Solving Wave Equations with Boundary Conditions

In summary, the conversation discusses a problem with matching boundary conditions for two different wave functions. There is a discrepancy between the given equations and the instructor's answer, leading to a debate over whether or not there is a mistake in the exercise. Some suggest there may be a typo and the instructor insists that the real part has a physical meaning. The conversation also touches on the importance of considering both the real and imaginary parts of a complex wave function and the role of taking the real part in obtaining a physical solution.
  • #1
dirk_mec1
761
13

Homework Statement


http://img685.imageshack.us/img685/5585/63862334.png [Broken]

Homework Equations


-

The Attempt at a Solution


[tex]
y_1(0,t)=y_2(0,t) \longrightarrow 1+\frac{B}{A}e^{2i \omega t} = \frac{C}{A}
[/tex][tex]
y_1_x(0,t)=y_2_x(0,t) \longrightarrow 1+\frac{B}{A}e^{2i \omega t} =\frac{k_2}{k_1} \frac{C}{A}
[/tex]

but I can not find an explicit expression for C/A and B/A expressed in v1 and v2 I can only find that k2/k1 =1. What am I doing wrong?
 
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  • #2
If k1=k2, the strings are identical, and you get no reflection.

Your wave function in region 1 looks a bit off. Typically, the second term is B e-i(k1+ωt). That still describes a wave moving to the left, but it let's the time dependence cancel out of the equations. Plus it'll result in equations you can actually solve.
 
  • #3
vela said:
Your wave function in region 1 looks a bit off. Typically, the second term is B e-i(k1+ωt).

You left out the x-dependence did you did this on purpose? Furthermore, the problem lies in the fact of the presence of the plus sign at [tex] \omega t[/tex]

But your conclusion is that there's a mistake in the question?
 
  • #4
Sorry, that was a typo. There should be an x in there.

It appears there's a mistake in the question. You either get k1=k2, which means there should be no reflection, or C=0, which means there's no transmission.
 
  • #5
I've send an e-mail to the instructor. Thanks for your help so far.
 
  • #6
The instructor repied back there isn't any mistake in the exercise, it is possible! But I don't know how!
 
  • #7
Well, if you take the problem as given, the only consistent solution I see is B/A=0 and C/A=1, which in turn requires k1=k2. I've asked others to take a look at this thread. Perhaps they'll spot something we're both overlooking.
 
  • #8
The instructor gave the answer if you take the real part of both equations you'll get a set of 2 equations which indeed lead to the right answer.
 
  • #9
dirk_mec1 said:
[tex]
y_1(0,t)=y_2(0,t) \longrightarrow 1+\frac{B}{A}e^{2i \omega t} = \frac{C}{A}
[/tex]
Here is my second opinion, for whatever it's worth. If C and A are constants, then the above expression says they are not because they are functions of time. The problem as stated implies that you may match boundary conditions at a specific time, but not at all times. To fix that, the correct form for the second term in y1(x,t) ought to be B*Exp[i(-kx-ωt)] which is still a wave traveling to the left. However, in this form, if you match boundary conditions at one time, you match them at all times because the e-iωt drop out.
 
  • #10
dirk_mec1 said:
The instructor gave the answer if you take the real part of both equations you'll get a set of 2 equations which indeed lead to the right answer.
It may lead to the [STRIKE]right[/STRIKE] intended answer, but the method isn't correct in my opinion. You can't just throw out the imaginary part of the equations because it's inconvenient.
 
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  • #11
dirk_mec1 said:
The instructor repied back there isn't any mistake in the exercise, it is possible! But I don't know how!

There has to be a typo. If [itex]y_1=Ae^{i(k_1x-\omega t)}+Be^{i(k_1x+\omega t)}[/itex] and [itex]y_2=Ce^{i(k_2x-\omega t)}[/itex] then the only way the two function can be equal at x=0 at all times (It wouldn't make sense for them to only be equal at one time unless the two strings were disconnected) is if [itex]Ae^{-i\omega t}+Be^{i\omega t}=Ce^{-i\omega t}[/itex] which tell you immediately B must be zero, and hence there is no reflected wave, and C=A.

If on the other hand [itex]y_1=Ae^{i(k_1x-\omega t)}+Be^{-i(k_1x+\omega t)}[/itex] , you don't have that problem, and there is a sensible solution.
 
  • #12
gabbagabbahey said:
There has to be a typo.

I think the same. Both the traveling and the reflected waves must have the same frequency, and omega is not the same as (-omega), although cos(wt) = cos(-wt).

ehild
 
  • #13
He told us that the real part has a physical meaning therefore only the real part of the functions have to be considered. But I don't get what you mean by the typo. The imaginary part of the function is irrelevant, isn't it?
 
  • #14
gabbagabbahey said:
(It wouldn't make sense for them to only be equal at one time unless the two strings were disconnected) is if [itex]Ae^{-i\omega t}+Be^{i\omega t}=Ce^{-i\omega t}[/itex] which tell you immediately B must be zero, and hence there is no reflected wave, and C=A.

You forgot the second equations which is the derative.
 
  • #15
dirk_mec1 said:
You forgot the second equations which is the derative.

That only makes things worse. The only way to also satisfy that condition is if [itex]k_1=k_2[/itex]...which says that the wave must travel with the same speed in both strings!

dirk_mec1 said:
He told us that the real part has a physical meaning therefore only the real part of the functions have to be considered. But I don't get what you mean by the typo. The imaginary part of the function is irrelevant, isn't it?

Considering only the real part will give you the correct answer since [itex]\cos(-\omega t)=\cos(\omega t)[/itex], however it makes no sense to consider waves as the real part of complex wavefunctions if the imaginary parts of the complex wavefunctions you choose don't satisfy the same equations as the real parts. Doing so means you have to work exclusivly with the real parts, which defeats the entire purpose of using complex exponentials (they are often easier to work with than sines and cosines) in the first place.

If, on the other hand, you choose your complex waverfunction so that both imaginary and real parts can satisfy the boundary conditions, you are free to do all the calculations with the complex wavefunctions and then simply take the real part at the very end to get your physical result.
 
  • #16
gabbagabbahey said:
If, on the other hand, you choose your complex waverfunction so that both imaginary and real parts can satisfy the boundary conditions, you are free to do all the calculations with the complex wavefunctions and then simply take the real part at the very end to get your physical result.
That sounds plausible. But what is the meaning of the complex part of the wave function? If I want to draw a picture I can only do so if I take the real part, agreed?
 

1. What are wave equations with boundary conditions?

Wave equations with boundary conditions are mathematical equations that describe the behavior of waves in a given system. They take into account the initial conditions of the system, as well as the boundaries of the system, to determine the behavior of the waves.

2. Why is it important to solve wave equations with boundary conditions?

Solving wave equations with boundary conditions allows us to accurately predict the behavior of waves in various systems. This is crucial in fields such as physics, engineering, and acoustics, where understanding and controlling wave behavior is essential.

3. What are some common techniques for solving wave equations with boundary conditions?

Some common techniques for solving wave equations with boundary conditions include separation of variables, the method of characteristics, and the finite difference method. Each method has its own advantages and is used in different situations.

4. How do boundary conditions affect the solution to a wave equation?

Boundary conditions play a crucial role in determining the behavior of waves in a system. They can affect the amplitude, frequency, and shape of the waves, and can even lead to the formation of standing waves or resonance.

5. What are some real-world applications of solving wave equations with boundary conditions?

Wave equations with boundary conditions have numerous applications in various fields, such as predicting the behavior of sound waves in a concert hall, determining the stability of structures during earthquakes, and designing efficient communication systems using electromagnetic waves.

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