks.
gabbagabbahey said:
Not sure why you have two different constants here.
You're right. After keeping track of the changes to the original ##c## I realized that in both cases the constant equals to ##c'=\left(-\frac{v_2}{v_1+v_2}\right)c##.
gabbagabbahey said:
Well, you can't "just add ##+z## into the brackets of each function and get the final solution", but you can make a substitution (carefully) to express the relationships between the functions in a different form if desired. Are you expected to provide your final answer in a specific format?
If not, I would personally write it in the form [itex]h_{R}(u) = ...[/itex] and [itex]g_T(u) = ...[/itex]
Well, in the case of ##h_R##, I want ##u=z+v_1t##. In order to get it, I would make the substitution ##t→t+\frac{z}{v_1}## and get [tex]h_R(z+v_1t)=h_R(u)=\frac{v_2-v_1}{v_2+v_1}g_I(-v_1t-z)+c'=\frac{v_2-v_1}{v_2+v_1}g_I(-u)+c'[/tex] In the case of ##g_T##, I want ##u=z-v_2t##. To get it, I would make the substitution ##t→t-\frac{z}{v_2}## to get [tex]g_T(z-v_2t)=g_T(u)=\frac{2v_2}{v_2+v_1}g_I(\frac{v_1}{v_2}z-v_1t)+c' = \frac{2v_2}{v_2+v_1}g_I(\frac{v_1}{v_2}u)+c'[/tex] I have to say that I feel more confident in these substitutions - are they correct?
 

1. What are the general equations for transmitted and reflected waves?

The general equations for transmitted and reflected waves describe the relationship between the incident wave and the transmitted or reflected wave at a boundary between two different media. They are as follows:

Transmitted wave: T = t1/t2 = 2Z2/(Z1 + Z2)

Reflected wave: R = r1/r2 = (Z2 - Z1)/(Z1 + Z2)

2. What do T and R represent in the general equations for transmitted and reflected waves?

T represents the transmission coefficient, which is the ratio of the amplitude of the transmitted wave to the amplitude of the incident wave. R represents the reflection coefficient, which is the ratio of the amplitude of the reflected wave to the amplitude of the incident wave.

3. How do the general equations for transmitted and reflected waves change with different media?

The general equations for transmitted and reflected waves remain the same, but the values for Z1 and Z2 will change depending on the properties of the media. These values represent the characteristic impedance of the media, which is determined by factors such as density, elasticity, and electrical conductivity.

4. Can the general equations for transmitted and reflected waves be used for all types of waves?

Yes, the general equations for transmitted and reflected waves can be used for all types of waves, including electromagnetic waves, sound waves, and water waves. However, the specific values for the characteristic impedance of the media will vary depending on the type of wave.

5. How are the general equations for transmitted and reflected waves used in practical applications?

The general equations for transmitted and reflected waves are used in fields such as acoustics, optics, and geophysics to understand and predict the behavior of waves at boundaries between different media. This information is crucial for designing and analyzing devices such as lenses, mirrors, and seismic sensors.

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